الجزء الثانى من استخدام برنامج ماثيماتيكا كلغة برمجة فى مجال الاحصاء الاستدلالى by Tharwat Abdelmonem

‫اﻟﻔﺻل اﻟﺳﺎدس‬ ‫ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن‬

‫‪٤٦٥‬‬

‫)‪ (١-٦‬ﻣﻘدﻣــﺔ‪:‬‬

‫‪Introduction‬‬

‫ذﻛرﻧﺎ ﻣﻣﺎ ﺳﺑق ان اﺧﺗﺑﺎر ‪ t‬واﻟذي ﯾﺧص اﻟﻔرق ﺑﯾن ﻣﺗوﺳطﻲ ﻣﺟﺗﻣﻌﯾن وذﻟك ﺗﺣت ﺷروط‬ ‫ﻣﻌﯾﻧﮫ‪ .‬ﻓﻲ ﻛﺛﯾر ﻣن اﻷﺣﯾﺎن ﯾﺣﺗﺎج اﻟﺑﺎﺣث إﻟﻰ ﻣﻘﺎرﻧﺔ ﻣﺗوﺳطﺎت ﺛﻼﺛﺔ ﻣﺟﺗﻣﻌﺎت ﻓﺄﻛﺛر‪ .‬ﻓﻌﻠﻰ ﺳﺑﯾل‬ ‫اﻟﻣﺛﺎل إذا ﻛﺎن ﻟدﯾﻧﺎ أرﺑﻊ طرق ﻟﻠﺗﻌﻠﯾم ‪ A , B , C , D‬ﯾﺣوي اﻟواﺣد ﻣﻧﮭﺎ ﻛل اﻷطﻔﺎل اﻟذﯾن ﯾﺗﻠﻘون‬ ‫ﺗﻌﻠﯾﻣﮭم ﺑﺈﺣدى ھذه اﻟطرق واﻟﻣطﻠوب ﻣﻘﺎرﻧﺔ ﻣﺗوﺳطﺎت اﻟﻣﻌرﻓﺔ اﻟﻣﻛﺗﺳﺑﺔ ﻓﻲ ﻛل ﻣن اﻟطرق‬ ‫اﻟﻣﺧﺗﻠﻔﺔ‪ .‬ﯾﻣﻛن اﺳﺗﺧدام اﺧﺗﺑﺎر ‪ t‬ﻟﻣﻘﺎرﻧﺔ ﻣﺗوﺳطﻲ ﻣﺟﺗﻣﻌﯾن ﻟﻛل زوج ﻣن اﻟﻣﺟﺗﻣﻌﺎت اﻷرﺑﻌﺔ ‪،‬‬ ‫أي اﺳﺗﺧدام اﺧﺗﺑﺎر ‪ t‬ﻟﻣﻘﺎرﻧﺔ اﻟطرﯾﻘﺔ ‪ A‬ﺑﺎﻟطرﯾﻘﺔ ‪ B‬ﺛم اﺳﺗﺧداﻣﮫ ﻣرة أﺧري ﻟﻣﻘﺎرﻧﺔ اﻟطرﯾﻘﺔ ‪A‬‬ ‫ﺑﺎﻟطرﯾﻘﺔ ‪ C‬وھﻛذا ‪ ،‬إﻻ أن ھذه اﻟطرﯾﻘﺔ ﻟﮭﺎ ﻣﺷﺎﻛل ﻛﺛﯾرة ﻣﻧﮭﺎ ‪:‬‬ ‫)أ( ﻏﯾر ﻋﻣﻠﯾﺔ ﺣﯾث ﯾزداد ﻋدد اﻟﻣﻘﺎرﻧﺎت ﺑﺳرﻋﺔ ﻛﻠﻣﺎ زاد ﻋدد اﻟﻣﺟﺗﻣﻌﺎت ‪.‬‬ ‫)ب( زﯾﺎدة اﺣﺗﻣﺎل اﻟوﻗوع ﻓﻲ ﺧطﺄ ﻣن اﻟﻧوع اﻷول أي رﻓض ﻓرض اﻟﻌدم وھو ﺻﺣﯾﺢ‪.‬‬ ‫ﻟﺣﺳ ن اﻟﺣ ظ ﻓﺈﻧ ﮫ ﯾﻣﻛ ن اﻟﺗﻐﻠ ب ﻋﻠ ﻰ اﻟﻣﺷ ﺎﻛل اﻟﺳ ﺎﺑﻘﺔ‪ ،‬وﻣﺷ ﺎﻛل أﺧ رى‪ ،‬ﺑﺎﺳ ﺗﺧدام اﺧﺗﺑ ﺎر إﺣﺻ ﺎﺋﻲ‬ ‫ﯾﺳﻣﻰ ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن واﻟذي ﯾﻌﺗﺑر واﺣد ﻣن أﻛﺛر اﻟط رق اﻹﺣﺻ ﺎﺋﯾﺔ اﺳ ﺗﺧداﻣﺎ‪ .‬ﺳ وف ﻧوﺿ ﺢ أﺳ ﻠوب‬ ‫ﺗﺣﻠﯾ ل اﻟﺗﺑ ﺎﯾن ﺑﺎﻟﻣﺛ ﺎل اﻟﺗ ﺎﻟﻲ‪ .‬إذا أﺟرﯾ ت ﺗﺟرﺑ ﺔ زراﻋﯾ ﺔ ﻟدراﺳ ﺔ ﺗ ﺄﺛﯾر اﻷوﻗ ﺎت اﻟﻣﺧﺗﻠﻔ ﺔ ﻟﻠزراﻋ ﺔ )‬ ‫ﻓﺑراﯾر – ﻣ ﺎرس – ﻧ وﻓﻣﺑر – أﻛﺗ وﺑر( ﻋﻠ ﻰ إﻧﺗﺎﺟﯾ ﮫ ﻣﺣﺻ ول اﻟﻘﺻ ب وإذا ﻛ ﺎن اھﺗﻣﺎﻣﻧ ﺎ ھ و اﺧﺗﺑ ﺎر‬ ‫ﻓ رض اﻟﻌ دم أن ﻣﺗوﺳ ط إﻧﺗﺎﺟﯾ ﺔ ﻣﺣﺻ ول اﻟﻘﺻ ب واﺣ د ﻟﻸوﻗ ﺎت اﻟﻣﺧﺗﻠﻔ ﺔ‪ .‬ﯾﻌﺗﻣ د أﺳ ﻠوب ﺗﺣﻠﯾ ل‬ ‫اﻟﺗﺑﺎﯾن‪ ،‬ﻓﻲ ھذه اﻟﺣﺎﻟﺔ‪ ،‬ﻋﻠ ﻰ ﺗﺟزﺋ ﺔ اﻻﺧ ﺗﻼف اﻟﻛﻠ ﻲ ﻟﻠﻣﺷ ﺎھدات إﻟ ﻰ ﻣﻛ وﻧﯾن ﻟﮭﻣ ﺎ ﻣﻌﻧ ﻲ ﯾﺳ ﺗﺧدﻣﺎن‬ ‫ﻓﻲ ﻗﯾﺎس اﻟﻣﺻﺎدر اﻟﻣﺧﺗﻠﻔﺔ ﻟﻼﺧﺗﻼف‪ .‬اﻟﻣﻛون اﻷول ﯾﻘﯾس اﻻﺧﺗﻼف اﻟ ذي ﯾرﺟ ﻊ إﻟ ﻰ ﺧط ﺄ اﻟﺗﺟرﺑ ﺔ‬ ‫واﻟﺛﺎﻧﻲ ﯾﻘﯾس اﻻﺧﺗﻼف اﻟذي ﯾرﺟﻊ إﻟﻰ ﺧطﺄ اﻟﺗﺟرﺑﺔ ﺑﺎﻹﺿﺎﻓﺔ إﻟﻰ اﻻﺧﺗﻼف اﻟذي ﯾرﺟﻊ إﻟ ﻰ أوﻗ ﺎت‬ ‫اﻟزراﻋ ﺎت اﻷرﺑﻌ ﺔ‪ .‬ﻋﻧ دﻣﺎ ﯾﻛ ون ﻓ رض اﻟﻌ دم ﺻ ﺣﯾﺢ‪ ،‬أي أن ﻣﺗوﺳ ط إﻧﺗﺎﺟﯾ ﺔ ﻣﺣﺻ ول اﻟﻘﺻ ب‬ ‫واﺣ دة ﻟﻸوﻗ ﺎت اﻟﻣﺧﺗﻠﻔ ﺔ‪ ،‬ﻓ ﺈن ﻛ ﻼ ﻣ ن اﻟﻣﻛ وﻧﯾن ﺳ وف ﯾﻣ دوﻧﻧﺎ ﺑﺗﻘ دﯾرﯾن ﻣﺳ ﺗﻘﻠﯾن ﻟﺧط ﺄ اﻟﺗﺟرﺑ ﺔ‪،‬‬ ‫وﻋﻠﻰ ذﻟك ﯾﻌﺗﻣد اﺧﺗﺑﺎرﻧﺎ ﻋﻠﻰ اﻟﻣﻘﺎرﻧﺔ ﺑﯾن اﻟﻣﻛوﻧﯾن ﺑﺎﺳﺗﺧدام ﺗوزﯾﻊ ‪.F‬‬ ‫ﺑﻔ رض أن اھﺗﻣﺎﻣﻧ ﺎ ﺳ وف ﯾﻛ ون ﻓ ﻲ ﻣﻘﺎرﻧ ﺔ ﻣﺗوﺳ ط إﻧﺗﺎﺟﯾ ﺔ ﻣﺣﺻ ول اﻟﻘﺻ ب ﻋﻧ د أوﻗ ﺎت‬ ‫ﻣﺧﺗﻠﻔﺔ ﻟﻠزراﻋﺔ وﺑﺎﺳﺗﺧدام ﺛﻼﺛﺔ طرق ﻟﻠزراﻋﺔ )‪ .( 1, 2, 3‬اھﺗﻣﺎﻣﻧﺎ ﻓﻲ ھذه اﻟﺣﺎﻟﺔ ﺳوف ﯾﻛ ون ﻓ ﻲ‬ ‫اﺧﺗﺑﺎر ﻣﺎ إذا ﻛﺎن اﻻﺧﺗﻼف ﻓﻲ إﻧﺗﺎﺟﯾﺔ ﻣﺣﺻول اﻟﻘﺻب ﯾرﺟ ﻊ إﻟ ﻰ اﻟﻔ روق ﻓ ﻲ ﻣواﻋﯾ د اﻟزراﻋ ﺔ أو‬ ‫اﻟﻔروق ﻓﻲ طرق اﻟزراﻋﺔ أو رﺑﻣﺎ اﻟﻔروق ﻓﻲ ﻛﻼھﻣﺎ‪ .‬ﯾﻌﺗﻣد ﺗﺣﻠﯾ ل اﻟﺗﺑ ﺎﯾن ‪ ،‬ﻓ ﻲ ھ ذه اﻟﺣﺎﻟ ﺔ ‪ ،‬ﻋﻠ ﻰ‬ ‫ﺗﺟزﺋﺔ اﻻﺧﺗﻼف اﻟﻛﻠﻲ ﻹﻧﺗﺎﺟﯾﺔ ﻣﺣﺻول اﻟﻘﺻب إﻟﻰ ﺛﻼﺛﺔ ﻣﻛوﻧﺎت ‪ ،‬اﻷول ﯾﻘﯾس ﺧط ﺄ اﻟﺗﺟرﺑ ﺔ ﻓﻘ ط‬ ‫واﻟﺛﺎﻧﻲ ﯾﻘﯾس ﺧطﺄ اﻟﺗﺟرﺑﺔ ﺑﺎﻹﺿﺎﻓﺔ إﻟﻰ أي اﺧﺗﻼف ﯾرﺟﻊ إﻟ ﻰ ﻣواﻋﯾ د اﻟزراﻋ ﺔ اﻟﻣﺧﺗﻠﻔ ﺔ ‪ ،‬واﻟﺛﺎﻟ ث‬ ‫ﯾﻘﯾس ﺧطﺄ اﻟﺗﺟرﺑﺔ ﺑﺎﻹﺿ ﺎﻓﺔ إﻟ ﻰ أي اﺧ ﺗﻼف ﯾرﺟ ﻊ إﻟ ﻰ ط رق اﻟزراﻋ ﺔ اﻟﻣﺧﺗﻠﻔ ﺔ ‪ .‬وﻋﻠ ﻰ ذﻟ ك ﻓ ﺈن‬ ‫ﻣﻘﺎرﻧ ﺔ اﻟﻣﻛ ون اﻷول ﺑﺎﻟﺛ ﺎﻧﻲ ﺳ وف ﯾﻣ دﻧﺎ ﺑﺎﺧﺗﺑ ﺎر اﻟﻔ رض أن ﻣﺗوﺳ ط إﻧﺗﺎﺟﯾ ﺔ ﻣﺣﺻ ول اﻟﻘﺻ ب‬ ‫واﺣدة ﻋﻧد ﻣواﻋﯾد اﻟزراﻋﺔ اﻟﻣﺧﺗﻠﻔﺔ‪ .‬ﺑﻧﻔس اﻟﺷﻛل ﯾﻣﻛن اﺧﺗﺑﺎر اﻟﻔرض أن ﻣﺗوﺳ ط إﻧﺗﺎﺟﯾ ﺔ ﻣﺣﺻ ول‬ ‫اﻟﻘﺻب واﺣد ﻟطرق اﻟزراﻋﺔ اﻟﻣﺧﺗﻠﻔﺔ ﻋن طرﯾق ﻣﻘﺎرﻧﺔ اﻟﻣﻛون اﻷول ﺑﺎﻟﺛﺎﻟث‪.‬‬ ‫إذا ﺻﻧﻔت اﻟﻣﺷﺎھدات وﻓﻘﺎ ً ﻟﺻﻔﺔ )ﺧﺎﺻﯾﺔ( واﺣدة ﻣﺛل اﻻﺧﺗﻼف ﻓ ﻲ ط رق اﻟزراﻋ ﺔ أو اﻟﺟ ﻧس‬ ‫أو اﻟﻌﻣ ر‪ ...‬اﻟ ﺦ ﻓﺳ وف ﯾﻛ ون ﻟ دﯾﻧﺎ ﺗﺻ ﻧﯾف أﺣ ﺎدي ‪ . one-way classification‬أﻣ ﺎ إذا ﺻ ﻧﻔت‬ ‫اﻟﻣﺷ ﺎھدات وﻓﻘ ﺎ ﻟﺻ ﻔﺗﯾن ﻣﺛ ل أﺻ ﻧﺎف اﻟﻘﻣ ﺢ وأﻧ واع اﻷﺳ ﻣدة ﻓﺳ وف ﯾﻛ ون ﻟ دﯾﻧﺎ ﺗﺻ ﻧﯾف ﺛﻧ ﺎﺋﻲ‬ ‫‪ .two-way classification‬ﻓﻲ اﻟﺑﻧود اﻟﺗﺎﻟﯾﺔ ﺳوف ﻧﺗﻧﺎول طرق ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﻓﻲ ﻛﻼ اﻟﺗﺻﻧﯾﻔﯾن‪.‬‬

‫)‪ (٢-٦‬اﻟﺗﺻﻧﯾف اﻷﺣﺎدي‪:‬‬

‫‪One-way Classification‬‬ ‫‪٤٦٦‬‬

‫ﺑﻔرض أن ﻋﯾﻧﺎت ﻋﺷواﺋﯾﺔ ﻣن اﻟﺣﺟم ‪ n‬ﺗم اﺧﺗﯾﺎرھﺎ ﻣن ‪ k‬ﻣن اﻟﻣﺟﺗﻣﻌﺎت‪ .‬ﺳوف ﻧﻔﺗرض أن‬ ‫اﻟﻣﺟﺗﻣﻌﺎت اﻟﺗﻲ ﻋددھﺎ ‪ k‬ﻣﺳﺗﻘﻠﺔ وﺗﺗﺑﻊ ﺗوزﯾﻌﺎت طﺑﯾﻌﯾﺔ ﺑﻣﺗوﺳطﺎت ‪ μ1 ,μ 2 ,,μ K‬وﺗﺑﺎﯾن‬ ‫ﻣﺷﺗرك ‪ . 2‬اﻟﻣطﻠوب اﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪:‬‬ ‫‪H 0 : 1   2  ...   k‬‬ ‫ﺿد اﻟﻔرض اﻟﺑدﯾل‪:‬‬ ‫واﺣد ﻋﻠﻰ اﻷﻗل ﻣن ‪ i‬ﯾﺧﺗﻠف ﻋن اﻟﺑﺎﻗﻲ ‪H1 :‬‬ ‫ﺑﻔرض أن ‪ xij‬ﺗرﻣز ﻟﻠﻣﺷﺎھدة رﻗم ‪ j‬اﻟﻣﺧﺗﺎرة ﻣ ن اﻟﻣﺟﺗﻣ ﻊ رﻗ م ‪ i‬وأن اﻟﻣﺷ ﺎھدات ﺗ م ﺗرﺗﯾﺑﮭ ﺎ ﻓ ﻲ‬ ‫اﻟﺟدول اﻟﺗﺎﻟﻰ ﺣﯾث ‪ Ti .‬ﺗرﻣز ﻟﻣﺟﻣوع ﻛل اﻟﻣﺷ ﺎھدات ﻓ ﻲ اﻟﻌﯾﻧ ﺔ اﻟﻣﺧﺗ ﺎرة ﻣ ن اﻟﻣﺟﺗﻣ ﻊ رﻗ م ‪ i‬و‬ ‫‪ x i .‬ﺗرﻣز ﻟﻣﺗوﺳط ﻛل اﻟﻣﺷﺎھدات ﻓﻲ اﻟﻌﯾﻧﺔ اﻟﻣﺧﺗ ﺎرة ﻣ ن اﻟﻣﺟﺗﻣ ﻊ رﻗ م ‪ i‬و ‪ T..‬ﺗرﻣ ز ﻟﻣﺟﻣ وع ﻛ ل‬ ‫اﻟﻣﺷﺎھدات اﻟﺗﻲ ﻋددھﺎ ‪ nk‬و ‪ x ..‬ﺗرﻣز ﻟﻣﺗوﺳط ﻛل اﻟﻣﺷﺎھدات اﻟﺗﻲ ﻋددھﺎ ‪. nk‬‬ ‫اﻟﻣﺟﺗﻣﻌﺎت‬ ‫…‪2‬‬ ‫…‪i‬‬ ‫‪x 21.... xi1...‬‬

‫‪x11‬‬

‫‪x 22.... xi2 ... x k2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬

‫‪x12‬‬ ‫‪‬‬

‫‪x 2n .... x in ... x kn‬‬

‫‪x1n‬‬

‫ﯾﻣﻛ ن‬ ‫‪k‬‬ ‫‪x k1‬‬

‫‪1‬‬

‫‪T..‬‬

‫‪Tk.‬‬

‫‪T2....‬‬

‫‪T1.‬‬

‫اﻟﻣﺟﻣوع‬

‫‪x ..‬‬

‫‪x 2.... x i.... x k.‬‬

‫‪x1.‬‬

‫اﻟﻣﺗوﺳط‬

‫‪Ti....‬‬

‫اﻟﺗﻌﺑﯾر ﻋن ﻛل ﻣﺷﺎھدة وﻓﻘﺎ ً ﻟﻠﻧﻣوذج اﻟرﯾﺎﺿﻲ اﻟﺗﺎﻟﻲ‪:‬‬

‫‪xij  i  ij ,‬‬ ‫ﺣﯾ ث ‪ ij‬ﯾﻘ ﯾس اﻧﺣ راف اﻟﻣﺷ ﺎھدة رﻗ م ‪ j‬ﻓ ﻲ اﻟﻌﯾﻧ ﺔ رﻗ م ‪ i‬ﻋ ن ﻣﺗوﺳ ط اﻟﻣﺟﺗﻣ ﻊ رﻗ م ‪ .i‬وﺑوﺿ ﻊ‬ ‫‪ i     i‬ﺣﯾث ‪:‬‬ ‫‪k‬‬

‫‪ i‬‬ ‫‪,‬‬

‫‪  i 1‬‬ ‫‪k‬‬

‫ﻓﺈﻧﮫ ﯾﻣﻛن ﻛﺗﺎﺑﺔ اﻟﻧﻣوذج أﻋﻼه ﻋﻠﻰ اﻟﺷﻛل ‪:‬‬

‫‪xij    i ij‬‬ ‫‪k‬‬

‫ﺗﺣت ﺷ رط أن ‪   i  0‬ﺣﯾ ث ‪ i‬ﺗﻌﺑ ر ﻋ ن ﺗ ﺄﺛﯾر اﻟﻣﺟﺗﻣ ﻊ رﻗ م ‪ . i‬وﺑﺎﺳ ﺗﻌﻣﺎل اﻟﻧﻣ وذج اﻷﺧﯾ ر‬ ‫‪i 1‬‬

‫ﯾﺻﺑﺢ ﻓرض اﻟﻌدم ‪H 0 : 1   2  ...   k‬‬ ‫ﻣﻛﺎﻓﺊ ﻟﻠﻔرض‪:‬‬ ‫‪٤٦٧‬‬

‫‪H 0 : 1   2  ...   k  0‬‬

‫ﺿد اﻟﻔرض اﻟﺑدﯾل‪:‬‬ ‫واﺣد ﻋﻠﻰ اﻷﻗل ﻣن ‪ i‬ﻻ ﯾﺳﺎوى ﺻﻔرا ً ‪H1 :‬‬ ‫‪2‬‬

‫اﺧﺗﺑﺎرﻧ ﺎ ﺳ وف ﯾﻌﺗﻣ د ﻋﻠ ﻰ ﻣﻘﺎرﻧ ﺔ ﺗﻘ دﯾرﯾن ﻣﺳ ﺗﻘﻠﯾن ﻟﺗﺑ ﺎﯾن اﻟﻣﺟﺗﻣ ﻊ ‪ . ‬ﯾ ﺗم اﻟﺣﺻ ول ﻋﻠ ﻰ‬ ‫اﻟﺗﻘدﯾرﯾن ﺑﺗﺟزﺋﮫ اﻻﺧﺗﻼف اﻟﻛﻠﻲ ﻟﻠﻣﺷﺎھدات إﻟﻰ ﻣﻛوﻧﯾن ‪ .‬ﻣن اﻟﻣﻌ روف أن اﻟﺗﺑ ﺎﯾن ﻟﻛ ل اﻟﻣﺷ ﺎھدات‬ ‫ﻣﺟﺗﻣﻌﮫ ﻓﻲ ﻋﯾﻧﺔ واﺣدة ﻣن اﻟﺣﺟم ‪ nk‬ﯾﻌطﻰ ﻣن اﻟﺻﯾﻐﺔ‪:‬‬ ‫‪2‬‬

‫‪k n‬‬

‫) ‪  (x ij  x..‬‬

‫‪i 1 j1‬‬

‫‪s2 ‬‬

‫‪,‬‬ ‫‪nk  1‬‬ ‫اﻟﺑﺳ ط ﻓ ﻲ اﻟﺻ ﯾﻐﺔ اﻟﺳ ﺎﺑﻘﺔ ﯾﺳ ﻣﻰ ﻣﺟﻣ وع اﻟﻣرﺑﻌ ﺎت اﻟﻛﻠ ﻲ )‪SSTO (total sum of squares‬‬ ‫واﻟذي ﯾﻘﯾس اﻻﺧﺗﻼف اﻟﻛﻠﻲ ﻟﻠﻣﺷﺎھدات ﺣﯾث ‪:‬‬ ‫‪SSTO = SSC + SSE‬‬ ‫ﺣﯾ ث ‪ SSC‬ﯾرﻣزﻟﻣﺟﻣ وع اﻟﻣرﺑﻌ ﺎت ﻟﻣﺗوﺳ طﺎت اﻻﻋﻣ دة و‪ SSE‬ﯾرﻣزﻟﻣﺟﻣ وع اﻟﻣرﺑﻌ ﺎت‬ ‫ﻟﻣﺗوﺳطﺎت ﻟﻠﺧطﺎ ‪ .‬ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت اﻟﻛﻠﻲ ﯾﻔﺿل ان ﯾﺣﺳب ﻣن اﻟﺻﯾﻐﺔ اﻟﺗﺎﻟﯾﺔ ‪:‬‬

‫‪x ij2  CF ،‬‬

‫‪k n‬‬ ‫‪ ‬‬ ‫‪i 1 j1‬‬

‫‪SSTO ‬‬

‫ﺣﯾث‬

‫‪T..2‬‬ ‫‪CF ‬‬ ‫‪nk‬‬ ‫ﯾﺳﻣﻲ ﻣﻌﺎﻣل اﻟﺗﺻﺣﯾﺢ ‪correction factor‬‬ ‫وﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻣﺗوﺳطﺎت اﻷﻋﻣدة ‪ sum of squares for columns means‬ھو ‪:‬‬

‫‪ CF‬‬

‫‪k‬‬ ‫‪2‬‬ ‫‪ Ti.‬‬ ‫‪i 1‬‬

‫‪n‬‬

‫‪SSC ‬‬

‫وﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻠﺧطﺄ ‪ error sum of squares‬ﯾﺣﺳب ﻣن اﻟﺻﯾﻐﺔ ﻟﺗﺎﻟﯾﺔ ‪:‬‬ ‫‪SSE = SSTO -SSC‬‬ ‫أﯾﺿﺎ ﺗﺟزئ درﺟﺎت اﻟﺣرﯾﺔ اﻟﻛﻠﯾﺔ ﻛﻣﺎ ﯾﻠﻲ ‪:‬‬ ‫‪nk-1= k-1 + k (n-1).‬‬ ‫ﻋ ﺎدة ﯾﺷ ﺎر ﻟﻣﺟﻣ وع اﻟﻣرﺑﻌ ﺎت ﻟﻣﺗوﺳ طﺎت اﻷﻋﻣ دة ﻣ ن ﻗﺑ ل ﻛﺛﯾ ر ﻣ ن اﻟﻣ ؤﻟﻔﯾن ﺑﻣﺟﻣ وع‬ ‫اﻟﻣرﺑﻌﺎت ﻟﻠﻣﻌﺎﻟﺟﺎت ‪ . treatment sum of squares‬وھذه اﻟﺗﺳﻣﯾﺔ ﺗرﺟ ﻊ إﻟ ﻰ اﻟﺣﻘﯾﻘ ﺔ أن ‪ k‬ﻣ ن‬ ‫اﻟﻣﺟﺗﻣﻌﺎت اﻟﻣﺧﺗﻠﻔﺔ ﻏﺎﻟﺑ ﺎ ً ﻣ ﺎ ﺗﺻ ﻧف ﺗﺑﻌ ﺎ ً ﻟﻣﻌﺎﻟﺟ ﺎت ﻣﺧﺗﻠﻔ ﺔ وﻋﻠ ﻲ ذﻟ ك ﻓ ﺈن اﻟﻣﺷ ﺎھدات = ‪xij ;(j‬‬ ‫)‪ 1,2,…,n‬ﺗﻣﺛ ل ‪ n‬ﻣ ن اﻟﻣﺷ ﺎھدات اﻟﻣﻘﺎﺑﻠ ﺔ ﻟﻠﻣﻌﺎﻟﺟ ﺔ رﻗ م ‪ . i‬اﻵن ﻛﻠﻣ ﺔ ﻣﻌﺎﻟﺟ ﺔ ﺗﺳ ﺗﺧدم أﻛﺛ ر‬ ‫ﻟﺗوﺿﯾﺢ اﻟﺗﺻﻧﯾﻔﺎت اﻟﻣﺧﺗﻠﻔﺔ ﺳواء أﺳﻣدة ﻣﺧﺗﻠﻔﺔ أو ﻣﺻﺎﻧﻊ ﻣﺧﺗﻠﻔﺔ أو ﻣﻧﺎطق ﻣﺧﺗﻠﻔﺔ ﻓﻲ ﻣدﯾﻧ ﺔ ﻣ ﺎ أو‬ ‫ﻣﺣﻠﻠﯾن ﻣﺧﺗﻠﻔﯾن‪.‬‬ ‫‪٤٦٨‬‬

‫ﻣﺗوﺳط ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻣﺗوﺳط اﻻﻋﻣدة ﯾﻌطﻲ ﻣن اﻟﺻﯾﻐﺔ ‪:‬‬ ‫‪SSC‬‬ ‫‪MSC ‬‬ ‫‪.‬‬ ‫‪k 1‬‬ ‫ﻣﺗوﺳط ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻠﺧطﺎ ﯾﻌطﻲ ﻣن اﻟﺻﯾﻐﺔ ‪:‬‬

‫‪SSE‬‬ ‫‪.‬‬ ‫)‪k (n  1‬‬

‫‪MSE ‬‬

‫اﻟﻧﺳﺑﺔ‪:‬‬

‫‪MSC‬‬ ‫‪,‬‬ ‫‪MSE‬‬ ‫ھ ﻲ ﻗﯾﻣ ﺔ ﻟﻣﺗﻐﯾ ر ﻋﺷ واﺋﻲ ‪ F‬ﯾﺗﺑ ﻊ ﺗوزﯾ ﻊ ‪ F‬ﺑ درﺟﺎت ﺣرﯾ ﺔ )‪ 1  k  1,  2  k(n  1‬ﻋﻧ دﻣﺎ‬ ‫‪ H 0‬ﺻ ﺣﯾﺢ‪ .‬ﻟﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ ‪ ‬ﻣﻧطﻘ ﺔ اﻟ رﻓض ) ‪ F  f (1,  2‬ﺣﯾ ث ) ‪ f  (1 ,  2‬ﺗﺳ ﺗﺧرج‬ ‫ﻣ ن ﺟ دول ﺗوزﯾ ﻊ ‪ F‬ﻓ ﻲ ﻣﻠﺣ ق )‪ (٤‬ﻋﻧ د ‪  = 0.05‬أو ﻓ ﻲ ﻣﻠﺣ ق )‪ (٥‬ﻋﻧ د ‪ . = 0.01‬إذا‬ ‫وﻗﻌت ‪ f‬ﻓﻲ ﻣﻧطﻘﺔ اﻟرﻓض ﻧرﻓض ‪. H 0‬‬ ‫ﻋ ﺎدةً اﻟﺣﺳ ﺎﺑﺎت ﻓ ﻲ ﺗﺣﻠﯾ ل اﻟﺗﺑ ﺎﯾن ﺗﻠﺧ ص ﻓ ﻲ ﺟ دول ﯾﺳ ﻣﻲ ﺟ دول ﺗﺣﻠﯾ ل اﻟﺗﺑ ﺎﯾن ‪Analysis of‬‬ ‫‪ ) Variance‬ﻋﺎدة ﯾﺳﻣﻰ ‪ ( ANOVA‬واﻟﻣوﺿﺢ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫‪f‬‬

‫‪f‬‬ ‫اﻟﻣﺣﺳوﺑﺔ‬

‫ﻣﺗوﺳط اﻟﻣرﺑﻌﺎت‬

‫‪MSC‬‬ ‫‪MSE‬‬

‫‪SSC‬‬ ‫‪k 1‬‬ ‫‪SSE‬‬ ‫‪MSE ‬‬ ‫)‪k (n  1‬‬ ‫‪MSC ‬‬

‫ﻣﺟﻣوع‬ ‫اﻟﻣرﺑﻌﺎت‬ ‫‪SSC‬‬

‫درﺟﺎت اﻟﺣرﯾﺔ‬

‫ﻣﺻدر اﻻﺧﺗﻼف‬

‫‪k-1‬‬

‫ﻣﺗوﺳطﺎت اﻷﻋﻣدة‬

‫‪SSE‬‬

‫)‪k(n-1‬‬

‫‪SSTO‬‬

‫‪nk-1‬‬

‫اﻟﺧطﺄ‬ ‫اﻟﻛﻠﻲ‬

‫ﻣﺛﺎل)‪( ١ -٦‬‬ ‫اﻟﺑﯾﺎﻧﺎت ﻓ ﻲ اﻟﺟ دول اﻟﺗ ﺎﻟﻰ ﺗﻣﺛ ل اﻟط ول ) ﻣﻘ ﺎس ﺑﺎﻟﺳ ﻧﺗﯾﻣﺗر ( ﻟﻧﺑﺎﺗ ﺎت ﺗ م زراﻋﺗﮭ ﺎ ﻓ ﻲ ﺛﻼﺛ ﺔ أوﺳ ﺎط‬ ‫ﻣﺧﺗﻠﻔﺔ ‪ 5 ) A, B, C‬ﻧﺑﺎﺗﺎت ﻓﻲ ﻛل وﺳ ط (‪ .‬أوﺟ د ﺟ دول ﺗﺣﻠﯾ ل اﻟﺗﺑ ﺎﯾن وأﺧﺗﺑ ر ﻓ رض اﻟﻌ دم أن‬ ‫‪ 1   2  3‬وذﻟك ﻋﻧد ﻣﺳﺗوي ﻣﻌﻧوﯾﺔ ‪.=0.05‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪13‬‬

‫‪15‬‬ ‫‪18‬‬ ‫‪10‬‬

‫‪18‬‬ ‫‪22‬‬ ‫‪8‬‬

‫‪14‬‬ ‫‪18‬‬ ‫‪12‬‬

‫‪٤٦٩‬‬

‫‪10‬‬ ‫‪16‬‬ ‫‪15‬‬

‫‪A‬‬ ‫‪B‬‬ ‫‪C‬‬

‫اﻷوﺳﺎط‬

‫اﻟﺣــل‪:‬‬ ‫اﻟﻣطﻠوب اﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪:‬‬

‫‪H 0 : 1   2   3‬‬ ‫ﺿد اﻟﻔرض اﻟﺑدﯾل‪:‬‬ ‫واﺣد ﻋﻠﻲ اﻷﻗل ﻣن ‪  i‬ﯾﺧﺗﻠف ﻋن اﻟﺑﺎﻗﻲ ‪H1 :‬‬ ‫‪  0.05 ‬‬ ‫‪ f.05 (2,12) = 3.89‬واﻟﻣﺳ ﺗﺧرﺟﺔ ﻣ ن ﺟ دول ﺗوزﯾ ﻊ ‪ F‬ﻓ ﻲ ﻣﻠﺣ ق )‪ (٤‬ﻋﻧ د درﺟ ﺎت ﺣرﯾ ﺔ‬ ‫‪ . 1  2,  2  12‬ﻣﻧطﻘﺔ اﻟرﻓض ‪. F > 3.89‬‬

‫‪x ij2  CF‬‬

‫‪k n‬‬ ‫‪ ‬‬ ‫‪i 1 j 1‬‬

‫‪SSTO ‬‬

‫‪(216) 2‬‬ ‫‪ 10  14  ...  10  13 ‬‬ ‫‪15‬‬ ‫‪ 3304  3110.4  193.6,‬‬ ‫‪2‬‬

‫‪2‬‬

‫‪k‬‬

‫‪2‬‬ ‫‪ Ti‬‬

‫‪SSC  i 1‬‬

‫‪ CF‬‬ ‫‪n‬‬ ‫‪692  892  582 (216)2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪5‬‬ ‫‪15‬‬ ‫‪ 3209.2  3110.4  98.8.‬‬ ‫ﺗﻠﺧص اﻟﻧﺗﺎﺋﺞ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫‪ f‬اﻟﻣﺣﺳوﺑﺔ‬ ‫*‪6.25316‬‬

‫ﻣﺗوﺳط‬ ‫اﻟﻣرﺑﻌﺎت‬ ‫‪49.4‬‬ ‫‪7.9‬‬

‫ﻣﺟﻣوع‬ ‫اﻟﻣرﺑﻌﺎت‬ ‫‪98.8‬‬ ‫‪94.8‬‬ ‫‪193.6‬‬

‫درﺟﺎت‬ ‫اﻟﺣرﯾﺔ‬ ‫‪2‬‬ ‫‪12‬‬ ‫‪14‬‬

‫ﻣﺻدر اﻻﺧﺗﻼف‬ ‫ﻣﺗوﺳطﺎت اﻷﻋﻣدة‬ ‫اﻟﺧطﺄ‬ ‫اﻟﻛﻠﻲ‬

‫وﺑﻣ ﺎ أن ‪ (6.25316) f‬ﺗﻘ ﻊ ﻓ ﻲ ﻣﻧطﻘ ﺔ اﻟ رﻓض ﻓﺈﻧﻧ ﺎ ﻧ رﻓض ‪ H 0‬وﻧﻌﺗﺑ ر أن ھﻧ ﺎك ﻓروﻗ ﺎ ً ﻣﻌﻧوﯾ ﺔ‬ ‫ﺑﯾن ﻣﺗوﺳطﺎت اﻷوﺳﺎط اﻟﻣﺧﺗﻠﻔﺔ‪ .‬اﻟﻧﺟﻣﺔ * ﺗﻌﻧﻲ أن اﻟﻔرق ﻣﻌﻧوي ﻋﻧد ‪.   0.05‬‬ ‫اﻵن ﺑﻔ رض أن اﻟﻌﯾﻧ ﺎت اﻟﺗ ﻲ ﻋ ددھﺎ ‪ k‬ذات أﺣﺟ ﺎم ‪) n1, n2, …,nK‬ﻋ دم ﺗﺳ ﺎوى ﺣﺟ وم اﻟﻌﯾﻧ ﺎت(‬ ‫‪k‬‬

‫ﺣﯾث ‪. N   n i‬‬ ‫‪i 1‬‬

‫درﺟﺎت اﻟﺣرﯾﺔ ﺳوف ﺗﺻﺑﺢ )‪ (N-1‬ﻟﻣﺟﻣوع اﻟﻣرﺑﻌ ﺎت اﻟﻛﻠﯾ ﺔ ‪ SSTO‬و )‪ (k-1‬ﻟﻣﺟﻣ وع ﻣرﺑﻌ ﺎت‬ ‫ﻣﺗوﺳطﺎت اﻷﻋﻣدة ‪ SSC‬و ‪ N-1-(k-1) = N-k‬ﻟﻣﺟﻣوع ﻣرﺑﻌﺎت اﻟﺧطﺄ‪.‬‬

‫‪٤٧٠‬‬

‫ وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬Mathematica ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﻣﻛﺗوب ﺑﻠﻐﺔ‬ . ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت‬ =0.05 0.05 a={{10,14,18,15,12.},{16,18,22,18,15.},{15,12,8,10,13.}} {{10,14,18,15,12.},{16,18,22,18,15.},{15,12,8,10,13.}} f[x_]:=Apply[Plus,x] h[x_]:=Length[x] k=h[a] 3 m=Table[h[a[[i]]],{i,1,k}] {5,5,5} n=f[m] 15 xy=Map[f,a] {69.,89.,58.} xp=xy/m {13.8,17.8,11.6} f[xy] 216. cf=((%)^2)/n 3110.4 x1=xy^2 {4761.,7921.,3364.} x2=x1/m {952.2,1584.2,672.8} sbet=f[x2] 3209.2 sbet=sbet-cf 98.8 xx=Map[f,a^2] {989.,1613.,702.} xx1=f[xx] 3304. ssto=xx1-cf 193.6 sser=ssto-sbet 94.8 k1=k-1 2 msb=sbet/k1 49.4 n1=n-1 14 sx=n1-k1 12 ٤٧١

msse=sser/sx 7.9 f1=msb/msse 6.25316 rt2=List[" df "," ss "," mss "," f "] { df , ss , mss , f } rt3=List[k1,sbet,msb,f1] {2,98.8,49.4,6.25316} rt4=List[sx,sser,msse,"-"] {12,94.8,7.9,-} rt5=List[n1,ssto,"-","-"] {14,193.6,-,-} a11=TableHeadings->{{ S.V,bet,within,total},{ANOVA}} TableHeadings{{S.V,bet,within,total},{ANOVA}} uu1=TableForm[{rt2,rt3,rt4,rt5},a11]

S.V bet within total

ANOVA df 2 12 14

ss 98.8 94.8 193.6

mss 49.4 7.9

 <<Statistics`ContinuousDistributions`

f 6.25316  

ff1=Quantile[FRatioDistribution[k1,sx],1-] 3.88529 If[f1>ff1,Print["RjectHo"],Print["AccpetHo"]] RjectHo

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬ ‫ﻣﺴﺘﻮى اﻟﻤﻌﻨﻮﯾﺔ ﻣﻦ اﻻﻣﺮ‬ =0.05 ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ ‫ وھﻰ‬a a={{10,14,18,15,12.},{16,18,22,18,15.},{15,12,8,10,13.}} (١- ٦) ‫وﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﯿﺎﻧﺎت اﻟﺨﺎﺻﺔ ﺑﻤﺜﺎل‬

‫ اﻟﻣﺧرﺟﺎت‬: ‫ﺛﺎﻧﯾﺎ‬

‫ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ uu1=TableForm[{rt2,rt3,rt4,rt5},a11]

: ‫ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم‬

H 0 : 1   2  3 :‫ﺿد اﻟﻔرض اﻟﺑدﯾل‬ ٤٧٢

H1 : ‫ ﯾﺧﺗﻠف ﻋن اﻟﺑﺎﻗﻲ‬i ‫واﺣد ﻋﻠﻲ اﻷﻗل ﻣن‬ ‫ اﻟﺟدوﻟﯾﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬f ff1=Quantile[FRatioDistribution[k1,sx],1-]

‫ اﻟﻣﺣﺳوﺑﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر‬f f1=msb/msse

‫اﻟﻘرار اﻟذى ﯾﺗﺧذ ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ If[f1>ff1,Print["RjectHo"],Print["AccpetHo"]]

‫واﻟﻣﺧرج ھو‬ Reject H0

‫اى رﻓض ﻓرض اﻟﻌدم‬

( ٢ -٦)‫ﻣﺛﺎل‬ Mathematica ‫ﺳوف ﯾﺗم ﺣل اﻟﻣﺛﺎل اﻟﺳﺎﺑق ﺑﺎﺳﺗﺧدام اﻟﺑرﻧﺎﻣﺞ اﻟﺟﺎھز اﻟﺗﺎﻟﻰ واﻟﻣﻛﺗوب ﺑﻠﻐﺔ‬ . ‫وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت‬ Off[General::spell1] <<Statistics`DataManipulation` <<Statistics`NormalDistribution` Clear[oneWayAnova,all,true] oneWayAnova::usage="oneWayAnova[dataset,catvar,quanvar] performs a one-way ANOVA comparing the categorical variable catvar to the quantitative variable quanva."; Options[oneWayAnova]={categories->all,means->true}; categories::usage="categories is an option for various statistical algorithms that specifies which values of the categorical variable (catvar) to use."; means::usage="means is an option for oneWayAnova that specifies if the means for the groups are displayed. The default is that means are displayed; if means->false is included, group means are not displayed."; oneWayAnova[dataset_,catvar_,quanvar_,opts___Rule]:=Module[{ k,r,n,data,groups,cats,responses,total,sqmeans,sumsqmeans,c, sst,dft,ssg,dfg,msg,sse,dfe,mse,fratio,pvalue,meantable,anov atable}, data=DropNonNumeric[Column[dataset,{quanvar,catvar}]]; all=Union[Map[#[[2]]&,data]]; cats=categories /. {opts} /. Options[oneWayAnova]; groups=Table[Column[Select[data,#[[2]]==cats[[i]]&],1], {i,1,Length[cats]}]; k=Length[groups]; ٤٧٣

n=groups//Flatten//Length; responses=Map[Apply[Plus,#]&,groups]; total=Apply[Plus,groups//Flatten]; squares=Apply[Plus,groups^2//Flatten]; sqmeans=responses^2.Table[1/Length[groups[[i]]],{i,1,Le ngth[groups]}]; sumsqmeans= Apply[Plus,sqmeans]; c=total^2/n; sst=squares-c; dft=n-1; ssg=sumsqmeans-c; dfg=k-1; msg=ssg/dfg; sse=sst-ssg; dfe=dft-dfg; mse=sse/dfe; fratio=msg/mse; pvalue=1-CDF[FRatioDistribution[dfg,dfe],fratio]; meantable=Table[{cats[[i]],Length[groups[[i]]],Apply[Pl us,groups[[i]]]/Length[groups[[i]]]//N},{i,1,Length[cats]}]; meantable=Join[{{"Group","Number","Mean"}},meantable]; anovatable={{"Source","Sum of Squares","DF","Mean Squares","Fratio","Pvalue"},{"Groups",ssg,dfg,ssg/dfg,fratio,pvalue},{"Error",ss e,dfe,sse/dfe,"",""},{"Total",sst,dft,"","",""}}//N; true={"ANOVA Table\n",TableForm[anovatable],"Means\n",TableForm[meantable ]}; false={"ANOVA Table\n",TableForm[anovatable]}; toprint=means/. {opts} /. Options[oneWayAnova]; Print[TableForm[toprint]]; {MSE->mse,DFE->dfe}; ] dataset1={{1,10},{1,14},{1,18},{1,15},{1,12.},{2,16},{2,18}, {2,22},{2,18},{2, 15},{3,15},{3,12},{3,8},{3,10},{3,13}}; oneWayAnova[dataset1,1,2]

٤٧٤

‫‪ANOVA Table‬‬ ‫‪Pvalue‬‬ ‫‪0.0137854‬‬

‫‪Fratio‬‬ ‫‪6.25316‬‬

‫‪DF‬‬ ‫‪2.‬‬ ‫‪12.‬‬ ‫‪14.‬‬

‫‪Mean Squares‬‬ ‫‪49.4‬‬ ‫‪7.9‬‬

‫‪Sum of Squares‬‬ ‫‪98.8‬‬ ‫‪94.8‬‬ ‫‪193.6‬‬

‫‪Source‬‬ ‫‪Groups‬‬ ‫‪Error‬‬ ‫‪Total‬‬ ‫‪Means‬‬

‫‪Mean‬‬ ‫‪13.8‬‬ ‫‪17.8‬‬ ‫‪11.6‬‬

‫‪Number‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬

‫‪Group‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬

‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ ‪:‬‬ ‫اوﻻ‪ :‬اﻟﻣدﺧﻼت‬ ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ ‫‪ dataset1‬وھﻰ‬ ‫‪dataset1={{1,10},{1,14},{1,18},{1,15},{1,12.},{2,16},{2,18},‬‬ ‫‪{2,22},{2,18},{2,‬‬ ‫;}}‪15},{3,15},{3,12},{3,8},{3,10},{3,13‬‬ ‫وﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﯿﺎﻧﺎت اﻟﺨﺎﺻﺔ ﺑﻤﺜﺎل )‪(٢- ٦‬‬

‫ﺛﺎﻧﯾﺎ ‪ :‬اﻟﻣﺧرﺟﺎت‬

‫ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﯾﺗم اﻟﺣﺻول ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ ‫]‪oneWayAnova[dataset1,1,2‬‬ ‫ﻛﻤﺎ ﯾﻌﻄﻰ ھﺬا اﻻﻣﺮ ﺟﺪول ﯾﺤﺘﻮى ﻋﻠﻰ رﻗﻢ اﻟﻤﻌﺎﻟﺠﺔ واﻟﻤﺘﻮﺳﻂ وﻋﺪد اﻟﻤﺸﺎھﺪات ﻟﻜﻞ ﻣﻌﺎﻟﺠﺔ‪.‬‬ ‫وﻣن اﻟﺟدول ﻓﺈن ‪ P=0137854‬وﺑﻣﺎ ان ‪ p  .05‬ﻓﺈﻧﻧﺎ ﻧرﻓض ﻓرض اﻟﻌدم‬

‫‪H 0 : 1   2  3‬‬

‫)‪ (١-٢-٦‬اﺧﺗﺑﺎرات ﺗﺟﺎﻧس ﻋدة ﺗﺑﺎﯾﻧﺎت ‪:‬‬ ‫‪Test for the Equality of Several Variances‬‬ ‫ھﻧﺎك اﻓﺗراﺿﺎت أﺳﺎﺳﯾﺔ وﺿرورﯾﺔ ﻹﺟراء ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن وھم ‪ :‬أن اﻟﻣﺟﺗﻣﻌﺎت اﻟﺗﻲ ﻋددھﺎ ‪k‬‬ ‫ﻣﺳﺗﻘﻠﺔ وﺗﺗﺑﻊ ﺗوزﯾﻌﺎت طﺑﯾﻌﯾﺔ ﺑﻣﺗوﺳطﺎت ‪ 1 ,  2 ,...,  k‬وﺗﺑﺎﯾن ﻣﺷﺗرك ‪.  2‬‬ ‫ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪:‬‬ ‫‪H 0 : σ12  σ 22  ...  σ 2k‬‬ ‫ﺿد اﻟﻔرض اﻟﺑدﯾل ‪:‬‬ ‫اﻟﺗﺑﺎﯾﻧﺎت ﻟﯾﺳت ﻛﻠﮭﺎ ﻣﺗﺳﺎوﯾﺔ ‪H1 :‬‬ ‫‪٤٧٥‬‬

‫ واﺧﺗﺑﺎر‬Modified Levene's ‫ و اﺧﺗﺑﺎر‬Bartlett ‫ھﻧﺎك ﻋدة طرق ﻻﺟراء ذﻟك ﻣﻧﮭﺎ اﺧﺗﺑﺎر‬ ‫ ودون اﻟدﺧول ﻓﻰ ﺗﻔﺎﺻﯾل ھذه اﻟطرق‬Hetetogeneity of Coefficients of Variation ‫ﺳوف ﻧﻘدم ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻻﺟراء ھذه اﻻﺧﺗﺑﺎرات ﻣﻊ ﺷرح ﻛﯾﻔﯾﺔ ﺗﻔﺳﯾر اﻟﻧﺗﺎﺋﺞ ﻟﻛل طرﯾﻘﺔ وذﻟك ﻣن‬ : ‫ﺧﻼل اﻟﻣﺛﺎل اﻟﺗﺎﻟﻰ‬

(٣-٦)‫ﻣﺛﺎل‬ ‫ وﺗﻐذي ﻋﻠﯾﮭﺎ أرﺑﻌﺔ ﻣﺟﻣوﻋﺎت ﻣن اﻷطﻔﺎل‬A, B, C, D ‫ﺑﻔرض أن أرﺑﻌﺔ أﻧواع ﻣن اﻟﻔﯾﺗﺎﻣﯾﻧﺎت‬ .‫ﻣﺗﺷﺎﺑﮭﯾن ﺗﻣﺎﻣﺎ ً ) أرﺑﻌﺔ ﻋﯾﻧﺎت ﻋﺷواﺋﯾﺔ ( وﻛﺎﻧت اﻟزﯾﺎدة ﻓﻲ وزن ﻛل ﻣﺟﻣوﻋﺔ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ‬

‫اﻟﻔﯾﺗﺎﻣﯾﻧﺎت‬ A 2 3 3

B 2 1 2 3

C 4 5 4 4

D 4 3 2 3

:‫اﻟﺣــل‬ : ‫اﻟﻣطﻠوب اﺧﺗﺑﺎر ﻓرض اﻟﻌدم‬

H0 : 12  22  32  42 : ‫ﺿد اﻟﻔرض اﻟﺑدﯾل‬

H1 : ‫اﻟﺗﺑﺎﯾﻧﺎت ﻟﯾﺳت ﻛﻠﮭﺎ ﻣﺗﺳﺎوﯾﺔ‬ . =0.01 ‫وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ‬ . ‫وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت‬ bartlettVariance Off[General::spell1] <<Statistics`DescriptiveStatistics` <<Statistics`ContinuousDistributions` Clear[bartlettVariance] bartlettVariance[data_]:=Module[{nus,s2,ss,sp2,b,c,bc,f1,f2, a,bcprime,dist,pvalue}, nus=Map[Length,data]-1; s2=Map[Variance,data]; ss=Table[nus[[i]]s2[[i]],{i,1,Length[data]}]; sp2=Apply[Plus,ss]/Apply[Plus,nus]; b=Log[sp2] Apply[Plus,nus]Sum[nus[[i]]Log[s2[[i]]],{i,1,Length[data]}]; c=1+1/(3 (Length[data]-1))(Apply[Plus,1/nus]1/Apply[Plus,nus])//N; bc=b/c; ٤٧٦

f1=Length[data]-1; f2=(Length[data]+1)/(c-1)^2; a=f2/(2-c+2/f2); bcprime=f2 bc c/(f1(a-bc c)); dist=FRatioDistribution[f1,f2]; pvalue=1-CDF[dist,bcprime]; Print["Bartlett/Box Test for Heterogeneity of Variance"]; {SampleVariances->s2,TestStatistic>bcprime,PValue->pvalue,Distribution->dist}//TableForm ]

leveneVariance groupedOneWayAnova Off[General::spell1] <<Statistics`DataManipulation` <<Statistics`NormalDistribution` Clear[groupedOneWayAnova] groupedOneWayAnova[groups_,variances_]:=Module[{k,r,n,respon ses,total,squares,data,sqmeans,sumsqmeans,c,sst,dft,ssg,dfg, msg,sse,dfe,mse,fratio,pvalue,meantable,anovatable}, k=Length[groups]; n=groups//Flatten//Length; responses=Map[Apply[Plus,#]&,groups]; total=Apply[Plus,groups//Flatten]; squares=Apply[Plus,groups^2//Flatten]; sqmeans=responses^2.Table[1/Length[groups[[i]]],{i,1,Le ngth[groups]}]; sumsqmeans= Apply[Plus,sqmeans]; c=total^2/n; sst=squares-c; dft=n-1; ssg=sumsqmeans-c; dfg=k-1; msg=ssg/dfg; sse=sst-ssg; dfe=dft-dfg; mse=sse/dfe; fratio=msg/mse; pvalue=1-CDF[FRatioDistribution[dfg,dfe],fratio]; {SampleVariances->variances,TestStatistic>fratio,PValue->pvalue,Distribution>FRatioDistribution[dfg,dfe]} ] Clear[leveneVariance] leveneVariance[dataset_]:=Module[{medians,variances,d,i,j,gr oups}, ٤٧٧

medians=Map[Median,dataset]; variances=Map[Variance,dataset]; d[i_,j_]:=Abs[dataset[[i,j]]-medians[[i]]]; groups=Table[d[i,j],{i,1,Length[dataset]},{j,1,Length[d ataset[[i]]]}]; Print["Modified Levene's Test for Heterogeneity of Variance"]; groupedOneWayAnova[groups,variances]//TableForm ]

coefficientsOfVariation Off[General::spell1] <<Statistics`DataManipulation` <<Statistics`NormalDistribution` Clear[coefficientsOfVariation] coefficientsOfVariation[dataset_]:=Module[{means,variances,s tnddevs,vs,nus,vp,chi2,dist,pvalue}, means=Map[Mean,dataset]; variances=Map[Variance,dataset]; stnddevs=Map[StandardDeviation,dataset]; vs=Table[stnddevs[[i]]/means[[i]],{i,1,Length[dataset]} ]; nus=Map[Length,dataset]-1; vp=Sum[nus[[i]]vs[[i]],{i,1,Length[dataset]}]/Apply[Plu s,nus]; chi2=(Sum[nus[[i]] (vs^2)[[i]],{i,1,Length[dataset]}]Sum[nus[[i]]vs[[i]],{i,1,Length[dataset]}]^2/Apply[Plus,nus] )/(vp^2 (0.5+vp^2)); dist=ChiSquareDistribution[Length[dataset]-1]; pvalue=1-CDF[dist,chi2]; Print["Test for Heterogeneity of Coefficients of Variation"]; {CoeffsOfVar->vs,TestStatistic->chi2,PValue>pvalue,Distribution->dist}//TableForm ] pig={{2,3.,3},{2.,1,2,3},{4.,5,4,4},{4.,3,2,3}}; bartlettVariance[pig] Bartlett/Box Test for Heterogeneity of Variance SampleVariances  0.333333, 0.666667, 0.25, 0.666667 TestStatistic  0.284557 PValue  0.836518 Distribution  FRatioDistribution3, 203.975 ٤٧٨

leveneVariance[pig] Modified Levene's Test for Heterogeneity of Variance SampleVariances  0.333333, 0.666667, 0.25, 0.666667 TestStatistic  0.196748 PValue  0.896426 Distribution  FRatioDistribution3, 11 coefficientsOfVariation[pig] Test for Heterogeneity of Coefficients of Variation CoeffsOfVar  0.216506, 0.408248, 0.117647, 0.272166 TestStatistic  3.49948 PValue  0.320829 Distribution  ChiSquareDistribution3

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬ ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ ‫ وھﻰ‬pig pig={{2,3.,3},{2.,1,2,3},{4.,5,4,4},{4.,3,2,3}} (٣-٦) ‫وﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﯿﺎﻧﺎت اﻟﺨﺎﺻﺔ ﺑﺎﻟﻤﺜﺎل‬

‫ اﻟﻣﺧرﺟﺎت‬: ‫ﺛﺎﻧﯾﺎ‬ : ‫اﺧﺗﺑﺎر ﻓرض اﻟﻌدم‬

H0 : 12  22  32  42 : ‫ﺿد اﻟﻔرض اﻟﺑدﯾل‬

H1 : ‫اﻟﺗﺑﺎﯾﻧﺎت ﻟﯾﺳت ﻛﻠﮭﺎ ﻣﺗﺳﺎوﯾﺔ‬ ‫ ﻣن ﺧﻼل اﻻﻣر اﻟﺗﺎﻟﻰ‬Bartlett‫ وذﻟك ﺑﺎﺳﺗﺧدام اﺧﺗﺑﺎر‬=0.01 ‫وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ‬ bartlettVariance[pig]

‫ﺣﯾث اﻟﻣﺧرج ھو‬ Bartlett/Box Test for Heterogeneity of Variance SampleVariances  0.333333, 0.666667, 0.25, 0.666667 TestStatistic  0.284557 PValue  0.836518 Distribution  FRatioDistribution3, 203.975 ‫ﺣﯾث‬ Sample Variances ‫ﯾﻌطﻰ ﺗﺑﺎﯾﻧﺎت اﻟﻌﯾﻧﺔ ﻟﻠﻣﻌﺎﻟﺟﺎت ﻛﻣﺎ ﯾﻌطﻰ اﻻﺣﺻﺎء اﻟﻣﻘدر ﻣن ﺧﻼل‬ TestStatistic ‫ ﻣن‬p‫و‬ ٤٧٩

Pvalue p ‫وﺑﻣﺎ ان ﻗﯾﻣﺔ‬

=0.01 ‫ ﻓﺈﻧﻧﺎ ﻧﻘﺑل ﻓرض اﻟﻌدم وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ‬.01 ‫اﻛﺑر ﻣن‬ ‫ ﺳوف ﻧﺳﺗﺧدم اﻻﻣر اﻟﺗﺎﻟﻰ‬Modified Levene's ‫و ﺑﺎﺳﺗﺧدام اﺧﺗﺑﺎر‬ leveneVariance[pig]

‫ﺣﯾث اﻟﻣﺧرج ھو‬ Modified Levene's Test for Heterogeneity of Variance SampleVariances  0.333333, 0.666667, 0.25, 0.666667 TestStatistic  0.196748 PValue  0.896426 Distribution  FRatioDistribution3, 11 ‫ﺣﯾث‬ Sample Variances ‫ﯾﻌطﻰ ﺗﺑﺎﯾﻧﺎت اﻟﻌﯾﻧﺔ ﻟﻠﻣﻌﺎﻟﺟﺎت ﻛﻣﺎ ﯾﻌطﻰ اﻻﺣﺻﺎء اﻟﻣﻘدر ﻣن ﺧﻼل‬ TestStatistic ‫اﯾﺿﺎ‬ ‫ ﻣن‬p Pvalue p ‫وﺑﻣﺎ ان ﻗﯾﻣﺔ‬

=0.01 ‫ ﻓﺈﻧﻧﺎ ﻧﻘﺑل ﻓرض اﻟﻌدم وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ‬.01 ‫اﻛﺑر ﻣن‬ ‫و ﺑﺎﺳﺗﺧدام اﺧﺗﺑﺎر‬ ‫ ﺳوف ﻧﺳﺗﺧدم اﻻﻣر اﻟﺗﺎﻟﻰ‬Hetetogeneity of Coefficients of Variation coefficientsOfVariation[pig]

‫ﺣﯾث اﻟﻣﺧرج ھو‬ Test for Heterogeneity of Coefficients of Variation CoeffsOfVar  0.216506, 0.408248, 0.117647, 0.272166 TestStatistic  3.49948 PValue  0.320829 Distribution  ChiSquareDistribution3 ‫ﺣﯾث اﻻﺣﺻﺎء اﻟﻣﻘدرﻧﺣﺻل ﻋﻠﯾﮫ ﻣن ﺧﻼل‬ TestStatistic ‫اﯾﺿﺎ‬ ‫ ﻣن‬p Pvalue p ‫وﺑﻣﺎ ان ﻗﯾﻣﺔ‬

=0.01 ‫ ﻓﺈﻧﻧﺎ ﻧﻘﺑل ﻓرض اﻟﻌدم وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ‬.01 ‫اﻛﺑر ﻣن‬ .‫وﻟﻠﻌﻠم ﻓﺈن ھﻧﺎك ﻣﺧرﺟﺎت اﺧري ﻻ داﻋﻰ ﻟذﻛرھﺎ‬

٤٨٠

‫)‪ (٢-٢-٦‬اﺧﺗﺑﺎر ﻧﯾوﻣن‪ -‬ﻛﻠز ﻟﻠﻣدى اﻟﻣﺗﻌدد‪:‬‬ ‫‪Multiple Range Test‬‬ ‫إذا ﻛﺎﻧ ت ﻗﯾﻣ ﺔ ‪ f‬اﻟﻣﺣﺳ وﺑﺔ ﻣ ن ﺟ دول ﺗﺣﻠﯾ ل اﻟﺗﺑ ﺎﯾن ﻏﯾ ر ﻣﻌﻧوﯾ ﺔ ﻓﮭ ذا ﯾ دل ﻋﻠ ﻰ أن اﻟﻔ روق ﺑ ﯾن‬ ‫ﻣﺗوﺳطﺎت اﻟﻣﻌﺎﻟﺟ ﺎت ﻟﯾﺳ ت ﻓ روق ﺣﻘﯾﻘﯾ ﺔ وإﻧﻣ ﺎ ﺗﻌ زى ﻟﻣﺟ رد اﻟﺻ دﻓﺔ ‪ ،‬وﺑﺎﻟﺗ ﺎﻟﻲ ﻓﺈﻧﻧ ﺎ ﻧﻘﺑ ل ﻓ رض‬ ‫اﻟﻌدم ‪ . H 0 : 1   2  ...   k‬إذا ﻛﺎﻧت ﻗﯾﻣﺔ ‪ f‬ﻣﻌﻧوﯾﺔ ﻓﮭذا ﯾ دل ﻋﻠ ﻰ أن ﺑﻌ ض اﻟﻔ روق ﺑ ﯾن‬ ‫ﻣﺗوﺳطﺎت اﻟﻣﻌﺎﻟﺟﺎت أو ﻛﻠﮭﺎ ﻣﻌﻧوﯾﺔ ‪ ،‬وﻟﻛن ھذا اﻻﺧﺗﺑﺎر ﻻ ﯾوﺿ ﺢ ﻟﻧ ﺎ أي ﻣ ن ھ ذه اﻟﻔ روق ﻣﻌﻧوﯾ ﺔ‬ ‫‪ ،‬وﻟذﻟك ﻓﺈن اﻟﺑﺎﺣث ﻻ ﺑ د أن ﯾﺟ ري ﻋ دة ﻣﻘﺎرﻧ ﺎت ﺑ ﯾن ھ ذه اﻟﻣﺗوﺳ طﺎت وھ ذا ﻣ ﺎ ﯾﺳ ﻣﻰ ﺑﺎﻟﻣﻘﺎرﻧ ﺎت‬ ‫اﻟﻣﺗﻌددة‪ .‬ھﻧﺎك ﻋدة ط رق ﺗﺳ ﺗﺧدم ﻟﮭ ذا اﻟﻐ رض ‪ .‬ﺳ وف ﺗﻘﺗﺻ ر دراﺳ ﺗﻧﺎ ﻓ ﻲ ھ ذا اﻟﺑﻧ د ﻋﻠ ﻰ اﺧﺗﺑ ﺎر‬ ‫ﻧﯾ وﻣن ﻟﻠﻣﻘﺎرﻧ ﺎت اﻟﻣﺗﻌ ددة ‪ .‬ﯾ ﺗﻠﺧص اﺧﺗﺑ ﺎر ﻧﯾ وﻣن ﻓ ﻲ إﯾﺟ ﺎد ﻋ دة ﻓ روق ﻣﻌﻧوﯾ ﺔ ذات ﻗ ﯾم ﻣﺗزاﯾ دة‬ ‫واﻟﺗﻲ ﺗﺗوﻗف ﺣﺟﻣﮭﺎ ﻋﻠﻰ ﻣدي اﻟﺑﻌد ﺑﯾن اﻟﻣﺗوﺳطﺎت ﺑﻌد ﺗرﺗﯾﺑﮭﺎ‪.‬‬ ‫وﺗﺗﻠﺧص ﺧطوات ﺗﻧﻔﯾذھﺎ ﻋﻠﻰ اﻟﻧﺣو اﻟﺗﺎﻟﻲ ‪:‬‬ ‫)أ( ﻧرﺗب ﻣﺗوﺳطﺎت اﻟﻣﻌﺎﻟﺟﺎت ﺗﻧﺎزﻟﯾﺎ ً‪.‬‬

‫‪MSE‬‬ ‫)ب( ﻧوﺟد اﻟﺧط ﺄ اﻟﻣﻌﯾ ﺎري ﻟﻠﻣﺗوﺳ ط‬ ‫‪n‬‬

‫‪ s x ‬ﺣﯾ ث ‪ MSE‬ھ و ﻣﺗوﺳ ط ﻣﺟﻣ وع ﻣرﺑﻌ ﺎت‬

‫اﻟﺧطﺄ واﻟذي ﯾﻌﺗﺑر ﺗﻘدﯾر ﻟﻠﺗﺑﺎﯾن ‪ ،  2‬وﻧﺣﺻ ل ﻋﻠﯾ ﮫ ﻣ ن ﺟ دول ﺗﺣﻠﯾ ل اﻟﺗﺑ ﺎﯾن ‪ .‬وإذا ﻛﺎﻧ ت أﺣﺟ ﺎم‬ ‫اﻟﻌﯾﻧ ﺎت ﻟﻠﻣﻌﺎﻟﺟ ﺎت ﻏﯾ ر ﻣﺗﺳ ﺎوﯾﺔ ﻓ ﺈن اﺧﺗﺑ ﺎر ﻧﯾ وﻣن ﯾﺳ ﻣﺢ ﺑﺎﺳ ﺗﺑدال ‪ n‬ﻓ ﻲ ﺻ ﯾﻐﺔ ‪ s x‬ﺑﺎﻟوﺳ ط‬ ‫اﻟﺗواﻓﻘﻲ ﻟﻠﻘﯾم ‪ n1, n2, …, nk‬ﺣﯾث اﻟوﺳط اﻟﺗواﻓﻘﻲ ‪:‬‬

‫‪k‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪ ... ‬‬ ‫‪n1 n 2‬‬ ‫‪nk‬‬

‫~‬ ‫‪n‬‬

‫ﺗﺣت ﺷرط أن أﺣﺟ ﺎم اﻟﻌﯾﻧ ﺎت ﺗﻛ ون ﻣﺗﻘﺎرﺑ ﺔ ﻣ ن ﺑﻌﺿ ﮭﺎ ‪ .‬ھ ذا وﯾﻣﻛ ن اﺳ ﺗﺑدال ‪ n‬ﻓ ﻲ ﺻ ﯾﻐﺔ ‪s x‬‬ ‫ﺑﺎﻟﻘﯾﻣﺔ ‪ n .‬ﺣﯾث ‪:‬‬

‫‪2‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪‬‬ ‫)‪n (1) n (k‬‬

‫‪n. ‬‬

‫و أن ‪:‬‬ ‫)‪ = n(1‬ﺣﺟم اﻟﻌﯾﻧﺔ اﻟﻣﻘﺎﺑل ﻷﺻﻐر ﻣﺗوﺳط ﻋﯾﻧﺔ ‪.‬‬ ‫)‪ = n(k‬ﺣﺟم اﻟﻌﯾﻧﺔ اﻟﻣﻘﺎﺑل ﻷﻛﺑر ﻣﺗوﺳط ﻋﯾﻧﺔ ‪.‬‬ ‫)ج( ﺗﺳﺗﺧرج ﻗﯾم ) ‪ ) q( p, ‬ﺗﺳﻣﻰ أﻗل ﻣدي ﻣﻌﻧوي ﻗﯾﺎﺳﻲ ‪least significant studentized‬‬ ‫‪ (range‬ﻣن ﺟ دول ﻧﯾ وﻣن ﻟﻠﻣ دى اﻟﻣﻌﻧ وي ﻓ ﻲ ﻣﻠﺣ ق )‪ (٩‬ﺣﯾ ث ‪ p = 2, 3,…, k‬و‬ ‫‪ ‬ھﻲ ﻣﺳﺗوى اﻟﻣﻌﻧوﯾﺔ و ‪ ‬ھﻲ درﺟﺎت ﺣرﯾﺔ ‪.MSE‬‬ ‫)د( ﻧﺣﺳب ﻗﯾﻣﺔ أﻗل ﻣدى ﻣﻌﻧوي ‪ Rp least significant range‬وذﻟك ﺑﺎﻟﻧﺳﺑﺔ ﻟﻛل ‪p = 2,3,‬‬ ‫‪ …, k‬ﻋﻠﻰ اﻟﻧﺣو اﻟﺗﺎﻟﻲ ‪:‬‬

‫‪R p  q  (p, )s x , p  2,3,..., k.‬‬ ‫‪٤٨١‬‬

‫)ھـ( ﻧﻘﺎرن اﻟﻔ روق ﺑ ﯾن ﻣﺗوﺳ طﺎت اﻟﻣﻌﺎﻟﺟ ﺎت وﻧﺑ دأ ﺑﻣﻘﺎرﻧ ﺔ اﻟﻔ رق ﺑ ﯾن أﻛﺑ ر ﻣﺗوﺳ ط وأﻗ ل ﻣﺗوﺳ ط‬ ‫ﺑﺎﻟﻘﯾﻣﺔ ‪ Rk‬ﺛم ﻧﻘﺎرن اﻟﻔرق ﺑﯾن أﻛﺑر ﻣﺗوﺳط وﺛ ﺎﻧﻲ أﺻ ﻐر ﻣﺗوﺳ ط ﺑﺎﻟﻘﯾﻣ ﺔ ‪ Rk-1‬وﻧواﺻل ھ ذه‬ ‫‪k‬‬ ‫اﻟﻌﻣﻠﯾ ﺔ وإﻟ ﻰ أن ﺗ ﺗم ﻣﻘﺎرﻧ ﺔ ﻛ ل اﻷزواج وﻋ ددھﺎ ‪ .    k(k  1) / 2‬إذا ﻛ ﺎن اﻟﻔ رق‬ ‫‪ 2‬‬ ‫اﻟﻣﺣﺳوب ﺑﯾن ﻣﺗوﺳطﯾن ﯾﺳﺎوى أو أﻋﻠﻰ ﻣن ‪ Rp‬ﻓﯾﻛون ذﻟك اﻟﻔرق ﻣﻌﻧوﯾﺎ‪.‬‬ ‫ﺗﻠﺧص ﻧﺗﺎﺋﺞ اﻻﺧﺗﺑﺎر ﺑوﺿﻊ ﺧطوط ﻣﺷﺗرﻛﺔ ﺗﺣت اﻟﻣﺗوﺳطﺎت اﻟﺗﻲ ﻟ م ﺗﻛ ن ﻓروﻗﮭ ﺎ ﻣﻌﻧوﯾ ﺔ ‪،‬‬ ‫ﻣﻊ اﻹﺑﻘﺎء ﻋﻠﻰ ﺗرﺗﯾب اﻟﻣﺗوﺳطﺎت ﺗﻧﺎزﻟﯾﺎ‪.‬‬

‫ﻣﺛﺎل)‪(٤-٦‬‬ ‫ﻟﺗوﺿﯾﺢ طرﯾﻘﺔ ﻧﯾوﻣن ﻟﻠﻣدى اﻟﻣﺗﻌدد ﻓﺳوف ﻧﺳﺗﺧدم اﻟﺑﯾﺎﻧﺎت اﻟﺗﺎﻟﯾﺔ وﻧﺗﺑﻊ اﻟﺧطوات اﻟﺗﺎﻟﯾﺔ ‪:‬‬

‫اﻟﺣــل‪:‬‬ ‫ﻧرﺗب ﻣﺗوﺳطﺎت اﻟﻣﻌﺎﻟﺟﺎت اﻟﺗﺎﻟﯾﺔ ﺗﻧﺎزﻟﯾﺎ ً ﻓﻰ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫‪C‬‬ ‫‪2.25‬‬

‫‪A‬‬ ‫‪3.75‬‬

‫‪B‬‬ ‫‪7.00‬‬

‫‪D‬‬ ‫‪9.43‬‬

‫اﻟﻣﺗوﺳط‬

‫)ب( ﻣن ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن اﻟﺧﺎص ﺑﮭذا اﻟﻣﺛﺎل ﻓ ﺈن ‪ MSE = 3.298‬ﺑ درﺟﺎت ﺣرﯾ ﺔ ‪  20‬‬

‫‪MSE‬‬ ‫‪ .‬ﻧوﺟ د اﻟﺧط ﺄ اﻟﻣﻌﯾ ﺎري ﻟﻠﻣﺗوﺳ ط‬ ‫‪n‬‬

‫‪ s x ‬وﺑﻣ ﺎ أن أﺣﺟ ﺎم اﻟﻣﻌﺎﻟﺟ ﺎت ﻏﯾ ر ﻣﺗﺳ ﺎوﯾﺔ ﻓﺈﻧﻧ ﺎ‬

‫ﻧﺣﺳب اﻟوﺳط اﻟﺗواﻓﻘﻲ ﻟﻠﻘﯾم ‪ n1, n2, …, nk‬ﻛﺎﻵﺗﻲ ‪:‬‬ ‫‪k‬‬ ‫‪n ‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪n1 n 2 n 3 n 4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪ 5.5721 ,‬‬ ‫‪1 1 1 1 .7178571‬‬ ‫‪  ‬‬ ‫‪4 5 8 7‬‬ ‫‪MSE‬‬ ‫‪3.298‬‬ ‫‪MSE  3.298,SX ‬‬ ‫‪‬‬ ‫‪ 0.7693.‬‬ ‫‪n‬‬ ‫‪5.5721‬‬ ‫ﯾﻣﻛن ﺗﻠﺧﯾص اﻟﻧﺗﺎﺋﺞ ﻟﻠﺣﺳﺎﺑﺎت اﻟﺳﺎﺑﻘﺔ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ﺣﯾث ﻗﯾم )‪ q 0.05 (p,20‬ﺗﺳﺗﺧرج ﻣن‬ ‫ﺟدول ﻧﯾوﻣن ‪-‬ﻛﻠز ﻓﻲ ﻣﻠﺣق )‪ (٥‬ﺣﯾث ‪. p  2,3,4,   20‬‬ ‫اﻟﻘﯾم )‪ R p , q 0.05 (p, 20‬ﻣﻌطﺎه ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫‪4‬‬ ‫‪3.96‬‬ ‫‪3.05‬‬

‫‪3‬‬ ‫‪3.58‬‬ ‫‪2.75‬‬

‫‪2‬‬ ‫‪2.95‬‬ ‫‪2.27‬‬ ‫‪٤٨٢‬‬

‫‪p‬‬

‫)‪q 0.05 (p, 20‬‬ ‫‪Rp‬‬

‫ﯾﻣﻛن ﺗﻠﺧﯾص اﻟﻧﺗﺎﺋﺞ اﻟﺳﺎﺑﻘﺔ ﻋﻠﻲ اﻟﻧﺣو اﻟﻣوﺿﺢ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫‪B‬‬ ‫‪A‬‬ ‫‪C‬‬ ‫‪p‬‬ ‫‪7.00‬‬ ‫‪3.75‬‬ ‫‪2.25‬‬ ‫‪Rp‬‬ ‫‪3.05‬‬ ‫‪2.75‬‬ ‫‪2.27‬‬

‫‪4‬‬ ‫‪3‬‬ ‫‪2‬‬

‫*‪7.18‬‬ ‫*‪4.75‬‬ ‫‪1.5‬‬ ‫‪-‬‬

‫*‪5.68‬‬ ‫*‪3.25‬‬ ‫‪-‬‬

‫*‪2.43‬‬ ‫‪-‬‬

‫‪D‬‬ ‫‪9.43‬‬ ‫‪-‬‬

‫اﻟﺗرﺗﯾب‬ ‫اﻟﻣﺗوﺳط‬ ‫اﻟﻣﻌﺎﻟﺟﺔ‬ ‫‪9.43‬‬ ‫‪7.00‬‬ ‫‪3.75‬‬ ‫‪2.25‬‬

‫ﺣﯾث وﺿﻌت ﻛل اﻟﻔروق اﻟﻣﻣﻛﻧﮫ ﺑﯾن اﻟﻣﺗوﺳطﺎت داﺧل اﻟﺟدول وﺗﻣ ت ﻣﻘﺎرﻧﺗﮭ ﺎ ﺑﻘ ﯾم ‪ R p‬اﻟﻣﻧﺎﺳ ﺑﺔ‪.‬‬ ‫ﯾﺗﺿﺢ ﻣن اﻟﺟدول اﻟﺳﺎﺑق أن اﻟﻔروق ﻋﻠﻰ ﻛل ﻗطر ﻗﯾﻣﮫ ﻣن اﻋﻠ ﻰ اﻟﯾﺳ ﺎر إﻟ ﻰ اﻋﻠ ﻰ اﻟﯾﻣ ﯾن ﻟﮭ ﺎ ﻧﻔ س‬ ‫ﻗﯾﻣﺔ ‪ . p‬ﻋﻠﻰ ﺳﺑﯾل اﻟﻣﺛﺎل اﻟﻔروق ‪ 2.43 , 3.25 , 1.5‬ﺗﻘﻊ ﻋﻠ ﻰ ﻗط ر واﺣ د وﻟﮭ ﺎ ‪ . p  2‬اﻟﻘﯾﻣ ﺔ‬ ‫اﻟﺣرﺟﮫ ﻟﮭذه اﻟﻔروق ھﻰ أﺧر ﻗﯾﻣﺔ ﻓﻰ اﻟﻌﻣود اﻷﺧﯾر )‪ . (2.27‬اﯾﺿ ﺎ اﻟﻔ روق ‪ 5.68 , 4.75‬ﺗﻘﻌ ﻊ‬ ‫ﻋﻠ ﻰ ﻗط ر واﺣ د وﻟﮭ ﺎ ‪ p  3‬وﺗﻘ ﺎرن ﺑﺎﻟﻘﯾﻣ ﺔ ‪) 2.75‬اﻟﻘﯾﻣ ﺔ اﻟﺛﺎﺑﺗ ﺔ ﻓ ﻰ اﻟﻌﻣ ود اﻷﺧﯾ ر( ‪ .‬أﺧﯾ را‬ ‫اﻟﻔ رق ‪ 7.18‬ﯾﻘ ﺎرن ﻋﻧ د ‪ p  4‬ﺑﺎﻟﻘﯾﻣ ﺔ اﻟﺣرﺟ ﺔ ‪ 3.05‬وھ ﻰ اﻟﻘﯾﻣ ﺔ اﻻوﻟ ﻰ ﻓ ﻰ اﻟﻌﻣ ود اﻻﺧﯾ ر‪.‬‬ ‫اﻟﻧﺟﻣﺔ * ﻓﻰ اﻟﺟدول ﺗدل ﻋﻠﻰ ﻓرق ﻣﻌﻧوى وذﻟك ﻋﻧد اﺳ ﺗﺧدام ‪ .   0.05‬ﻟﻠﺳ ﮭوﻟﺔ ﯾﻣﻛ ن ﺗﻠﺧ ﯾص‬ ‫ﻧﺗﺎﺋﺞ اﻟﺟدول اﻟﺳﺎﺑق وذﻟك ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ‪ .‬ﻧﻼﺣظ أﻧﻧﺎ ﻟ م ﻧرﺻ د ﻗﯾﻣ ﺔ ﻟﻠﻔ رق ﺑ ﯾن أي اﻟﻣﺗوﺳ طﯾن‬ ‫ﻣوﺿﻊ اﻟﻣﻘﺎرﻧﺔ ﻛﻣﺎ ﻛﻧﺎ ﻧﻔﻌل ﻣن ﻗﺑل ﺑل رﺻدﻧﺎ ﻓﻘط ﻧﺟﻣﺔ‪.‬‬ ‫‪4‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫*‬ ‫*‬ ‫*‬ ‫‪2‬‬ ‫*‬ ‫*‬ ‫‪1‬‬ ‫‪3‬‬ ‫ﺑﺎﻻﺿﺎﻓﺔ ﻻﺧﺗﺑﺎر ﻧﯾوﻣن ھﻧﺎك ﻋدة طرق ودون اﻟدﺧول ﻓﻰ ﺗﻔﺎﺻﯾل ھذه اﻟطرق ﺳوف ﻧﻘدم ﺑرﻧﺎﻣﺞ‬ ‫ﺟﺎھز ﻟذﻟك ﻣﻊ ﺷرح ﻛﯾﻔﯾﺔ ﺗﻔﺳﯾر اﻟﻧﺗﺎﺋﺞ ﻟﻛل طرﯾﻘﺔ ‪.‬‬

‫ﻣﺛﺎل)‪(٥-٦‬‬ ‫ﺑﺈﺳﺗﺧدام ﺑﯾﺎﻧﺎت اﻟﻣﺛﺎل )‪ (١-٦‬ﺳوف ﻧﺳﺗﺧدم اﻟطرﯾﻘﺔ اﻻوﻟﻰ واﻟﻣﺳﻣﺎه‬ ‫‪mcmTukeyPairs‬‬ ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ ‪:‬‬ ‫]‪Off[General::spell1‬‬ ‫`‪<<Statistics`DataManipulation‬‬ ‫`‪<<Statistics`ContinuousDistributions‬‬ ‫‪Ptrng-Preliminaries‬‬ ‫‪Ptrng‬‬ ‫‪mcmTukeyPairs‬‬ ‫]‪Clear[mcmTukeyPairs,all‬‬ ‫‪٤٨٣‬‬

mcmTukeyPairs::usage="mcmTukeyPairs[dataset,catvar,quanvar,a lpha,options] performsm ultiple pairwise comparisons using the Tukey-Kramer Procedure."; Options[mcmTukeyPairs]={categories->all}; categories::usage="categories is an option for various statistical algorithms that specifies which values of the categorical variable (catvar) to use.";

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m c m T u k e y P a i r s  d a t a s e t _ , c a t v a r _ , q u a n v a r _ ,  _ , o p t s _ _ _ R u l e  :  M o d u l e   i , d a t a , c a t s , g r o u p s , p a i r s , m u s , k , n , r e s p o n s e s , t o t a l , s q u a r e s , s q m e a n s , s u m s q m e a n s , m e a n s , n s , c , s s t , d f t , s s g , d f g , m s g , s s e , d f e , m s e ,  p r i m e , t a l p h a , m e a n d i f f e r e n c e s , l e f t , r i g h t , t o p r i n t , l 2 , l 3 , q a l p h a n u k  , d a t a  D r o p N o n N u m e r i c  C o l u m n  d a t a s e t ,  q u a n v a r , c a t v a r    ; a l l  U n i o n  M a p  #   2   & , d a t a   ; c a t s  c a t e g o r i e s  .  o p t s   . O p t i o n s  m c m T u k e y P a i r s  ; g r o u p s  T a b l e  C o l u m n  S e l e c t  d a t a , #   2     c a t s   i   &  , 1  ,  i , 1 , L e n g t h  c a t s    ; p a i r s  F l a t t e n  T a b l e   c a t s   i   , c a t s   j    ,  j , 2 , L e n g t h  c a t s   ,  i , 1 , j  1   , 1  ; m u s  T a b l e   pairs  i,1    pairs  i,2  ,  i , 1 , L e n g t h  p a i r s    ; k  L e n g t h  g r o u p s  ; n  g r o u p s   F l a t t e n   L e n g t h ; r e s p o n s e s  M a p  A p p l y  P l u s , #  & , g r o u p s  ; t o t a l  A p p l y  P l u s , g r o u p s   F l a t t e n  ; s q u a r e s  A p p l y  P l u s , g r o u p s ^ 2   F l a t t e n  ; s q m e a n s  r e s p o n s e s ^ 2 . T a b l e  1  L e n g t h  g r o u p s   i    ,  i , 1 , L e n g t h  g r o u p s    ; s u m s q m e a n s  A p p l y  P l u s , s q m e a n s  ; m e a n s  T a b l e  r e s p o n s e s   i    L e n g t h  g r o u p s   i    ,  i , 1 , L e n g t h  g r o u p s    ; n s  T a b l e  L e n g t h  g r o u p s   i    ,  i , 1 , L e n g t h  g r o u p s    ; c  t o t a l ^ 2  n ; s s t  s q u a r e s  c ; d f t  n  1 ; s s g  s u m s q m e a n s  c ; d f g  k  1 ; m s g  s s g  d f g ; s s e  s s t  s s g ; d f e  d f t  d f g ; m s e  s s e  d f e ; q a l p h a n u k  q t r  1   , d f e , k  ; m e a n d i f f e r e n c e s  F l a t t e n  T a b l e   m e a n s   i    m e a n s   j   , 1  n s   i    1  n s   j    ,  j , 2 , L e n g t h  m e a n s   ,  i , 1 , j  1   , 1  ; l e f t  i _  :  m e a n d i f f e r e n c e s   i , 1    q a l p h a n u k S q r t  m s e  2 m e a n d i f f e r e n c e s   i , 2    ; r i g h t  i _  :  m e a n d i f f e r e n c e s   i , 1    q a l p h a n u k S q r t  m s e  2 m e a n d i f f e r e n c e s   i , 2    ; t o p r i n t  T a b l e   p a i r s   i , 1   , " , " , p a i r s   i , 2   , m e a n d i f f e r e n c e s   i , 1   , l e f t  i  , "  " , m u s   i   , "  " , r i g h t  i  , I f  l e f t  i   0  r i g h t  i  , " N o " , " Y e s "   ,  i , 1 , L e n g t h  p a i r s    ; t o p r i n t    J o i n    " " , " i , j " , " " , " Y i  Y j" , " " , " " , " C o n f i d e n c e I n t e r v a l " , " " , " " , " S i g n i f i c a n t D i f f e r e n c e "   , t o p r i n t  ; P r i n t  " M u l t i p l e P a i r w i s e C o m p a r i s o n s u s i n g t h e T u k e y  K r a m e r P r o c e d u r e . "  ; P r i n t  T a b l e F o r m  t o p r i n t , T a b l e A l i g n m e n t s    C e n t e r  , T a b l e S p a c i n g    1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1    ; l 2    " T h e f a m i l y w i s e c o n f i d e n c e l e v e l i s a t l e a s t " , 1 0 0  1    , " p e r c e n t . "   ; P r i n t  T a b l e F o r m  l 2 , T a b l e S p a c i n g    1 , 1    ; l 3    " T h e f a m i l y w i s e l e v e l o f s i g n i f i c a n c e i s a t m o s t " , 1 0 0  , " p e r c e n t . "   ; P r i n t  T a b l e F o r m  l 3 , T a b l e S p a c i n g    1 , 1    ;

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

dataset1={{1,10.},{1,14},{1,18},{1,15},{1,12.},{2,16},{2,18} ,{2,22.},{2,18},{2, 15},{3,15},{3,12},{3,8},{3,10},{3,13.}}; mcmTukeyPairs[dataset1,1,2,0.05] Multiple Pairwise Comparisons using the Tukey-Kramer Procedure.   i,j YiYj ConfidenceInterval SignificantDifference

1 , 2 4. 8.80637  12  1 , 3 2.2 2.60637  13  2 , 3 6.2 1.39363  23  The familywise confidence level is

0.80637 No 7.00637 No 11.0064 Yes at least 95. percent.

The familywise level of significance is at most 5. percent.

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬: ‫اوﻻ‬ ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ ‫ وھﻰ‬dataset1 dataset1={{1,10.},{1,14},{1,18},{1,15},{1,12.},{2,16},{2,18} ,{2,22.},{2,18},{2, 15},{3,15},{3,12},{3,8},{3,10},{3,13.}};

‫ اﻟﻣﺧرﺟﺎت‬: ‫ﺛﺎﻧﯾﺎ‬ ‫ﺳوف ﻧﺳﺗﺧدم اﻻﻣر‬ mcmTukeyPairs[dataset1,1,2,0.05]

: ‫وذﻟك ﻟﻠﺣﺻول ﻋﻠﻰ اﻟﻣﺧرج اﻟﺗﺎﻟﻰ‬ Multiple Pairwise Comparisons using the Tukey-Kramer Procedure.

 

i,j YiYj Confidence Interval Significant Difference 1 , 2 4. 8.80637  1 2  0.80637 No 1 , 3 2.2 2.60637  1 3  7.00637 No 2 , 3 6.2 1.39363  2 3  11.0064 Yes The familywise confidence level is at least 95. percent. The familywise level of significance is at most 5. percent.

‫ﯾوﺿ ﺢ اﻟﻌﻣ ود اﻻﺧﯾ ر ﻋ دم وﺟ ود ﻓ رق‬. ‫واﻟ ذى ﯾﺣﺗ وى ﻋﻠ ﻰ ﻓﺗ رات ﺛﻘ ﺔ ﻟﻠﻔ رق ﺑ ﯾن ﻛ ل ﻣﺗوﺳ طﯾن‬ ‫ﻣﻌﻧ وى ﺑ ﯾن اﻟﻣﻌﺎﻟﺟ ﺔ اﻻوﻟ ﻰ واﻟﺛﺎﻧﯾ ﺔ واﻻوﻟ ﻰ واﻟﺛﺎﻟﺛ ﺔ واﯾﺿ ﺎ وﺟ ود ﻓ رق ﻣﻌﻧ وى ﺑ ﯾن اﻟﻣﻌﺎﻟﺟ ﺔ‬ . ‫اﻟﺛﺎﻧﯾﺔ واﻟﻣﻌﺎﻟﺟﺔ اﻟﺛﺎﻟﺛﺔ‬

(٦-٦)‫ﻣﺛﺎل‬ ٤٨٦

‫( ﺳوف ﻧﺳﺗﺧدم اﻟطرﯾﻘﺔ اﻟﺛﺎﻧﯾﺔ واﻟﻣﺳﻣﺎه‬١-٦) ‫ﺑﺈﺳﺗﺧدام ﺑﯾﺎﻧﺎت اﻟﻣﺛﺎل‬ mcmBonferroniPairs ‫وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات اﻟﺑرﻧﺎﻣﺞ ﻟﮭذه اﻟطرﯾﻘﺔ‬ Off[General::spell1] <<Statistics`DataManipulation` <<Statistics`NormalDistribution` Clear[mcmBonferroniPairs,all] mcmBonferroniPairs::usage="mcmBonferroniPairs[dataset,catvar ,quanvar,alpha,options] performsm ultiple pairwise comparisons using the Bonferroni inequality."; Options[mcmBonferroniPairs]={categories->all}; categories::usage="categories is an option for various statistical algorithms that specifies which values of the categorical variable (catvar) to use."; m

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s , t o t a l , s q u a r e s , s q m e a n s , s u m s q m e a n s , s , c , s s t , d f t , s s g , d f g , m s g , s s e , d f e , i m e , t a l p h a , m e a n d i f f e r e n c e s , l e f t , t , o p r i n t , l 2 , l 3  , a  D r o p N o n N u m e r i c  u m n  d a t a s e t ,  q u a n v a r , c a t v a r    ; i o n  M a p  #   2   & , d a t a   ; s  c a t e g o r i e s  .  o p t s   . i o n s  m c m B o n f e r r o n i P a i r s  ; u p s  e  C o l u m n  S e l e c t  d a t a , #   2     c a t s   i   & 1 , L e n g t h  c a t s    ; r s  t e n  T a b l e   c a t s   i   , c a t s   j    , , 2 , L e n g t h  c a t s   ,  i , 1 , j  1   , 1  ;  T a b l e     , p a i r s   i , 1   p a i r s   i , 2   1 , L e n g t h  p a i r s    ; g t h  g r o u p s  ; n  g r o u p s   F l a t t e n   L e n g t h ; r e s p o n s e s  M a p  A p p l y  P l u s , #  & , g r o u p s  ; t o t a l  A p p l y  P l u s , g r o u p s   F l a t t e n  ; s q u a r e s  A p p l y  P l u s , g r o u p s ^ 2   F l a t t e n  s q m e a n s  p o n s e s ^ 2 .  T a b l e  1  L e n g t h  g r o u p s   i    , i , 1 , L e n g t h  g r o u p s    ; u m s q m e a n s  A p p l y  P l u s , s q m e a n s  ; m e a n s  l e  r e s p o n s e s   i    L e n g t h  g r o u p s   i    , , 1 , L e n g t h  g r o u p s    ; n s  T a b l e  L e n g t h  g r o u p s   i    , , 1 , L e n g t h  g r o u p s    ; c  t o t a l ^ 2  n ; s s t  s q u a r e s  c ; d f t  n  1 ; s s g  s u m s q m e a n s  c ; d f g  k  1 ; m s g  s s g  d f g ; s s e  s s t  s s g ; d f e  d f t  d f g ; m s e  s s e  d f e ;  p r i m e     k  k  1   ; t a l p h a  Q u a n t i l e  S t u d e n t T D i s t r i b u t i o n  d f  p r i m e  ; m e a n d i f f e r e n c e s  t t e n  b l e   m e a n s   i    m e a n s   j   , 1  n s   i    1  n s   j    ,  j , 2 , L e n g t h  m e a n s i , 1 , j  1   , 1  ; f t  i _  :  m e a n d i f f e r e n c e s   i , 1    l p h a S q r t  m s e m e a n d i f f e r e n c e s   i , 2    ; i _  :  m e a n d i f f e r e n c e s   i , 1    l p h a S q r t  m s e m e a n d i f f e r e n c e s   i , 2    ; p r i n t  l e   p a i r s   i , 1   , " , " , p a i r s   i , 2   , e a n d i f f e r e n c e s   i , 1   , l e f t  i  , "  " , u s   i   , "  " , r i g h t  i  , f  l e f t  i   0  r i g h t  i  , " N o " , " Y e s "   , , 1 , L e n g t h  p a i r s    ; p r i n t    n    " " , " i , j " , " " , " Y  Y j " , " " , " " , i " C o n f i d e n c e I n t e r v a l " , " " , " " , " S i g n i f i c a n t  D i f f e r e n c e "   , t o p r i n t  ; 

w "

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f i d e n c e . "   ; l 2 , T a b l

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s 1

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

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s i g n i f t . "   ; p a c i n g

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;

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dataset1={{1,10.},{1,14},{1,18},{1,15},{1,12.},{2,16},{2,18} ,{2,22.},{2,18},{2, 15},{3,15},{3,12},{3,8},{3,10},{3,13.}}; mcmBonferroniPairs[dataset1,1,2,0.05] Multiple Pairwise Comparisons using the Bonferroni inequality.  

i,j YiYj Confidence Interval Significant Difference 1 , 2 4. 8.9409  1 2  0.940899 No 1 , 3 2.2 2.7409  1 3  7.1409 No 2 , 3 6.2 1.2591  2 3  11.1409 Yes ٤٨٧

The familywise confidence level is at least 95. percent. The familywise level of significance is at most 5. percent.

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬ ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ ‫ وھﻰ‬dataset1 dataset1={{1,10.},{1,14},{1,18},{1,15},{1,12.},{2,16},{2,18} ,{2,22.},{2,18},{2, 15},{3,15},{3,12},{3,8},{3,10},{3,13.}};

‫ اﻟﻣﺧرﺟﺎت‬: ‫ﺛﺎﻧﯾﺎ‬ ‫ﺳوف ﻧﺳﺗﺧدم اﻻﻣر‬ mcmBonferroniPairs[dataset1,1,2,0.05]

: ‫وذﻟك ﻟﻠﺣﺻول ﻋﻠﻰ اﻟﻣﺧرج اﻟﺗﺎﻟﻰ‬ Multiple Pairwise Comparisons using the Bonferroni inequality.   i,j YiYj Confidence Interval Significant Difference

1 , 2 4. 8.9409  12 1 , 3 2.2 2.7409  13 2 , 3 6.2 1.2591  23 The familywise confidence level is

 0.940899  7.1409  11.1409

No No Yes at least 95. percent.

The familywise level of significance is at most 5. percent.

(٧-٦)‫ﻣﺛﺎل‬ ‫( ﺳوف ﻧﺳﺗﺧدم اﻟطرﯾﻘﺔ اﻟﺛﺎﻟﺛﺔ واﻟﻣﺳﻣﺎه‬١-٦) ‫ﺑﺈﺳﺗﺧدام ﺑﯾﺎﻧﺎت اﻟﻣﺛﺎل‬ mcmDunnSidakPairs ‫وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات اﻟﺑرﻧﺎﻣﺞ ﻟﮭذه اﻟطرﯾﻘﺔ‬ Off[General::spell1] <<Statistics`DataManipulation` <<Statistics`NormalDistribution` Clear[mcmDunnSidakPairs,all] mcmDunnSidakPairs::usage="mcmDunnSidakPairs[dataset,catvar,q uanvar] performs selected Multiple comparisons using the Dunn-Sidak inequality."; Options[mcmDunnSidakPairs]={categories->all};

٤٨٨

categories::usage="categories is an option for various statistical algorithms that specifies which values of the categorical variable (catvar) to use."; m

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T a b l e   c a t s   i   , c a t s   j    , L e n g t h  c a t s   ,  i , 1 , j  1   , 1  ; b l e     , p a i r s   i , 1   p a i r s   i , 2   , e n g t h  p a i r s    ; h  g r o u p s  ;  g r o u p s   F l a t t e n   L e n g t h ; e s p o n s e s  M a p  A p p l y  P l u s , #  & , g r o u p s o t a l  A p p l y  P l u s , g r o u p s   F l a t t e n  ; q u a r e s  A p p l y  P l u s , g r o u p s ^ 2   F l a t t e q m e a n s  s e s ^ 2 .  T a b l e  1  L e n g t h  g r o u p s   i    , 1 , L e n g t h  g r o u p s    ; q m e a n s  A p p l y  P l u s , s q m e a n s  ; e a n s  r e s p o n s e s   i    L e n g t h  g r o u p s   i    , , L e n g t h  g r o u p s    ; s  T a b l e  L e n g t h  g r o u p s   i    , , L e n g t h  g r o u p s    ;  t o t a l ^ 2  n ; s t  s q u a r e s  c ; f t  n  1 ; s g  s u m s q m e a n s  c ; f g  k  1 ; s g  s s g  d f g ; s e  s s t  s s g ; f e  d f t  d f g ; s e  s s e  d f e ; p r i m e   1   1    ^  1   k   k  1   2     a l p h a  Q u a n t i l e  S t u d e n t T D i s t r i b u t i o n  r i m e  ; e a n d i f f e r e n c e s  n    m e a n s   i    m e a n s   j   , n s   i    1  n s   j    ,  j , 2 , L e n g t h  m e a 1 , j  1   , 1  ; i _  :  m e a n d i f f e r e n c e s   i , 1    a S q r t  m s e m e a n d i f f e r e n c e s   i , 2    ;  :  m e a n d i f f e r e n c e s   i , 1    a S q r t  m s e m e a n d i f f e r e n c e s   i , 2    ; n t   p a i r s   i , 1   , " , " , p a i r s   i , 2   , d i f f e r e n c e s   i , 1   , l e f t  i  , "  " ,  i   , "  " , r i g h t  i  , e f t  i   0  r i g h t  i  , " N o " , " Y e s "   , , L e n g t h  p a i r s    ; n t     " " , " i , j " , " " , " Y  Y j " , " " , " " , i n f i d e n c e I n t e r v a l " , " " , " " , g n i f i c a n t  D i f f e r e n c e "   , t o p r i n t  ;

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dataset1={{1,10.},{1,14},{1,18},{1,15},{1,12.},{2,16},{2,18} ,{2,22.},{2,18},{2, 15},{3,15},{3,12},{3,8},{3,10},{3,13.}}; mcmDunnSidakPairs[dataset1,1,2,0.05] Multiple Pairwise Comparisons using the Dunn-Sidak inequality.  

i,j YiYj Confidence Interval Significant Difference 1 , 2 4. 8.9246  1 2  0.924596 No 1 , 3 2.2 2.7246  1 3  7.1246 No 2 , 3 6.2 1.2754  2 3  11.1246 Yes The familywise confidence level is at least 95. percent. The familywise level of significance is at most 5. percent.

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬ ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ ‫ وھﻰ‬dataset1 dataset1={{1,10.},{1,14},{1,18},{1,15},{1,12.},{2,16},{2,18} ,{2,22.},{2,18},{2, 15},{3,15},{3,12},{3,8},{3,10},{3,13.}};

‫ اﻟﻣﺧرﺟﺎت‬: ‫ﺛﺎﻧﯾﺎ‬ ‫ﺳوف ﻧﺳﺗﺧدم اﻻﻣر‬ mcmDunnSidakPairs[dataset1,1,2,0.05]

: ‫وذﻟك ﻟﻠﺣﺻول ﻋﻠﻰ اﻟﻣﺧرج اﻟﺗﺎﻟﻰ‬ ٤٨٩

Multiple Pairwise Comparisons using the Dunn-Sidak inequality.  i,j YiYj ConfidenceInterval Significant Difference

1 , 1 , 2 , The

2 4. 8.9246  12  0.924596 No 3 2.2 2.7246  13  7.1246 No 3 6.2 1.2754  23  11.1246 Yes familywise confidence level is at least 95. percent.

The familywise level of significance is at most 5. percent.

(٨-٦)‫ﻣﺛﺎل‬ ‫( ﺳوف ﻧﺳﺗﺧدم اﻟطرﯾﻘﺔ اﻟراﺑﻌﺔ واﻟﻣﺳﻣﺎه‬١-٦) ‫ﺑﺈﺳﺗﺧدام ﺑﯾﺎﻧﺎت اﻟﻣﺛﺎل‬ mcmScheffePairs ‫وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات اﻟﺑرﻧﺎﻣﺞ ﻟﮭذه اﻟطرﯾﻘﺔ‬ Off[General::spell1] <<Statistics`DataManipulation` <<Statistics`NormalDistribution` Clear[mcmScheffePairs,all] mcmScheffePairs::usage="mcmScheffePairs[dataset,catvar,quanv ar,,options] performs multiplem pairwise comparisons using the Scheffe procedure."; Options[mcmScheffePairs]={categories->all}; categories::usage="categories is an option for various statistical algorithms that specifies which values of the categorical variable (catvar) to use.";

٤٩٠

m o o

S p d

c t u

h s l

r m m t

e e s o d C

s a e p a o U a p r b i a a

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dataset1={{1,10.},{1,14},{1,18},{1,15},{1,12.},{2,16},{2,18} ,{2,22.},{2,18},{2, 15},{3,15},{3,12},{3,8},{3,10},{3,13.}}; mcmScheffePairs[dataset1,1,2,0.05] Multiple Pairwise Comparisons using the Scheffe procedure.

 

i,j YiYj Confidence Interval 1 , 2 4. 8.95531  1 2 1 , 3 2.2 2.75531  1 3 2 , 3 6.2 1.24469  2 3 The familywise confidence level is at

Significant Difference  0.955306 No  7.15531 No  11.1553 Yes least 95. percent.

The familywise level of significance is at most 5. percent.

‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬ ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ ‫ وھﻰ‬dataset1 dataset1={{1,10.},{1,14},{1,18},{1,15},{1,12.},{2,16},{2,18} ,{2,22.},{2,18},{2, 15},{3,15},{3,12},{3,8},{3,10},{3,13.}};

‫ اﻟﻣﺧرﺟﺎت‬: ‫ﺛﺎﻧﯾﺎ‬ ‫ﺳوف ﻧﺳﺗﺧدم اﻻﻣر‬ mcmScheffePairs[dataset1,1,2,0.05]

: ‫وذﻟك ﻟﻠﺣﺻول ﻋﻠﻰ اﻟﻣﺧرج اﻟﺗﺎﻟﻰ‬ Multiple Pairwise Comparisons using the Bonferroni inequality.

٤٩١

‫‪Significant Difference‬‬ ‫‪ 0.955306‬‬ ‫‪No‬‬ ‫‪ 7.15531‬‬ ‫‪No‬‬ ‫‪ 11.1553‬‬ ‫‪Yes‬‬ ‫‪least 95. percent.‬‬

‫‪ ‬‬

‫‪i,j‬‬ ‫‪YiYj‬‬ ‫‪Confidence Interval‬‬ ‫‪1 , 2 4. 8.95531 ‬‬ ‫‪1 2‬‬ ‫‪1 , 3 2.2 2.75531 ‬‬ ‫‪1 3‬‬ ‫‪2 , 3 6.2 1.24469 ‬‬ ‫‪2 3‬‬ ‫‪The familywise confidence level is at‬‬

‫‪The familywise level of significance is at most 5. percent.‬‬

‫واﻟ ذى ﯾﺣﺗ وى ﻋﻠ ﻰ ﻓﺗ رات ﺛﻘ ﺔ ﻟﻠﻔ رق ﺑ ﯾن ﻛ ل ﻣﺗوﺳ طﯾن ‪.‬ﯾوﺿ ﺢ اﻟﻌﻣ ود اﻻﺧﯾ ر ﻋ دم وﺟ ود ﻓ رق‬ ‫ﻣﻌﻧ وى ﺑ ﯾن اﻟﻣﻌﺎﻟﺟ ﺔ اﻻوﻟ ﻰ واﻟﺛﺎﻧﯾ ﺔ واﻻوﻟ ﻰ واﻟﺛﺎﻟﺛ ﺔ واﯾﺿ ﺎ وﺟ ود ﻓ رق ﻣﻌﻧ وى ﺑ ﯾن اﻟﻣﻌﺎﻟﺟ ﺔ‬ ‫اﻟﺛﺎﻧﯾﺔ واﻟﻣﻌﺎﻟﺟﺔ اﻟﺛﺎﻟﺛﺔ ‪.‬‬

‫) ‪ (٣-٦‬اﻟﺗﺻﻧﯾف اﻟﺛﻧﺎﺋﻲ ‪ ،‬ﻣﺷﺎھدة واﺣدة ﻓﻲ ﻛل ﺧﻠﯾﺔ‪:‬‬ ‫‪Two-Way Classification, Single Observation Per Cell‬‬ ‫ﻗد ﺗﺻﻧف ﻓﺋﺔ ﻣن اﻟﻣﺷﺎھدات ﺗﺑﻌﺎ ً ﻟﺻﻔﺗﯾن ﻣﻌﺎ‪ .‬ﻋﻠﻰ ﺳﺑﯾل اﻟﻣﺛ ﺎل ﻋﻧ دﻣﺎ ﯾرﻏ ب اﻟﺑﺎﺣ ث ﻓ ﻲ ﻣﺟ ﺎل‬ ‫اﻟزراﻋﺔ ﻓﻲ دراﺳﺔ ﺗﺄﺛﯾر اﻟطرق اﻟﻣﺧﺗﻠﻔﺔ ﻟﻠزراﻋﺔ ) ﺛﻼﺛﺔ طرق ( وﻛذﻟك اﻷوﻗ ﺎت اﻟﻣﺧﺗﻠﻔ ﺔ ﻟﻠزراﻋ ﺔ‬ ‫) ﻣ ﺎرس وﻓﺑراﯾ ر وﻧ وﻓﻣﺑر وأﻛﺗ وﺑر ( ﻋﻠ ﻰ إﻧﺗﺎﺟﯾ ﺔ ﻣﺣﺻ ول اﻟﻘﺻ ب‪ .‬اﻟﻣﺷ ﺎھدات ﻓ ﻲ ھ ذه اﻟﺣﺎﻟ ﺔ‬ ‫ﯾﻣﻛن وﺿﻌﮭﺎ ﻓﻲ ﺟدول ﻣن ﺛﻼﺛﺔ ﺻﻔوف وأرﺑﻌﺔ أﻋﻣدة ﺣﯾث ﺗﻣﺛل اﻟﺻ ﻔوف ط رق اﻟزراﻋ ﺔ ‪1, 2,‬‬ ‫‪ 3‬وﺗﻣﺛل اﻷﻋﻣدة أوﻗﺎت اﻟزراﻋﺔ ) ﻣﺎرس وﻓﺑراﯾر وﻧ وﻓﻣﺑر وأﻛﺗ وﺑر (‪ .‬ﯾطﻠ ق ﻋﻠ ﻰ ﺗﻘﺎطﻊ أي ﺻ ف‬ ‫ﻣ ﻊ أي ﻋﻣ ود ﺑﺎﻟﺧﻠﯾ ﺔ وﻛ ل ﺧﻠﯾ ﺔ ﺗﺣﺗ وي ﻋﻠ ﻰ ﻣﺷ ﺎھدة واﺣ دة‪ .‬ﻋﻣوﻣ ﺎ ً ‪ ،‬ﻓ ﻲ ﺣﺎﻟ ﺔ اﻟﺗﺻ ﻧﯾف اﻟﺛﻧ ﺎﺋﻲ‬ ‫ﻟﻣﺷﺎھدة واﺣدة ﯾﻣﻛن وﺿﻊ اﻟﻣﺷﺎھدات ﻓ ﻲ ﺟ دول ﯾﺗﻛ ون ﻣ ن ‪ r‬ﻣ ن اﻟﺻ ﻔوف و ‪ c‬ﻣ ن اﻷﻋﻣ دة ﻛﻣ ﺎ‬ ‫ھو ﻣوﺿﺢ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ﺣﯾث أن ‪ x ij‬ﺗرﻣز ﻟﻠﻣﺷﺎھدة ﻓﻲ اﻟﺻف رﻗ م ‪ i‬واﻟﻌﻣ ود رﻗ م ‪ . j‬ﺳ وف‬ ‫ﻧﻔﺗرض أن ‪ x ij‬ﻗﯾم ﻟﻣﺗﻐﯾرات ﻋﺷواﺋﯾﺔ ﻣﺳ ﺗﻘﻠﺔ ﻟﮭ ﺎ ﺗوزﯾﻌ ﺎت طﺑﯾﻌﯾ ﺔ ﺑﻣﺗوﺳ ط ‪  ij‬وﺗﺑ ﺎﯾن ﻣﺷ ﺗرك‬ ‫‪ .  2‬ﻓ ﻲ اﻟﺟ دول اﻟﺗ ﺎﻟﻰ ‪ Ti.‬و ‪ x i.‬ﺗرﻣ ز ﻟﻠﻣﺟﻣ وع واﻟﻣﺗوﺳ ط ﻋﻠ ﻰ اﻟﺗ واﻟﻲ ﻟﻛ ل اﻟﻣﺷ ﺎھدات ﻓ ﻲ‬ ‫اﻟﺻ ف رﻗ م ‪ i‬و ‪ T.j‬و ‪ x. j‬ﺗرﻣ ز ﻟﻠﻣﺟﻣ وع واﻟﻣﺗوﺳ ط ﻟﻛ ل اﻟﻣﺷ ﺎھدات ﻓ ﻲ اﻟﻌﻣ ود رﻗ م ‪ j‬و‬ ‫‪ x .. ,T..‬ﺗرﻣز ﻟﻠﻣﺟﻣوع‬ ‫واﻟﻣﺗوﺳط ﻋﻠﻰ اﻟﺗواﻟﻲ ﻟﻛ ل اﻟﻣﺷ ﺎھدات اﻟﺗ ﻲ ﻋ ددھﺎ ‪ . rc‬اﻟﻣﺗوﺳ ط ﻟﻣﺗوﺳ طﺎت اﻟﻣﺟﺗﻣﻌ ﺎت ﻟﻠﺻف‬ ‫رﻗم ‪ ، i. ، i‬ﯾﻌرف ﻛﺎﻵﺗﻲ ‪:‬‬

‫‪٤٩٢‬‬

‫‪c‬‬

‫‪ ij‬‬

‫‪j1‬‬

‫‪.‬‬ ‫اﻟﻣﺗوﺳط‬

‫اﻟﻣﺟﻣوع‬

‫‪x1.‬‬ ‫‪x 2.‬‬

‫‪T1.‬‬ ‫‪T2.‬‬

‫‪‬‬ ‫‪xi.‬‬

‫‪‬‬ ‫‪Ti.‬‬

‫‪‬‬ ‫‪x r.‬‬

‫‪‬‬ ‫‪Tr.‬‬

‫‪c‬‬

‫‪T..‬‬ ‫‪x..‬‬

‫‪c‬‬

‫…‬

‫‪i. ‬‬ ‫اﻟﺻف‬

‫اﻷﻋﻣــدة‬ ‫… ‪2‬‬ ‫‪j‬‬

‫‪1‬‬

‫‪x11  x12  x1 j  x1c‬‬

‫‪1‬‬

‫‪x 21‬‬

‫‪2‬‬ ‫‪‬‬

‫‪x 22  x 2 j  x 2c‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬

‫‪‬‬

‫‪x i1  x i 2  x ij ‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪x r1  x r 2  x rj ‬‬

‫‪x ic‬‬ ‫‪‬‬ ‫‪x rc‬‬

‫‪T.2  T. j  T.c‬‬ ‫‪x.c‬‬

‫‪x.j‬‬

‫‪T.1‬‬

‫‪x.1 x.2‬‬

‫‪i‬‬ ‫‪‬‬ ‫‪r‬‬ ‫اﻟﻣﺟﻣوع‬ ‫اﻟﻣﺗوﺳط‬

‫وﺑﻧﻔس اﻟﺷﻛل ‪ ،‬اﻟﻣﺗوﺳط ﻟﻣﺗوﺳطﺎت اﻟﻣﺟﺗﻣﻌﺎت ﻟﻠﻌﻣود رﻗم ‪ j‬و ‪ . j‬ﯾﻌرف ﻛﺎﻵﺗﻲ ‪:‬‬ ‫‪r‬‬

‫‪ ij‬‬

‫‪.j  i 1‬‬

‫‪.‬‬ ‫‪r‬‬ ‫واﻟﻣﺗوﺳط ﻟﻣﺗوﺳطﺎت اﻟﻣﺟﺗﻣﻌﺎت اﻟﺗﻲ ﻋددھﺎ ‪ ،  ، rc‬ﯾﻌرف ﻛﺎﻵﺗﻲ ‪:‬‬

‫‪.‬‬

‫‪r c‬‬ ‫‪   ij‬‬ ‫‪i 1 j1‬‬

‫‪rc‬‬

‫‪‬‬

‫ﻟﺗﻘدﯾر ﻣﺎ إذا ﻛﺎن ﺟزء ﻣن اﻻﺧﺗﻼف ﺑﯾن اﻟﻣﺷﺎھدات ﯾرﺟﻊ إﻟﻰ اﻻﺧﺗﻼف ﺑ ﯾن اﻟﺻ ﻔوف‪ ،‬ﻓﺈﻧﻧ ﺎ ﻧﺧﺗﺑ ر‬ ‫ﻓرض اﻟﻌدم‪:‬‬ ‫‪H '0 : μ1.  μ 2.    μ r .  μ,‬‬

‫ﺿد اﻟﻔرض اﻟﺑدﯾل ‪:‬‬

‫واﺣد ﻋﻠﻰ اﻷﻗل ﻣن ‪ μ i.‬ﯾﺧﺗﻠف ﻋن اﻟﺑﺎﻗﻲ ‪H1 : :‬‬ ‫وﺑﻧﻔس اﻟﺷﻛل ﻟﺗﻘدﯾر ﻣﺎ إذا ﻛﺎن ﺟزء ﻣن اﻻﺧ ﺗﻼف ﺑ ﯾن اﻟﻣﺷ ﺎھدات ﯾرﺟ ﻊ إﻟ ﻰ اﻷﻋﻣ دة‪ ،‬ﻓﺈﻧﻧ ﺎ ﻧﺧﺗﺑ ر‬ ‫ﻓرض اﻟﻌدم ‪:‬‬

‫‪H '0' : .1  .2    .c  ‬‬ ‫ﺿد اﻟﻔرض اﻟﺑدﯾل ‪:‬‬

‫‪٤٩٣‬‬

‫واﺣد ﻋﻠﻰ اﻷﻗل ﻣن ‪ . j‬ﯾﺧﺗﻠف ﻋن اﻟﺑﺎﻗﻲ‪H1" :‬‬ ‫ﯾﻣﻛن ﻛﺗﺎﺑﺗﮫ ﻛل ﻣﺷﺎھدة ﻋﻠﻰ اﻟﺷﻛل ‪:‬‬ ‫‪xij  ij ij‬‬ ‫ﺣﯾ ث ‪ ij‬ﯾﻘ ﯾس اﻧﺣ راف ﻗﯾﻣ ﺔ اﻟﻣﺷ ﺎھدة ‪ x ij‬ﻋ ن ﻣﺗوﺳ ط اﻟﻣﺟﺗﻣ ﻊ ‪. ij‬اﻟﺷ ﻛل اﻟﻣﻔﺿ ل واﻟﺷ ﺎﺋﻊ‬ ‫اﻻﺳ ﺗﺧدام ﻟﮭ ذه اﻟﻣﻌﺎدﻟ ﺔ ) أو اﻟﻧﻣ وذج ( ﯾﻣﻛ ن اﻟﺣﺻ ول ﻋﻠﯾ ﮫ ﺑوﺿ ﻊ ‪  ij     i   j‬ﺣﯾ ث‬ ‫‪ i‬ﺗرﻣز ﻟﺗﺄﺛﯾر اﻟﺻف رﻗ م ‪ i‬و ‪  j‬ﺗرﻣ ز ﻟﺗ ﺄﺛﯾر اﻟﻌﻣ ود رﻗ م ‪ . j‬ﺳ وف ﻧﻔﺗ رض أن ﺗ ﺄﺛﯾر اﻟﺻ ﻔوف‬ ‫واﻷﻋﻣ دة ﺗﺟﻣﯾﻌ ﻲ ‪ ) additive‬ﺳ وف ﻧﺷ رح ذﻟ ك ﺑﺎﻟﺗﻔﺻ ﯾل ﻓ ﻲ اﻟﺑﻧ د اﻟﺗ ﺎﻟﻲ (‪ .‬وﻋﻠ ﻰ ذﻟ ك ﯾﻣﻛ ن‬ ‫إﻋﺎدة ﻛﺗﺎﺑﺔ ‪ x ij‬ﻋﻠﻰ اﻟﺷﻛل ‪:‬‬

‫‪x ij     i   j  ij ،‬‬ ‫وذﻟك ﺗﺣت اﻟﻘﯾود اﻟﺗﺎﻟﯾﺔ ‪:‬‬ ‫‪c‬‬

‫‪r‬‬

‫‪j1‬‬

‫‪i 1‬‬

‫‪ i  0 ,   j  0‬‬ ‫وﻋﻠﻰ ذﻟك ‪:‬‬ ‫‪c‬‬

‫) ‪ (   i   j‬‬ ‫‪   i ,‬‬

‫‪j1‬‬

‫‪c‬‬

‫‪i. ‬‬

‫‪r‬‬

‫) ‪ (  i   j‬‬ ‫‪    j,‬‬

‫‪r‬‬

‫‪.j  i 1‬‬

‫اﻵن اﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪:‬‬ ‫‪H0 : 1.   2.     r.  ,‬‬

‫ﺿد اﻟﻔرض اﻟﺑدﯾل ‪:‬‬

‫واﺣد ﻋﻠﻰ اﻷﻗل ﻣن ‪  i.‬ﯾﺧﺗﻠف ﻋن اﻟﺑﺎﻗﻲ‪H1 :‬‬ ‫ﯾﻛﺎﻓﺊ اﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪:‬‬

‫‪H'0 : 1  2  ...  r  0‬‬ ‫ﺿد اﻟﻔرض اﻟﺑدﯾل ‪:‬‬

‫واﺣد ﻋﻠﻰ اﻷﻗل ﻣن ‪ i‬ﻻ ﯾﺳﺎوي ﺻﻔرا ً ‪H1 :‬‬ ‫وﺑﻧﻔس اﻟﺷﻛل اﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪:‬‬ ‫‪H0 : .1  .2    .c  ‬‬ ‫ﺿد اﻟﻔرض اﻟﺑدﯾل ‪:‬‬

‫واﺣد ﻋﻠﻰ اﻷﻗل ﻣن ‪ . j‬ﯾﺧﺗﻠف ﻋن اﻟﺑﺎﻗﻲ ‪H1 :‬‬ ‫ﯾﻛﺎﻓﺊ اﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪:‬‬

‫‪H0 : 1  2  ...  c  0‬‬ ‫‪٤٩٤‬‬

‫ﺿد اﻟﻔرض اﻟﺑدﯾل ‪:‬‬

‫واﺣد ﻋﻠﻰ اﻷﻗل ﻣن ‪  j‬ﻻ ﯾﺳﺎوي ﺻﻔرا ً ‪H1 :‬‬ ‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت اﻟﻛﻠﻲ ھو ‪:‬‬ ‫‪r‬‬

‫‪c‬‬

‫‪SSTO    x ij2  CF,‬‬ ‫‪i 1 j1‬‬

‫ﺣﯾث ‪:‬‬

‫‪‬‬

‫‪T..2‬‬ ‫‪rc‬‬

‫‪CF ‬‬

‫وﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻣﺗوﺳطﺎت اﻟﺻﻔوف ‪ sum of squares for rows means :‬ھو ‪:‬‬ ‫‪r‬‬

‫‪2‬‬ ‫‪ Ti.‬‬

‫‪ CF, ،‬‬

‫‪c‬‬

‫‪SSR  i 1‬‬

‫وﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻣﺗوﺳطﺎت اﻷﻋﻣدة ھو ‪:‬‬ ‫‪c‬‬

‫‪2‬‬ ‫‪ T.j‬‬

‫‪j1‬‬

‫‪ CF,‬‬

‫‪r‬‬

‫‪،‬‬ ‫وﻣﺟﻣوع ﻣرﺑﻌﺎت اﻟﺧطﺄ ھو ‪ SSE‬ﺣﯾث ‪:‬‬

‫‪SSC ‬‬

‫‪SSTO  SSR  SSC  SSE ,‬‬

‫ﻋﻣﻠﯾ ﺎ ً أوﻻ ﻧﺣﺳ ب ‪ SSTO‬و ‪ SSR‬و ‪ SSC‬ﺛ م ﻧﺣﺻ ل ﻋﻠ ﻰ ‪ SSE‬ﺑط رح ﻛ ل ﻣ ن ‪ SSR‬و‬ ‫‪ SSC‬ﻣن ‪ . SST‬أي أن ‪. SSE = SSTO – SSR – SSC‬‬ ‫ﻣﺗوﺳط ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻣﺗوﺳطﺎت اﻟﺻﻔوف ﯾﻌطﻰ ﻛﺎﻵﺗﻲ ‪:‬‬

‫‪SSR‬‬ ‫‪.‬‬ ‫‪r 1‬‬

‫‪MSR ‬‬

‫ﻣﺗوﺳط ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻣﺗوﺳطﺎت اﻟﺻﻔوف ﯾﻌطﻰ ﻛﺎﻵﺗﻲ ‪:‬‬

‫‪SSC‬‬ ‫‪.‬‬ ‫‪c 1‬‬

‫‪MSC ‬‬

‫ﻣﺗوﺳط ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻣﺗوﺳطﺎت اﻻﻋﻣدة ﯾﻌطﻰ ﻛﺎﻵﺗﻲ ‪:‬‬

‫‪SSE‬‬ ‫‪,‬‬ ‫)‪(r  1)(c  1‬‬ ‫ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪ H 0‬ﻓﺈﻧﻧﺎ ﻧﺣﺳب اﻟﻧﺳﺑﺔ ‪:‬‬ ‫‪٤٩٥‬‬

‫‪MSE ‬‬

‫‪MSR‬‬ ‫‪MSE‬‬

‫‪f1 ‬‬

‫ﺳوف ﻧرﻓض ﻓرض اﻟﻌدم ‪ ،‬ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪ ،‬ﻋﻧدﻣﺎ ‪:‬‬ ‫‪F1  f  (r  1, (r  1)(c  1)) .‬‬ ‫ﺑﻧﻔس اﻟﺷﻛل ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪ H 0‬ﻓﺈﻧﻧﺎ ﻧﺣﺳب اﻟﻧﺳﺑﺔ ‪:‬‬

‫‪MSC‬‬ ‫‪.‬‬ ‫‪MSE‬‬

‫‪f2 ‬‬

‫ﺳوف ﻧرﻓض ﻓرض اﻟﻌدم ‪ ،‬ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪ ، ‬ﻋﻧدﻣﺎ ‪:‬‬ ‫‪F2  f  (c  1,(r  1)(c  1)) ‬‬ ‫درﺟﺎت اﻟﺣرﯾﺔ اﻟﺧﺎﺻﺔ ﺑـ ‪ SSE‬ھﻲ‪:‬‬ ‫‪(r  1)(c  1).‬‬

‫ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﻓﻲ ﺣﺎﻟﺔ اﻟﺗﺻﻧﯾف اﻟﺛﻧﺎﺋﻲ ﺑﻣﺷﺎھدة واﺣدة ﻟﻛل ﺧﻠﯾﺔ ﻣوﺿﺢ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫‪ f‬اﻟﻣﺣﺳوﺑﺔ‬

‫‪MSR‬‬ ‫‪MSE‬‬ ‫‪MSC‬‬ ‫‪f2 ‬‬ ‫‪MSE‬‬ ‫‪f1 ‬‬

‫ﻣﺗوﺳط ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت‬

‫ﻣﺟﻣوع‬ ‫اﻟﻣرﺑﻌﺎت‬

‫درﺟﺎت اﻟﺣرﯾﺔ‬

‫‪r 1‬‬

‫‪SSR‬‬ ‫‪r 1‬‬ ‫‪SSC‬‬ ‫‪MSC ‬‬ ‫‪c 1‬‬ ‫‪SSE‬‬ ‫‪MSE ‬‬ ‫)‪(r  1)(c  1‬‬ ‫‪MSR ‬‬

‫‪SSR‬‬ ‫‪SSC‬‬ ‫‪SSE‬‬ ‫‪SST‬‬

‫‪c 1‬‬ ‫)‪( r  1)(c  1‬‬ ‫‪rc  1‬‬

‫ﻣﺻدر‬ ‫اﻻﺧﺗﻼف‬ ‫ﻣﺗوﺳطﺎت‬ ‫اﻟﺻﻔوف‬ ‫ﻣﺗوﺳطﺎت‬ ‫اﻷﻋﻣدة‬ ‫اﻟﺧطﺄ‬ ‫اﻟﻣﺟﻣوع‬

‫ﻣﺛﺎل)‪(٩-٦‬‬ ‫ﺗﻌطﻲ اﻟﺑﯾﺎﻧﺎت ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ اﻟدرﺟﺎت اﻟﺗﻲ ﺣﺻل ﻋﻠﯾﮭﺎ ﺳﺗﺔ ﻣن اﻟطﻠب ﻓﻲ ﺛﻼﺛﺔ ﻣﻘررات‬ ‫واﻟﻣطﻠوب‪):‬أ( ھل ھﻧﺎك ﺗﻔﺎوت ﻓﻲ ﻣﻘدرة اﻟطﻠﺑﺔ ؟ )ب( ھل ھﻧﺎك ﺗﻔﺎوت ﻓﻲ ﺻﻌوﺑﺔ اﻟﻣﻘررات )‬ ‫اﺳﺗﺧدم ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪.(  =0.05‬‬

‫اﻟﻠﻐﺔ اﻟﻔرﻧﺳﯾﺔ‬ ‫‪15‬‬ ‫‪14‬‬ ‫‪12‬‬ ‫‪14‬‬ ‫‪12‬‬ ‫‪9‬‬

‫اﻟﻣﻘرر‬ ‫اﻟﻠﻐﺔ اﻹﻧﺟﻠﯾزﯾﺔ‬ ‫‪18‬‬ ‫‪16‬‬ ‫‪17‬‬ ‫‪19‬‬ ‫‪12‬‬ ‫‪13‬‬

‫اﻟرﯾﺎﺿﯾﺎت‬ ‫‪14‬‬ ‫‪12‬‬ ‫‪16‬‬ ‫‪15‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪٤٩٦‬‬

‫اﻟطﺎﻟب‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬

‫اﻟﺣــل‪:‬‬ ‫ﻓروض اﻟﻌدم ﺳوف ﺗﻛون ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬ ‫)أ( ‪H0 : α1  α 2  ...  α 6  0‬‬

‫)ب( ‪H0 : 1  2  3  0‬‬ ‫اﻟﻔروض اﻟﺑدﯾﻠﺔ ﺳوف ﺗﻛون ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬ ‫)أ( ﻋﻠﻰ اﻷﻗل واﺣد ﻣن ‪  i‬ﻻ ﯾﺳﺎوي ﺻﻔرا ً ‪H1 :‬‬ ‫)ب( ﻋﻠﻰ اﻷﻗل واﺣد ﻣن ‪ j‬ﻻ ﯾﺳﺎوي ﺻﻔرا ً ‪H1 :‬‬ ‫‪ f.05(5,10)=3.33‬واﻟﻣﺳ ﺗﺧرﺟﺔ ﻣ ن ﺟ دول ﺗوزﯾ ﻊ ‪ F‬ﻓ ﻲ ﻣﻠﺣ ق )‪ (٤‬ﺑ درﺟﺎت ﺣرﯾ ﺔ‬ ‫‪ . 1  5 ,  2  10‬أﯾﺿﺎ ‪ . f.05(2,10) = 4.1‬ﻣﻧطﻘﺔ اﻟرﻓض ‪:‬‬ ‫)أ( ‪F1 > 3.33‬‬ ‫)ب( ‪F2 > 4.1‬‬ ‫‪6‬‬

‫‪3‬‬

‫‪SSTO    x ij2  CF‬‬ ‫‪i 1 j1‬‬

‫‪(249)2‬‬ ‫‪ 14  12  ...  12  9 ‬‬ ‫‪18‬‬ ‫‪= 3571 – 3444.5 = 126.5,‬‬ ‫‪2‬‬

‫‪2‬‬

‫‪6‬‬

‫‪2‬‬ ‫‪ T i.‬‬

‫‪ CF‬‬

‫‪c‬‬

‫‪SSR  i 1‬‬

‫‪47 2  422  452  482  342  332 (249) 2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪18‬‬ ‫‪= 3515.67 – 3444.5 = 71.17,‬‬

‫‪ CF‬‬

‫‪3‬‬ ‫‪2‬‬ ‫‪ T. j‬‬ ‫‪j1‬‬

‫‪r‬‬

‫‪SSC ‬‬

‫‪782  952  762 (249)2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪6‬‬ ‫‪18‬‬ ‫‪= 3480.83 – 3444.5 = 36.33.‬‬ ‫ﻓﯾﻣﺎ ﯾﻠﻰ ﻓﻲ ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ‪:‬‬ ‫‪ f‬اﻟﻣﺣﺳوﺑﺔ‬ ‫‪f1=7.492‬‬ ‫‪f2= 9.561‬‬

‫ﻣﺗوﺳط‬ ‫اﻟﻣرﺑﻌﺎت‬ ‫‪14.234‬‬ ‫‪18.165‬‬

‫ﻣﺟﻣوع‬ ‫اﻟﻣرﺑﻌﺎت‬ ‫‪71.17‬‬ ‫‪36.33‬‬ ‫‪٤٩٧‬‬

‫درﺟﺎت‬ ‫اﻟﺣرﯾﺔ‬ ‫‪5‬‬ ‫‪2‬‬

‫ﻣﺻدر اﻻﺧﺗﻼف‬ ‫ﻣﺗوﺳطﺎت اﻟﺻﻔوف‬ ‫ﻣﺗوﺳطﺎت اﻷﻋﻣدة‬

‫‪1.9‬‬

‫‪19‬‬ ‫‪126.5‬‬

‫‪10‬‬ ‫‪17‬‬

‫اﻟﺧطﺄ‬ ‫اﻟﻛﻠـﻲ‬

‫ﺑﻣ ﺎ أن ‪ f1  7.492‬ﺗﻘ ﻊ ﻓ ﻲ ﻣﻧطﻘ ﺔ اﻟ رﻓض ﻧ رﻓض ‪ H 0‬أي أن ھﻧ ﺎك ﺗﻔ ﺎوت ﻓ ﻲ ﻣﻘ درة‬ ‫اﻟطﻠﺑ ﺔ‪.‬أﯾﺿ ﺎ ﺑﻣ ﺎ أن ‪ f 2  9.561‬ﺗﻘ ﻊ ﻓ ﻲ ﻣﻧطﻘ ﺔ اﻟ رﻓض ﻧ رﻓض ‪ H0‬أي أن ھﻧ ﺎك ﺗﻔ ﺎوت ﻓ ﻲ‬ ‫ﺻﻌوﺑﺔ اﻟﻣﻘررات ‪.‬‬ ‫ﺳوف ﯾﺗم اﺳﺗﺧدام ]‪ [Mathematica 5.0‬ﻓﻲ ﺣل اﻟﻣﺛﺎل اﻟﺳﺎﺑق ﺣﯾث ‪:‬‬ ‫ﻣﺻ در اﻷﺧ ﺗﻼف ﺗﻌﻧ ﻲ )‪ ، (S.V‬ﻣﺗوﺳ طﺎت اﻟﺻ ﻔوف ﺗﻌﻧ ﻰ )‪ ، (A‬ﻣﺗوﺳ طﺎت اﻷﻋﻣ دة ﺗﻌﻧ ﻲ‬ ‫)‪، (B‬اﻟﺧط ﺄ ﺗﻌﻧ ﻲ )‪، (error‬اﻟﻛﻠ ﻲ ﺗﻌﻧ ﻲ )‪ ، (total‬درﺟ ﺎت اﻟﺣرﯾ ﺔ ﺗﻌﻧ ﻲ )‪ ، (df‬ﻣﺟﻣ وع‬ ‫اﻟﻣرﺑﻌﺎت ﺗﻌﻧﻲ )‪ ، (ss‬ﻣﺗوﺳط اﻟﻣرﺑﻌﺎت ﺗﻌﻧﻲ )‪ f ، (mss‬اﻟﻣﺣﺳوﺑﺔ ﺗﻌﻧﻲ )‪.(f‬‬ ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﻣﻛﺗوب ﺑﻠﻐﺔ ‪ Mathematica‬وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬ ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫‪=0.05‬‬ ‫‪0.05‬‬ ‫‪a={{14,18,15.},{12,16,14.},{16,17,12.},‬‬ ‫}}‪{15,19,14.},{10,12,12.},{11,13.,9‬‬ ‫}‪{{14,18,15.},{12,16,14.},{16,17,12.},{15,19,14.},{10,12,12.‬‬ ‫}}‪,{11,13.,9‬‬ ‫]‪Dimensions[a‬‬ ‫}‪{6,3‬‬ ‫]]‪b=N[Transpose[a‬‬ ‫‪{{14.,12.,16.,15.,10.,11.},{18.,16.,17.,19.,12.,13.},{15.,14‬‬ ‫}}‪.,12.,14.,12.,9.‬‬ ‫]‪f[x_]:=Apply[Plus,x‬‬ ‫]‪g[x_]:=Length[x‬‬ ‫]]]‪na=g[a[[1‬‬ ‫‪3‬‬ ‫]]]‪nb=g[b[[2‬‬ ‫‪6‬‬ ‫‪k1=na-1‬‬ ‫‪2‬‬ ‫‪k2=nb-1‬‬ ‫‪5‬‬ ‫‪N1=na*nb‬‬ ‫‪18‬‬ ‫]‪xxx=Flatten[a‬‬ ‫}‪{14,18,15.,12,16,14.,16,17,12.,15,19,14.,10,12,12.,11,13.,9‬‬ ‫]‪xx2=f[xxx‬‬ ‫‪٤٩٨‬‬

249. sxx22=f[xxx^2] 3571. cf=(xx2^2)/N1 3444.5 x1=Map[f,a] {47.,42.,45.,48.,34.,33.} aaaa1=f[x1^2]/na 3515.67 aaaa=aaaa1-cf 71.1667 ssa=aaaa/k2 14.2333 xx2=Map[f,b] {78.,95.,76.} bbb1=f[xx2^2]/nb 3480.83 bbbb=bbb1-cf 36.3333 sbb=bbbb/k1 18.1667 tot=sxx22-cf 126.5 sser=tot-aaaa-bbbb 19. v1=N1-1-k1-k2 10 msser=sser/v1 1.9 v1=N1-1-k1-k2 10 ff1=ssa/msser 7.49123 ff2=sbb/msser 9.5614 rt2=List[" df "," ss "," mss "," f "] { df , ss , mss , f } rt1=List[k2,aaaa,ssa,ff1] {5,71.1667,14.2333,7.49123} rt3=List[k1,bbbb,sbb,ff2] {2,36.3333,18.1667,9.5614} rt9=List[v1,sser,msser,"-"] {10,19.,1.9,-} rr9=List[N1-1,tot,"_","_"] {17,126.5,_,_} a11=TableHeadings->{{ S.V,A,B,error,total},{ANOVA}} ٤٩٩

‫}}‪TableHeadings{{S.V,A,B,error,total},{ANOVA‬‬ ‫]‪uu1=TableForm[{rt2,rt1,rt3,rt9,rr9},a11‬‬

‫‪f‬‬ ‫‪7.49123‬‬ ‫‪9.5614‬‬ ‫‪‬‬

‫_‬

‫‪mss‬‬ ‫‪14.2333‬‬ ‫‪18.1667‬‬ ‫‪1.9‬‬ ‫_‬

‫‪ss‬‬ ‫‪71.1667‬‬ ‫‪36.3333‬‬ ‫‪19.‬‬ ‫‪126.5‬‬

‫‪ANOVA‬‬ ‫‪df‬‬ ‫‪5‬‬ ‫‪2‬‬ ‫‪10‬‬ ‫‪17‬‬

‫‪S.V‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪error‬‬ ‫‪total‬‬

‫`‪<<Statistics`ContinuousDistributions‬‬ ‫]‪f1=Quantile[FRatioDistribution[k2,v1],1-‬‬ ‫‪3.32583‬‬ ‫]]"‪If[ff1>f1,Print["RjectHo"],Print["AccpetHo‬‬ ‫‪RjectHo‬‬

‫]‪f2=Quantile[FRatioDistribution[k1,v1],1-‬‬ ‫‪4.10282‬‬ ‫]]"‪If[ff2>f2,Print["RjectHo"],Print["AccpetHo‬‬ ‫‪RjectHo‬‬ ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ ‪:‬‬ ‫اوﻻ ‪ :‬اﻟﻣدﺧﻼت‬ ‫ﻣﺴﺘﻮى اﻟﻤﻌﻨﻮﯾﺔ ﻣﻦ اﻻﻣﺮ‬ ‫‪=0.05‬‬

‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ ‪ a‬وﻫﻰ‬ ‫‪a={{14,18,15.},{12,16,14.},{16,17,12.},‬‬ ‫}}‪{15,19,14.},{10,12,12.},{11,13.,9‬‬ ‫وﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﻴﺎﻧﺎت اﻟﺨﺎﺻﺔ ﻓﻰ ﻣﺜﺎل )‪ (٩-٦‬ﺣﻴﺚ ﻳﺘﻢ ادﺧﺎل اﻟﺒﻴﺎﻧﺎت ﺻﻒ ﺻﻒ‪.‬‬ ‫ﺛﺎﻧﯾﺎ ‪ :‬اﻟﻣﺧرﺟﺎت‬

‫ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﯾﺗم اﻟﺣﺻول ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ ‫]‪=TableForm[{rt2,rt1,rt3,rt9,rr9},a11‬‬ ‫ﻓروض اﻟﻌدم ﺳوف ﺗﻛون ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬ ‫)أ( ‪H0 : α1  α 2  ...  α 6  0‬‬

‫)ب( ‪H0 : 1  2  3  0‬‬ ‫اﻟﻔروض اﻟﺑدﯾﻠﺔ ﺳوف ﺗﻛون ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬ ‫)أ( ﻋﻠﻰ اﻷﻗل واﺣد ﻣن ‪  i‬ﻻ ﯾﺳﺎوي ﺻﻔرا ً ‪H1 :‬‬ ‫)ب( ﻋﻠﻰ اﻷﻗل واﺣد ﻣن ‪ j‬ﻻ ﯾﺳﺎوي ﺻﻔرا ً ‪H1 :‬‬ ‫)‪ f.05 (5,10‬واﻟﻣﺳﺗﺧرﺟﺔ ﻋﻧد درﺟﺎت ﺣرﯾﺔ ‪ 1  5, 2  10‬ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪f1=Quantile[FRatioDistribution[k2,v1],1-‬‬ ‫‪ f1‬اﻟﻣﺣﺳوﺑﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر‬ ‫‪ff1=ssa/msser‬‬

‫اﻟﻘرار اﻟذى ﯾﺗﺧذ ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ ‫‪٥٠٠‬‬

If[ff1>f1,Print["RjectHo"],Print["AccpetHo"]]

‫واﻟﻣﺧرج ھو‬ Reject H0

.‫اى رﻓض ﻓرض اﻟﻌدم أي أن ھﻧﺎك ﺗﻔﺎوت ﻓﻲ ﻣﻘدرة اﻟطﻠﺑﺔ‬ ‫اﯾﺿﺎ‬ ‫ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬1  2, 2  10 ‫ واﻟﻣﺳﺗﺧرﺟﺔ ﻋﻧد درﺟﺎت ﺣرﯾﺔ‬f.05 (2,10) f2=Quantile[FRatioDistribution[k1,v1],1-] ‫ اﻟﻣﺣﺳوﺑﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر‬f 2 ff2=sbb/msser

‫اﻟﻘرار اﻟذى ﯾﺗﺧذ ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ If[ff2>f2,Print["RjectHo"],Print["AccpetHo"]]

‫واﻟﻣﺧرج ھو‬ Reject H0

‫اى رﻓض ﻓرض اﻟﻌدم أي أن ھﻧﺎك ﺗﻔﺎوت ﻓﻲ ﺻﻌوﺑﺔ اﻟﻣﻘررات‬

(١٠-٦)‫ﻣﺛﺎل‬ : ‫ﺳوف ﻧﻘدم ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻟﺣل اﻟﻣﺛﺎل اﻟﺳﺎﺑق‬ Off[General::spell1]; <<Statistics`NormalDistribution` <<Statistics`DataManipulation` twoWayAnova - Data entered as Matrix twoWayANOVAarray - Data entered as an array Model I Model II Model III Model IV Model IV modelFour[twoLists_]:=Module[{a,b,n,capN,sumOfVals, sumOfValsSq,sumOverA,sqValsA,factorASumOfSq,sumOverB,sq ValsB,factorBSumOfSq,c,totalSS,totalDF,sumOverColumnAndSquar e,flatList,num,cellsSS,cellsDF,withinCellsSS,withinCellsDF,f actorADF,factorBDF,ACrossBinteractionSS,ACrossBinteractionDF ,factorAMS,factorBMS,ACrossBMS,withinCellsMS,fA,pA,fB,pB,fAB ,pAB,cd,lineOne,lineTwo,lineFour,lastLine}, a=Length[twoLists]; b=Length[twoLists[[1]]]; n=Length[twoLists[[1,1]]]; capN=a b n;

٥٠١

sumOfVals=Sum[twoLists[[i,j,l]],{i,1,a},{j,1,b},{l,1,n} ]; sumOfValsSq=Sum[twoLists[[i,j,l]]^2,{i,1,a},{j,1,b},{l, 1,n}]; sumOverA=Table[Sum[twoLists[[i,j,l]],{j,1,b},{l,1,n}],{ i,1,a}]; sqValsA=Table[sumOverA[[i]]^2,{i,1,Length[sumOverA]}]; factorASumOfSq=Apply[Plus,sqValsA]/(b n)-c; sumOverB=Table[Sum[twoLists[[i,j,l]],{i,1,a},{l,1,n}],{ j,1,b}]; sqValsB=Table[sumOverB[[j]]^2,{j,1,Length[sumOverB]}]; factorBSumOfSq=Apply[Plus,sqValsB]/(a n)-c; c=sumOfVals^2/capN; totalSS=sumOfValsSq-c; totalDF=a b-1; sumOverColumnAndSquare[v_]:=Sum[v[[k]],{k,1,Length[v]}] ^2; flatList=Flatten[twoLists,1]; num=Map[sumOverColumnAndSquare,flatList]; cellsSS=Apply[Plus,num]/n-c; cellsDF=a b-1; withinCellsSS=totalSS-cellsSS; withinCellsDF=(a-1)(b-1); factorADF=a-1; factorBDF=b-1; ACrossBinteractionDF=factorADF factorBDF; factorAMS=factorASumOfSq/factorADF; factorBMS=factorBSumOfSq/factorBDF; SSAB=totalSS-factorASumOfSq-factorBSumOfSq; MSAB=SSAB/ACrossBinteractionDF; withinCellsMS=withinCellsSS/withinCellsDF; fA=factorAMS/MSAB; pA=1CDF[FRatioDistribution[factorADF,ACrossBinteractionDF ],fA]; fB=factorBMS/MSAB; pB=1CDF[FRatioDistribution[factorBDF,withinCellsDF ],fB]; cd=N[(totalSS-withinCellsSS)/totalSS]; Print["Analysis of Variance Table"]; lineOne={"Factor A",factorASumOfSq//N,factorADF,factorAMS//N,fA//N,pA//N}; ٥٠٢

r A r B

lineTwo={"Factor B",factorBSumOfSq//N,factorBDF,factorBMS//N,fB//N,pB//N}; lineFour={"Error",SSAB//N,withinCellsDF,MSAB//N," "," "}; lastLine={"Total",totalSS//N,totalDF," "," "," "}; Print[TableForm[{lineOne,lineTwo,lineFour,lastLine},Tab leHeadings->{None,{"Source","Sum of Squares","DF","Mean Squares","Fratio","P-value"}}]]; ] temperature={{{14.},{12},{16},{15},{10},{11}},{{18},{16.},{1 7},{19},{12},{13}},{{15.},{14},{12},{14},{12},{9}}}; twoWayAnova[temperature,model->singleValuePerCell] Analysis of Variance Table

Sum of Squares 36.3333 71.1667 19. 126.5

DF 2 5 10 17

Mean Squares 18.1667 14.2333 1.9

Fratio 9.5614 7.49123

Pvalue 0.00477347 0.0036498

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬ : ‫ وھﻰ‬temperature ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ temperature={{{14.},{12},{16},{15},{10},{11}},{{18},{16.},{1 7},{19},{12},{13}},{{15.},{14},{12},{14},{12},{9}}};

.(١٠-٦) ‫وﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﻴﺎﻧﺎت اﻟﺨﺎﺻﺔ ﻓﻰ ﻣﺜﺎل‬ ‫ اﻟﻣﺧرﺟﺎت‬: ‫ﺛﺎﻧﯾﺎ‬

‫ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﯾﺗم اﻟﺣﺻول ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ twoWayAnova[temperature,model->singleValuePerCell] ‫واﻟﻣﺧرج ھو‬

Analysis of Variance Table

Source Factor A Factor B Error Total

Sum of Squares 36.3333 71.1667 19. 126.5

DF 2 5 10 17

Mean Squares 18.1667 14.2333 1.9

Fratio 9.5614 7.49123

Pvalue 0.00477347 0.0036498

. ‫ ﺗﻣﺛل اﻟﺻﻔوف‬B ‫ ھﻧﺎ ﺗﻣﺛل اﻻﻋﻣدة و‬A : ‫ﻣﻠﺣوظﺔ‬

٥٠٣

‫)‪ (٤-٦‬اﻟﺗﺻﻧﯾف اﻟﺛﻧﺎﺋﻲ ‪ ،‬ﻋدة ﻣﺷﺎھدات ﻟﻛل ﺧﻠﯾﺔ‪:‬‬ ‫‪Two-Way Classification , Several Observations Per Cell‬‬ ‫ﻟﻠﺗوﺿ ﯾﺢ ﻟﮭ ذه اﻟطرﯾﻘ ﺔ وﺑ ﺎﻟرﺟوع إﻟ ﻰ اﻟﺗﺟرﺑ ﺔ اﻟزراﻋﯾ ﺔ اﻟﺧﺎﺻ ﺔ ﺑدراﺳ ﺔ ﺗ ﺄﺛﯾر اﻷوﻗ ﺎت اﻟﻣﺧﺗﻠﻔ ﺔ‬ ‫ﻟﻠزراﻋ ﺔ ) ﻓﺑراﯾ ر – ﻣ ﺎرس – أﻛﺗ وﺑر – ﻧ وﻓﻣﺑر ( وط رق اﻟزراﻋ ﺔ اﻟﻣﺧﺗﻠﻔ ﺔ ‪ 1, 2, 3‬وﺑﻔ رض أن‬ ‫ﻧﺗﯾﺟ ﺔ اﻟﺗﺟرﺑ ﺔ أوﺿ ﺣت أﻧ ﮫ ﻋﻧ د وﻗ ت اﻟزراﻋ ﺔ ﻓﺑراﯾ ر ﻛ ﺎن أﻋﻠ ﻲ ﻣﺗوﺳ ط ﻟﻣﺣﺻ ول اﻟﻘﺻ ب ﻋﻧ د‬ ‫اﺳ ﺗﺧدام اﻟطرﯾﻘ ﺔ ‪ 1‬وأﻗ ل ﻣﺗوﺳ ط ﻟﻣﺣﺻ ول اﻟﻘﺻ ب ﻋﻧ د اﺳ ﺗﺧدام اﻟطرﯾﻘ ﺔ ‪ 2‬ﺑﯾﻧﻣ ﺎ ﻋﻧ د وﻗ ت‬ ‫اﻟزراﻋ ﺔ ﻣ ﺎرس ﻛ ﺎن أﻋﻠ ﻲ ﻣﺗوﺳ ط ﻟﻣﺣﺻ ول اﻟﻘﺻ ب ﻋﻧ د اﺳ ﺗﺧدام اﻟطرﯾﻘ ﺔ ‪ 2‬واﻗ ل ﻣﺗوﺳ ط‬ ‫ﻟﻣﺣﺻ ول اﻟﻘﺻ ب ﻋﻧ د اﺳ ﺗﺧدام اﻟطرﯾﻘ ﺔ ‪ . 1‬ھ ذه اﻟﺧﺎﺻ ﯾﺔ ﺗﻌ رف ﺑﺎﻟﺗﻔﺎﻋ ل ﺑ ﯾن أوﻗ ﺎت اﻟزراﻋ ﺔ‬ ‫وطرق اﻟزراﻋﺔ وھﻲ ﺗﻛﺷف ﻋﻣﺎ إذا ﻛﺎن ﻷوﻗﺎت اﻟزراﻋﺔ آﺛﺎر ﻣﺧﺗﻠﻔﺔ ﻋﻠ ﻰ ط رق اﻟزراﻋ ﺔ ‪ .‬ﻛﻣ ﺎ ﻻ‬ ‫ﯾﻛ ون ﻟﻠﺗﻔﺎﻋ ل أﺛ ر إذا اﺗﺿ ﺢ أن ط رق اﻟزراﻋ ﺔ ﻣوﺿ ﺢ اﻟﺑﺣ ث ﻣﺗﻧ ﺎظرة ﻟ دي اﻷوﻗ ﺎت اﻟﻣﺧﺗﻠﻔ ﺔ‬ ‫ﻟﻠزراﻋﺔ ‪.‬‬ ‫ﻻﺧﺗﺑ ﺎر اﻟﻔ روق ﺑ ﯾن اﻟﺻ ﻔوف واﻷﻋﻣ دة ﻋﻧ دﻣﺎ ﻻ ﯾﺗﺣﻘ ق اﻟﺷ رط اﻟﺗﺟﻣﯾﻌ ﻲ ‪ ،‬أي ﻓ ﻲ وﺟ ود ﺗﻔﺎﻋ ل‬ ‫‪ interaction‬ﺑﯾن اﻟﺻﻔوف واﻷﻋﻣدة وﻟﻠﺣﺻول ﻋﻠﻰ اﻟﺻ ﯾﻎ اﻟﻌﺎﻣ ﺔ ﻟﺗﺣﻠﯾ ل اﻟﺗﺑ ﺎﯾن ﻓ ﻲ ھ ذه اﻟﺣﺎﻟ ﺔ‬ ‫ﺳوف ﻧﻔﺗرض اﻟﺣﺎﻟﺔ اﻟﺗ ﻲ ﺗﻛ ون ﻓﯾﮭ ﺎ ﻋ دد اﻟﻣﺷ ﺎھدات ) اﻟﻣﻛ ررات ( ‪ replications‬ﻓ ﻲ ﻛ ل ﺧﻠﯾ ﺔ‬ ‫ﺗﺳ ﺎوي ‪ . n‬ﺑﻔ رض أن اﻟﻣﺷ ﺎھدات اﻟﺗ ﻲ ﺳ وف ﻧﺣﺻ ل ﻋﻠﯾﮭ ﺎ ﻣ ن اﻟﺗﺟرﺑ ﺔ ‪ ،‬ﻓ ﻲ ھ ذه اﻟﺣﺎﻟ ﺔ ‪ ،‬ﯾﻣﻛ ن‬ ‫ﺗرﺗﯾﺑﮭﺎ ﻓﻲ ﺟدول ‪ ،‬ﻣﺛل اﻟﺟدول اﻟﺗﺎﻟﻰ ‪ ،‬واﻟذي ﯾﺗﻛون ﻣن ‪ r‬ﻣن اﻟﺻﻔوف و ‪ c‬ﻣ ن اﻷﻋﻣ دة ‪ .‬وﻛ ل‬ ‫ﺧﻠﯾ ﺔ ﺗﺣﺗ وي ﻋﻠ ﻰ ‪ n‬ﻣ ن اﻟﻣﺷ ﺎھدات ‪ .‬ﺑﻔ رض أن ‪ xijk‬ﺗرﻣ ز ﻟﻠﻣﺷ ﺎھدة رﻗ م ‪ k‬ﻓ ﻲ اﻟﺻ ف رﻗ م ‪i‬‬ ‫واﻟﻌﻣود رﻗم ‪ . j‬اﻟﻣﺷﺎھدات اﻟﺗﻲ ﻋددھﺎ ‪ rcn‬ﻣوﺿﺣﺔ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪.‬‬ ‫اﻟﻣﺗوﺳط‬

‫اﻟﻣﺟﻣوع‬

‫‪x ..2‬‬

‫‪T1..‬‬

‫‪x ..1‬‬

‫‪T2..‬‬

‫‪c‬‬ ‫‪x1c1‬‬ ‫‪x1c2‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪x1cn‬‬ ‫‪x2c1‬‬ ‫‪x2c2‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪x2cn‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪xrc1‬‬ ‫‪xrc2‬‬

‫‪.‬‬

‫اﻷﻋﻣـــدة‬ ‫‪2‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪x121‬‬ ‫‪x122‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪x12n‬‬ ‫‪x221‬‬ ‫‪x222‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪x22n‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪xr21‬‬ ‫‪xr22‬‬ ‫‪٥٠٤‬‬

‫اﻟﺻﻔوف‬ ‫‪1‬‬ ‫‪x111‬‬ ‫‪x112‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪x11n‬‬ ‫‪x211‬‬ ‫‪x212‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪x21n‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪xr11‬‬ ‫‪xr12‬‬

‫‪1‬‬

‫‪2‬‬

‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬

‫‪x ..r‬‬

‫‪Tr..‬‬

‫‪.‬‬

‫‪xrcn‬‬ ‫‪x ...‬‬

‫…‪T‬‬

‫‪‬‬

‫‪T.c.‬‬ ‫‪x .2.‬‬

‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪xr2n‬‬

‫‪xr1n‬‬

‫‪T.2.‬‬

‫‪T.1.‬‬

‫‪x .1.‬‬

‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬

‫‪r‬‬

‫‪x .c.‬‬

‫اﻟﻣﺟﻣوع‬ ‫اﻟﻣﺗوﺳط‬

‫اﻟﻣﺷﺎھدات ﻓﻲ اﻟﺧﻠﯾﺔ رﻗم ‪ ij‬ﺗﻣﺛل ﻋﯾﻧ ﺔ ﻋﺷ واﺋﯾﺔ ﻣ ن اﻟﺣﺟ م ‪ n‬ﻣ ن ﻣﺟﺗﻣ ﻊ ﯾﻔﺗ رض أﻧ ﮫ ﯾﺗﺑ ﻊ ﺗوزﯾﻌ ﺎ ً‬ ‫طﺑﯾﻌﯾﺎ ً ﺑﻣﺗوﺳط ‪  ij‬وﺗﺑﺎﯾن ‪ .  2‬ﻛل اﻟﻣﺟﺗﻣﻌﺎت اﻟﺗﻲ ﻋددھﺎ ‪ rc‬ﯾﻔﺗرض أن ﻟﮭﺎ ﺗﺑ ﺎﯾن ﻣﺷ ﺗرك ‪ 2‬‬ ‫‪ .‬ﺑﻘﯾﺔ اﻟرﻣوز اﻟﻣﻔﯾدة ‪ ،‬ﺑﻌﺿﮭﺎ ﻣﻌطﻰ ﻓﻲ اﻟﺟدول اﻟﺳﺎﺑق‪ ،‬ﯾﻣﻛن ﺗوﺿﯾﺣﮭﺎ ﻛﺎﻵﺗﻲ‪:‬‬ ‫‪ = Tij.‬ﻣﺟﻣوع اﻟﻣﺷﺎھدات ﻓﻲ اﻟﺧﻠﯾﺔ رﻗم ‪i,j‬‬ ‫‪ = Ti..‬ﻣﺟﻣوع اﻟﻣﺷﺎھدات ﻓﻲ اﻟﺻف رﻗم ‪i‬‬ ‫‪ = T.j.‬ﻣﺟﻣوع اﻟﻣﺷﺎھدات ﻓﻲ اﻟﻌﻣود رﻗم ‪j‬‬ ‫…‪ = T‬ﻣﺟﻣوع ﻛل اﻟﻣﺷﺎھدات اﻟﺗﻲ ﻋددھﺎ ‪rcn‬‬ ‫‪ = xij.‬ﻣﺗوﺳط ﻛل اﻟﻣﺷﺎھدات ﻓﻲ اﻟﺧﻠﯾﺔ رﻗم ‪ij‬‬ ‫‪ = x i..‬ﻣﺗوﺳط ﻛل اﻟﻣﺷﺎھدات ﻓﻲ اﻟﺻف رﻗم ‪i‬‬ ‫‪ = x.j.‬ﻣﺗوﺳط ﻛل اﻟﻣﺷﺎھدات ﻓﻲ اﻟﻌﻣود رﻗم ‪j‬‬ ‫‪ = x ...‬ﻣﺗوﺳط ﻛل اﻟﻣﺷﺎھدات اﻟﺗﻲ ﻋدد ھﺎ ‪rcn‬‬ ‫ﻛل ﻣﺷﺎھدة ﻓﻲ اﻟﺟدول اﻟﺳﺎﺑق ﯾﻣﻛن ﻛﺗﺎﺑﺗﮭﺎ ﻋﻠﻰ اﻟﺷﻛل ‪:‬‬ ‫‪xijk  ij ijk ,‬‬ ‫ﺣﯾث ‪ ij‬ﯾﻘﯾس اﻧﺣراف ﻗﯾﻣ ﺔ اﻟﻣﺷ ﺎھدة ‪ xijk‬ﻋ ن ﻣﺗوﺳ ط اﻟﻣﺟﺗﻣ ﻊ ‪.‬ﺑﻔ رض أن ‪ ()ij‬ﯾرﻣ ز ﻟﺗ ﺄﺛﯾر‬ ‫اﻟﺗﻔﺎﻋل ﻟﻠﺻف ‪ i‬ﻣﻊ اﻟﻌﻣود رﻗم ‪ j‬و ‪  i‬ﯾرﻣز ﻟﺗﺄﺛﯾر اﻟﺻف رﻗ م ‪ i‬و ‪ j‬ﯾرﻣ ز ﻟﺗ ﺄﺛﯾر اﻟﻌﻣ ود رﻗ م ‪j‬‬ ‫و ‪ ‬ﺗﻣﺛل اﻟﻣﺗوﺳط ﻟﻛل اﻟﻣﺗوﺳطﺎت ﻓﺈن ‪:‬‬

‫‪μij  μ  αi  β j  (αβ)ij‬‬ ‫وﻋﻠﻰ ذﻟك‪:‬‬

‫‪x ijk     i   j  ()ij  ijk ,‬‬ ‫ﺗﺣت اﻟﻘﯾود اﻟﺗﺎﻟﯾﺔ ‪:‬‬ ‫‪c‬‬

‫‪r‬‬

‫‪c‬‬

‫‪r‬‬

‫‪j1‬‬

‫‪i 1‬‬

‫‪j1‬‬

‫‪i 1‬‬

‫‪ i  0,   j  0,  ()ij  0,  ()ij  0‬‬ ‫اﻟﻔروض اﻟﺛﻼﺛﺔ ﺳوف ﻧﺧﺗﺑرھﺎ ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬ ‫)أ( ‪:  1   2  ...   r  0,‬‬ ‫ﻋﻠﻰ اﻷﻗل واﺣد ﻣن ‪ i‬ﻻ ﯾﺳﺎوي ﺻﻔرا ً ‪H1 :‬‬

‫‪H '0‬‬

‫)ب( ‪H "0 :  1   2  ...   c  0,‬‬ ‫‪٥٠٥‬‬

‫ﻋﻠﻰ اﻷﻗل واﺣد ﻣن ‪ j‬ﻻ ﯾﺳﺎوي ﺻﻔرا ً ‪H1:‬‬

‫)ج( ‪H0 : ()11  ()12  ...  () rc  0,‬‬ ‫ﻋﻠﻰ اﻷﻗل واﺣد ﻣن ‪ ()ij‬ﻻ ﯾﺳﺎوي ﺻﻔرا ً ‪H1 :‬‬ ‫ﻛل اﺧﺗﺑﺎر ﻣن اﻻﺧﺗﺑﺎرات اﻟﺳﺎﺑﻘﺔ ﯾﻌﺗﻣد ﻋﻠﻰ ﺣﺳﺎب ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت اﻟﺗﺎﻟﯾﺔ ‪:‬‬ ‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت اﻟﻛﻠﻲ ھو ‪:‬‬ ‫‪n‬‬

‫‪r‬‬

‫‪c‬‬

‫‪SSTO     x ijk 2  CF ,‬‬ ‫‪i 1 j1 k 1‬‬

‫ﺣﯾث ‪:‬‬ ‫‪2‬‬ ‫‪...‬‬

‫‪T‬‬ ‫‪rcn‬‬

‫‪ ) CF ‬ﻣﻌﺎﻣل اﻟﺗﺻﺣﯾﺢ (‬

‫وﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻣﺗوﺳطﺎت اﻟﺻﻔوف ھو ‪:‬‬

‫‪ CF,‬‬

‫‪2‬‬ ‫‪r‬‬ ‫‪ Ti..‬‬ ‫‪i 1‬‬

‫‪cn‬‬

‫‪SSR ‬‬

‫وﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻣﺗوﺳطﺎت اﻷﻋﻣدة ھو ‪:‬‬

‫‪ CF,‬‬

‫‪2‬‬ ‫‪c‬‬ ‫‪ T. j.‬‬ ‫‪j1‬‬

‫‪rn‬‬

‫‪SSC ‬‬

‫وﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻠﺗﻔﺎﻋل ﺑﯾن اﻟﺻﻔوف واﻷﻋﻣدة ھو ‪:‬‬ ‫‪c‬‬

‫‪2‬‬ ‫‪ T.j.‬‬

‫‪j1‬‬

‫‪Ti..2‬‬

‫‪c‬‬

‫‪r‬‬

‫‪2‬‬ ‫‪  Tij.‬‬

‫‪‬‬ ‫‪i 1‬‬

‫‪r‬‬

‫‪i 1 j1‬‬

‫‪ CF ،‬‬ ‫‪n‬‬ ‫‪cn‬‬ ‫‪rn‬‬ ‫وﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻠﺧطﺄ ‪ SSE‬ﯾﻣﻛن اﻟﺣﺻول ﻋﻠﯾﮭﺎ ﻣن اﻟﺻﯾﻐﺔ اﻟﺗﺎﻟﯾﺔ‬ ‫‪SSE= SSTO – SSR – SSC – SS(RC).‬‬ ‫أﯾﺿﺎ ﺗﺟزأ درﺟﺎت اﻟﺣرﯾﺔ إﻟﻰ ‪:‬‬ ‫‪‬‬

‫‪‬‬

‫‪SS(RC) ‬‬

‫‪rcn  1  (r  1)  (c  1)  ( r  1)(c  1)  rc( n  1).‬‬ ‫اﻵن ﯾﻣﻛن اﻟﺣﺻول ﻋﻠﻰ اﻟﻘﯾم اﻟﺗﺎﻟﯾﺔ ‪:‬‬

‫‪SSR‬‬ ‫‪SSC‬‬ ‫‪, MSC ‬‬ ‫‪,‬‬ ‫‪r 1‬‬ ‫‪c 1‬‬ ‫)‪SS(RC‬‬ ‫‪SSE‬‬ ‫‪MS(RC) ‬‬ ‫‪, MSE ‬‬ ‫‪.‬‬ ‫)‪(r  1)(c  1‬‬ ‫)‪rc(n  1‬‬ ‫ﻻﺧﺗﺑﺎر اﻟﻔرض ‪ H 0‬ﻧﺣﺳب اﻟﻧﺳﺑﺔ ‪:‬‬ ‫‪MSR‬‬ ‫‪f1 ‬‬ ‫‪,‬‬ ‫‪MSE‬‬ ‫‪MSR ‬‬

‫‪٥٠٦‬‬

‫واﻟﺗﻲ ﺗﻣﺛل ﻗﯾﻣﺔ ﻟﻠﻣﺗﻐﯾ ر اﻟﻌﺷ واﺋﻲ ‪ F1‬واﻟ ذي ﯾﺗﺑ ﻊ ﺗوزﯾ ﻊ ‪ F‬ﺑ درﺟﺎت ﺣرﯾ ﺔ )‪( r  1), rc( n  1‬‬ ‫ﻋﻧ دﻣﺎ ﺗﻛ ون ‪ H 0‬ﺻ ﺣﯾﺢ‪ .‬ﻧ رﻓض ‪ ، H 0‬ﻋﻧ د ﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ ‪ ، ‬ﻋﻧ دﻣﺎ‬ ‫))‪. F1  f  ((r  1), rc(n  1‬ﺑﻧﻔس اﻟﺷﻛل ﻻﺧﺗﺑﺎر اﻟﻔرض ‪ H 0‬ﻧﺣﺳب اﻟﻧﺳﺑﺔ ‪:‬‬ ‫‪MSC‬‬ ‫‪f2 ‬‬ ‫‪MSE‬‬ ‫واﻟﺗﻲ ﺗﻣﺛل ﻗﯾﻣﺔ ﻟﻠﻣﺗﻐﯾر اﻟﻌﺷواﺋﻲ ‪ F2‬واﻟذي ﯾﺗﺑ ﻊ ﺗوزﯾ ﻊ ‪ F‬ﺑ درﺟﺎت ﺣرﯾ ﺔ )‪(c  1), rc( n  1‬‬ ‫ﻋﻧ دﻣﺎ ﺗﻛ ون ‪ H 0‬ﺻ ﺣﯾﺢ‪ .‬ﻧ رﻓض ‪ ، H 0‬ﻋﻧ د ﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ ‪ ، ‬ﻋﻧ‬ ‫))‪ . F2  f  ((c  1), rc(n  1‬أﺧﯾرا ً ﻻﺧﺗﺑﺎر اﻟﻔرض ‪ H 0‬ﻓﺈﻧﻧﺎ ﻧﺣﺳب اﻟﻧﺳﺑﺔ‪:‬‬

‫دﻣﺎ‬

‫)‪MS(RC‬‬ ‫‪MSE‬‬ ‫واﻟ ذي ﯾﺗﺑ ﻊ ﺗوزﯾ ﻊ ‪ F‬ﺑ درﺟﺎت ﺣرﯾ ﺔ‬ ‫واﻟﺗ ﻲ ﺗﻣﺛ ل ﻗﯾﻣ ﺔ ﻟﻠﻣﺗﻐﯾ ر اﻟﻌﺷ واﺋﻲ ‪F3‬‬ ‫‪ H ‬ﺻﺣﯾﺢ ‪ .‬ﻧرﻓض ‪ ، H 0‬ﻋﻧ د ﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ ‪‬‬ ‫)‪ ( r  1)(c  1), rc( n  1‬ﻋﻧدﻣﺎ ﺗﻛون‬ ‫‪0‬‬ ‫‪ ،‬ﻋﻧدﻣﺎ ))‪. F3  f  ((r  1)(c  1), rc(n  1‬‬ ‫‪f3 ‬‬

‫أﻣﺎ ‪ SSE‬ﻓﯾﻣﻛن اﻟﺣﺻول ﻋﻠﯾﮭﺎ ﻣن اﻟﺻﯾﻐﺔ اﻟﺗﺎﻟﯾﺔ‬ ‫اﻟﺣﺳﺎﺑﺎت ﻓﻲ ﻣﺷﻛﻠﺔ ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ‪ ،‬ﻓ ﻲ اﻟﺗﺻ ﻧﯾف اﻟﺛﻧﺎﺋﯾ ﺔ ﺑﻌ دة ﻣﺷ ﺎھدات ﻓ ﻲ ﻛ ل ﺧﻠﯾ ﺔ ‪ ،‬ﻣوﺿ ﺣﺔ‬ ‫ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪.‬‬ ‫‪ f‬اﻟﻣﺣﺳوﺑﺔ‬ ‫ﺎت ﻣﺻدر اﻻﺧﺗﻼف‬ ‫ﻣﺟﻣ وع درﺟ‬ ‫ﻣﺗوﺳط اﻟﻣرﺑﻌﺎت‬ ‫اﻟﻣرﺑﻌﺎت اﻟﺣرﯾﺔ‬ ‫‪r-1‬‬ ‫‪SSR‬‬ ‫‪SSR‬‬ ‫ﻣﺗوﺳطﺎت‬ ‫‪MSR‬‬ ‫‪MSR ‬‬ ‫‪f1 ‬‬ ‫اﻟﺻﻔوف‬ ‫‪r 1‬‬ ‫‪MSE‬‬ ‫‪c-1‬‬ ‫‪SSC‬‬ ‫ﻣﺗوﺳطﺎت اﻷﻋﻣدة‬ ‫‪SSC‬‬ ‫‪MSC‬‬ ‫‪MSC ‬‬ ‫‪f2 ‬‬ ‫‪c 1‬‬ ‫‪MSE‬‬ ‫اﻟﺗﻔﺎﻋـل‬ ‫)‪SS(RC‬‬ ‫‪(r-1)(c-1) SS(RC) MS(RC) ‬‬ ‫)‪MS(RC‬‬ ‫‪f‬‬ ‫‪‬‬ ‫‪(r‬‬ ‫‪‬‬ ‫‪1)(c‬‬ ‫‪‬‬ ‫)‪1‬‬ ‫‪3‬‬ ‫اﻟﺧطـﺄ‬ ‫‪MSE‬‬ ‫)‪rc(n-1‬‬ ‫‪SSE‬‬ ‫‪SSE‬‬ ‫‪MSE ‬‬ ‫)‪rc(n  1‬‬ ‫‪rcn-1‬‬ ‫‪SSTO‬‬ ‫اﻟﻛﻠـﻲ‬

‫ﻣﺛﺎل)‪(١١-٦‬‬ ‫اﺳﺗﺧدﻣت ﺛﻼﺛﺔ ﻣﺳﺗوﯾﺎت ﻣن ﻣﺑﯾد ﻣﺎ ﻟﻣﻘﺎوﻣﺔ ﺛﻼﺛﺔ أﺟﻧﺎس ﻣن ﺣﺷرة ‪Drosophila‬‬ ‫‪ .Pseudoobscura‬ﺗﻌطﻲ اﻟﺑﯾﺎﻧﺎت ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ﻣﻌدﻻت اﻟوﻓﯾﺎت ﺧﻼل ﻓﺗرة ﻣن اﻟزﻣن ‪.‬‬ ‫وﺗﻌﺗﻣد اﻟﺗﺟرﺑﺔ ﻋﻠﻰ ﺧﻣﺳﺔ ﻣﺷﺎھدات ﻓﻲ ﻛل ﺧﻠﯾﺔ‪.‬‬ ‫اﻟﻣطﻠوب ‪) :‬أ( اﺧﺗﺑﺎر ﻣﻌﻧوﯾﺔ اﻟﻔروق ﺑﯾن ﻣﺳﺗوﯾﺎت اﻟﻣﺑﯾد‪.‬‬ ‫)ب( اﺧﺗﺑﺎر ﻣﻌﻧوﯾﺔ اﻟﻔروق ﺑﯾن اﻷﺟﻧﺎس اﻟﻣﺧﺗﻠﻔﺔ ﻣن اﻟﺣﺷرات‬ ‫‪٥٠٧‬‬

‫)ج( اﻟﺗﻔﺎﻋل ﺑﯾن ﻣﺳﺗوﯾﺎت اﻟﻣﺑﯾد و اﻷﺟﻧﺎس )ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪.(=0.05‬‬

‫‪a3‬‬

‫اﻟﺟﻧـس‬ ‫‪a2‬‬

‫‪a1‬‬

‫‪37, 43, 57, 60, 66‬‬ ‫‪59,51,53,62,71‬‬

‫‪58,53,50,35,30‬‬ ‫‪63,59,54,38,38‬‬

‫‪60,55,52,38,31‬‬ ‫‪44,37,54,57,65‬‬

‫‪2‬‬

‫‪51,80,68,71,55‬‬

‫‪63,44,46,66,71‬‬

‫‪46,51,63,66,74‬‬

‫‪3‬‬

‫اﻟﻣﺳﺗوي‬ ‫‪1‬‬

‫اﻟﺣــل‪:‬‬ ‫ﻓروض اﻟﻌدم ﺳوف ﺗﻛون ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬ ‫‪H0 : α1  α2  α3  0,‬‬ ‫)أ (‬

‫‪H0 : 1  2  3  0,‬‬ ‫‪H0 : ()11  ()12  ...  ()33  0,‬‬

‫)ب(‬ ‫) ج(‬ ‫اﻟﻔروض اﻟﺑدﯾﻠﺔ ﺳوف ﺗﻛون ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬ ‫ﻋﻠﻰ اﻷﻗل واﺣد ﻣن ‪  i‬ﻻ ﯾﺳﺎوي ﺻﻔرا ً ‪H1 :‬‬ ‫)أ (‬ ‫ﻋﻠﻰ اﻷﻗل واﺣد ﻣن ‪ j‬ﻻ ﯾﺳﺎوي ﺻﻔرا ً ‪H1 :‬‬ ‫)ب(‬ ‫)ج( ﻋﻠﻰ اﻷﻗل واﺣد ﻣن ‪ ( ) ij‬ﻻ ﯾﺳﺎوي ﺻﻔرً ‪H1:‬‬

‫‪ f.05 (2,36)  3.23‬و ‪ =0.05‬اﻟﻣﺳﺗﺧرﺟﺔ ﻣن ﺟدول ﺗوزﯾﻊ ‪ F‬ﻓ ﻲ ﻣﻠﺣ ق )‪ .(٤‬ﻣﻧطﻘ ﺔ اﻟ رﻓض ‪F1‬‬ ‫‪. > 3.23‬‬ ‫‪ f.05(2, 36) ~ 3.23‬ﻣﻧطﻘﺔ اﻟرﻓض ‪F2 > 3.23‬‬ ‫‪ f.05(4, 36) ~ 2.61‬ﻣﻧطﻘﺔ اﻟرﻓض ‪F3 > 2.61‬‬ ‫اﻟﺑﯾﺎﻧﺎت ﻓﻲ اﻟﺟدول اﻟﺳﺎﺑق ﯾﻣﻛن ﺗﻠﺧﯾﺻﮭﺎ ﻓﻲ ﺟدول اﻟﺗﺎﻟﻰ ‪ .‬اﻵن ‪:‬‬ ‫‪2‬‬

‫‪x ijk  CF‬‬

‫‪n‬‬

‫‪r‬‬

‫‪c‬‬

‫‪‬‬ ‫‪j1‬‬

‫‪k 1‬‬

‫‪SSTO  ‬‬ ‫‪i 1‬‬

‫‪(2445) 2‬‬ ‫‪ 60  55  ...  71  55 ‬‬ ‫‪45‬‬ ‫‪ 139307  132845  6462,‬‬ ‫‪2‬‬

‫‪2‬‬

‫‪c‬‬

‫‪2‬‬ ‫‪ Ti..‬‬

‫‪ CF‬‬

‫‪SSC  i 1‬‬ ‫‪cn‬‬

‫‪٥٠٨‬‬

‫‪2‬‬

‫‪c‬‬

‫‪2‬‬ ‫‪ T.j.‬‬

‫‪ CF‬‬

‫‪j1‬‬

‫‪rn‬‬

‫‪Ti..2‬‬ ‫‪‬‬

‫‪c‬‬

‫‪r‬‬

‫‪‬‬ ‫‪i 1‬‬

‫‪cn‬‬

‫‪r‬‬

‫‪2‬‬ ‫‪  Tij.‬‬

‫‪i 1 j1‬‬

‫‪‬‬

‫‪n‬‬

‫‪SS(RC) ‬‬

‫‪2362  2262  ...  3252‬‬ ‫‪‬‬ ‫‪ 134058.33‬‬ ‫‪5‬‬ ‫‪133341.93  132845  11.74,‬‬ ‫)‪SSE  SSTO  SSR  SSC SS(RC‬‬ ‫‪ 6462  1213.33  496.93  11.74  4740.‬‬ ‫اﻟﺟﻧـس‬ ‫اﻟﻣﺟﻣوع‬ ‫‪725‬‬ ‫‪805‬‬ ‫‪915‬‬ ‫‪2445‬‬

‫‪a3‬‬ ‫‪263‬‬ ‫‪296‬‬ ‫‪325‬‬ ‫‪884‬‬

‫‪a2‬‬ ‫‪226‬‬ ‫‪252‬‬ ‫‪290‬‬ ‫‪768‬‬

‫‪a1‬‬ ‫‪236‬‬ ‫‪257‬‬ ‫‪300‬‬ ‫‪793‬‬

‫اﻟﻣﺳﺗوي‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫اﻟﻣﺟﻣوع‬

‫ﻓﯾﻣﺎ ﯾﻠﻰ ﺟدول اﻟﺗﺑﺎﯾن ‪:‬‬

‫‪ f‬اﻟﻣﺣﺳوﺑﺔ‬ ‫‪f1=4.608‬‬ ‫‪f2= 1.887‬‬ ‫‪f3= 0.022‬‬

‫ﻣﺗوﺳط‬ ‫اﻟﻣرﺑﻌﺎت‬ ‫‪606.665‬‬ ‫‪248.465‬‬ ‫‪2.935‬‬ ‫‪131.666‬‬

‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت‬ ‫‪1213.33‬‬ ‫‪496.93‬‬ ‫‪11.74‬‬ ‫‪4740‬‬ ‫‪6462‬‬

‫درﺟﺎت‬ ‫اﻟﺣرﯾﺔ‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪36‬‬ ‫‪44‬‬

‫ﻣﺻدر اﻻﺧﺗﻼف‬ ‫ﻣﺗوﺳطﺎت اﻟﺻﻔوف‬ ‫ﻣﺗوﺳطﺎت اﻷﻋﻣدة‬ ‫اﻟﺗﻔﺎﻋـل‬ ‫اﻟﺧطـﺄ‬ ‫اﻟﻛﻠـﻲ‬

‫ﻣن ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﯾﻣﻛن اﺳﺗﻧﺗﺎج ‪:‬‬ ‫)أ( ﻧرﻓض ‪ H0‬ﻷن ‪ f1‬ﺗﻘﻊ ﻓﻲ ﻣﻧطﻘﺔ اﻟرﻓض ‪ ،‬أي أن ھﻧﺎك ﻓروق ﻣﻌﻧوﯾﺔ ﺑﯾن ﺑﻣﺳﺗوﯾﺎت اﻟﻣﺑﯾد‪.‬‬ ‫)ب( ﻧﻘﺑل ‪ H 0‬ﻷن ‪ f2‬ﺗﻘﻊ ﻓﻲ ﻣﻧطﻘﺔ اﻟﻘﺑول ‪ ،‬أي أﻧﮫ ﻻ ﯾوﺟد ﻓروق ﻣﻌﻧوﯾﺔ ﺑﯾن اﻷﺟﻧﺎس‪.‬‬ ‫)ج( ﻧﻘﺑ ل ‪ H 0‬ﻷن ‪ f3‬ﺗﻘ ﻊ ﻓ ﻲ ﻣﻧطﻘ ﺔ اﻟﻘﺑ ول ‪ ،‬أي أﻧ ﮫ ﻻ ﯾوﺟ د ﺗﻔﺎﻋ ل ﺑ ﯾن ﻣﺳ ﺗوﯾﺎت اﻟﻣﺑﯾ د و‬ ‫أﺟﻧﺎس اﻟﺣﺷرة‪.‬‬ ‫ﺳوف ﯾﺗم اﺳﺗﺧدام ]‪ [Mathematica 5.0‬ﻓﻲ ﺣل اﻟﻣﺛﺎل اﻟﺳﺎﺑق ﺣﯾث ‪:‬‬

‫‪٥٠٩‬‬

‫ﻣﺻ در اﻷﺧ ﺗﻼف ﺗﻌﻧ ﻲ )‪ ، (S.V‬ﻣﺗوﺳ طﺎت اﻟﺻ ﻔوف )‪ ، (A‬ﻣﺗوﺳ طﺎت اﻷﻋﻣ دة ﺗﻌﻧ ﻲ )‪B‬‬ ‫(‪،‬اﻟﺧطﺄ ﺗﻌﻧﻲ )‪ ، (error‬ﺗﻌﻧ ﻰ)‪ (AB‬اﻟﺗﻔﺎﻋ ل‪ ،‬اﻟﻛﻠ ﻲ ﺗﻌﻧ ﻲ )‪ ، (total‬درﺟ ﺎت اﻟﺣرﯾ ﺔ ﺗﻌﻧ ﻲ )‪، (df‬‬ ‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﺗﻌﻧﻲ )‪ (ss‬ﺗﻌﻧﻰ ﻣﺗوﺳط اﻟﻣرﺑﻌﺎت ﺗﻌﻧﻲ )‪ f ، (mss‬اﻟﻣﺣﺳوﺑﺔ ﺗﻌﻧﻲ )‪(f‬‬ ‫وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫‪=0.05‬‬ ‫‪0.05‬‬ ‫{‪a={{{60.,55,52,38,31},{58.,53,50,35,30},{37,43,57.,60,66}},‬‬ ‫‪{44,37,54,57,65},{63.,59.,54,38,38},{59,51,53,62,71}},{{46,5‬‬ ‫}}}‪1,63,66,74},{63,44,46.,66,71},{51,80,68,71,55‬‬ ‫‪{{{60.,55,52,38,31},{58.,53,50,35,30},{37,43,57.,60,66}},{{4‬‬ ‫‪4,37,54,57,65},{63.,59.,54,38,38},{59,51,53,62,71}},{{46,51,‬‬ ‫}}}‪63,66,74},{63,44,46.,66,71},{51,80,68,71,55‬‬ ‫]‪dd=Dimensions[a‬‬ ‫}‪{3,3,5‬‬ ‫]]‪p=dd[[1‬‬ ‫‪3‬‬ ‫]]‪q=dd[[2‬‬ ‫‪3‬‬ ‫]]‪n=dd[[3‬‬ ‫‪5‬‬ ‫‪k1=q-1‬‬ ‫‪2‬‬ ‫‪k2=p-1‬‬ ‫‪2‬‬ ‫‪N1=n*p*q‬‬ ‫‪45‬‬ ‫‪k4=n-1‬‬ ‫‪4‬‬ ‫‪v1=N1-1-k1-k2-k4‬‬ ‫‪36‬‬ ‫]‪b=Flatten[a‬‬ ‫‪{60.,55,52,38,31,58.,53,50,35,30,37,43,57.,60,66,44,37,54,57‬‬ ‫‪,65,63.,59.,54,38,38,59,51,53,62,71,46,51,63,66,74,63,44,46.‬‬ ‫}‪,66,71,51,80,68,71,55‬‬ ‫]‪b=Flatten[a‬‬ ‫‪{60.,55,52,38,31,58.,53,50,35,30,37,43,57.,60,66,44,37,54,57‬‬ ‫‪,65,63.,59.,54,38,38,59,51,53,62,71,46,51,63,66,74,63,44,46.‬‬ ‫}‪,66,71,51,80,68,71,55‬‬ ‫‪0000‬‬ ‫‪0‬‬ ‫]‪g[x_]:=Length[x‬‬ ‫]‪f[x_]:=Apply[Plus,x‬‬ ‫‪٥١٠‬‬

x1=f[b] 2445. x2=f[b^2] 139307. cf=x1^2./N1 132845. tot=x2-cf 6462. xx=Table[f[a[[i,j]]],{i,1,p},{j,1,q}] {{236.,226.,263.},{257,252.,296},{300,290.,325}} Flatten[xx] {236.,226.,263.,257,252.,296,300,290.,325} xx3=f[%^2] 672835. xx4=%/n 134567. a3=Map[f,xx] {725.,805.,915.} sa=f[a3^2]/(n*q) 134058. aaaa=sa-cf 1213.33 ssa=aaaa/k1 606.667 ww=Transpose[xx] {{236.,257,300},{226.,252.,290.},{263.,296,325}} a33=Map[f,ww] {793.,768.,884.} sb=N[f[a33^2]/(q*n)] 133342. bbbb=sb-cf 496.933 sbb=bbbb/k2 248.467 sser=x2-xx4 4740. msse=sser/v1 131.667 ss12ab=tot-aaaa-bbbb-sser 11.7333 abss=ss12ab/k4 2.93333 ff1=ssa/msse 4.60759 ff2=sbb/msse 1.88709 ff3=abss/msse ٥١١

0.0222785 rt2=List[" df "," ss "," mss "," f "] { df , ss , mss , f } rt1=List[k1,aaaa,ssa,ff1] {2,1213.33,606.667,4.60759} rt3=List[k2,bbbb,sbb,ff2] {2,496.933,248.467,1.88709} rt5=List[k4,ss12ab,abss,ff3] {4,11.7333,2.93333,0.0222785} {k4,ss12ab,abss,ff4} rt9=List[v1,sser,msse,"-"] {36,4740.,131.667,-} rr9=List[N1-1,tot,"_","_"] {44,6462.,_,_} a11=TableHeadings->{{ S.V,A,B,AB,error,total},{ANOVA}} TableHeadings{{S.V,A,B,AB,error,total},{ANOVA}} uu1=TableForm[{rt2,rt1,rt3,rt5,rt9,rr9},a11]

S.V A B AB error total

ANOVA df 2 2 4 36 44

ss 1213.33 496.933 11.7333 4740. 6462.

mss 606.667 248.467 2.93333 131.667 _

f 4.60759 1.88709 0.0222785 

<<Statistics`ContinuousDistributions` f1=Quantile[FRatioDistribution[k1,v1],1-] 3.25945 If[ff1>f1,Print["RjectHo"],Print["AccpetHo"]] RjectHo

f2=Quantile[FRatioDistribution[k2,v1],1-] 3.25945 If[ff2>f2,Print["RjectHo"],Print["AccpetHo"]] AccpetHo

f3=Quantile[FRatioDistribution[k4,v1],1-] 2.63353 If[ff3>f3,Print["RjectHo"],Print["AccpetHo"]] AccpetHo

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬ ‫ﻣﺴﺘﻮى اﻟﻤﻌﻨﻮﯾﺔ ﻣﻦ اﻻﻣﺮ‬ =0.05

‫ وﻫﻰ‬a ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ ٥١٢

‫{‪a={{{60.,55,52,38,31},{58.,53,50,35,30},{37,43,57.,60,66}},‬‬ ‫‪{44,37,54,57,65},{63.,59.,54,38,38},{59,51,53,62,71}},{{46,5‬‬ ‫}}}‪1,63,66,74},{63,44,46.,66,71},{51,80,68,71,55‬‬

‫وﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﻴﺎﻧﺎت اﻟﺨﺎﺻﺔ ﻓﻰ ﻣﺜﺎل )‪ (١١-٦‬ﺣﻴﺚ ﻳﺘﻢ ادﺧﺎل اﻟﺒﻴﺎﻧﺎت ﺻﻒ ﺻﻒ‪.‬‬ ‫ﺛﺎﻧﯾﺎ ‪ :‬اﻟﻣﺧرﺟﺎت‬

‫ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﯾﺗم اﻟﺣﺻول ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ ‫]‪uu1=TableForm[{rt2,rt1,rt3,rt5,rt9,rr9},a11‬‬

‫ﻓروض اﻟﻌدم ﺳوف ﺗﻛون ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬ ‫‪H0 : α1  α2  α3  0,‬‬ ‫)أ (‬

‫‪H0 : 1  2  3  0,‬‬ ‫‪H0 : ()11  ()12  ...  ()33  0,‬‬

‫)‪ f.05 (2,36‬واﻟﻣﺳﺗﺧرﺟﺔ ﻋﻧد درﺟﺎت ﺣرﯾﺔ ‪ 1  2, 2  36‬ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪f1=Quantile[FRatioDistribution[k1,v1],1-‬‬

‫‪ f1‬اﻟﻣﺣﺳوﺑﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر‬ ‫‪ff1=ssa/msse‬‬

‫اﻟﻘرار اﻟذى ﯾﺗﺧذ ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ ‫]]"‪If[ff1>f1,Print["RjectHo"],Print["AccpetHo‬‬

‫واﻟﻣﺧرج ھو‬ ‫‪Reject H0‬‬

‫اى رﻓض ﻓرض اﻟﻌدم‪.‬‬ ‫اﯾﺿﺎ‬ ‫)‪ f.05 (2,36‬واﻟﻣﺳﺗﺧرﺟﺔ ﻋﻧد درﺟﺎت ﺣرﯾﺔ ‪ 1  2, 2  10‬ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪f2=Quantile[FRatioDistribution[k2,v1],1-‬‬

‫‪ f 2‬اﻟﻣﺣﺳوﺑﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر‬ ‫‪ff2=sbb/msse‬‬

‫اﻟﻘرار اﻟذى ﯾﺗﺧذ ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ ‫]]"‪If[ff2>f2,Print["RjectHo"],Print["AccpetHo‬‬

‫واﻟﻣﺧرج ھو‬ ‫‪AccpetHo‬‬

‫اى ﻗﺑول ﻓرض اﻟﻌدم ‪.‬‬ ‫)‪ f.05 (4,36‬واﻟﻣﺳﺗﺧرﺟﺔ ﻋﻧد درﺟﺎت ﺣرﯾﺔ ‪ 1  4, 2  36‬ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]]"‪If[ff3>f3,Print["RjectHo"],Print["AccpetHo‬‬ ‫‪٥١٣‬‬

‫ اﻟﻣﺣﺳوﺑﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر‬f3 ff3=abss/msse

‫اﻟﻘرار اﻟذى ﯾﺗﺧذ ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ If[ff3>f3,Print["RjectHo"],Print["AccpetHo"]]

‫واﻟﻣﺧرج ھو‬ AccpetHo

. ‫اى ﻗﺑول ﻓرض اﻟﻌدم‬

(١٢-٦)‫ﻣﺛﺎل‬ : ‫ﺳوف ﻧﻘدم ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻟﺣل اﻟﻣﺛﺎل اﻟﺳﺎﺑق‬ Off[General::spell1]; <<Statistics`NormalDistribution` <<Statistics`DataManipulation` twoWayAnova - Data entered as Matrix ] twoWayANOVAarray - Data entered as an array Model I Model II Model III Model IV modelFour[twoLists_]:=Module[{a,b,n,capN,sumOfVals, sumOfValsSq,sumOverA,sqValsA,factorASumOfSq,sumOverB,sq ValsB,factorBSumOfSq,c,totalSS,totalDF,sumOverColumnAndSquar e,flatList,num,cellsSS,cellsDF,withinCellsSS,withinCellsDF,f actorADF,factorBDF,ACrossBinteractionSS,ACrossBinteractionDF ,factorAMS,factorBMS,ACrossBMS,withinCellsMS,fA,pA,fB,pB,fAB ,pAB,cd,lineOne,lineTwo,lineFour,lastLine}, a=Length[twoLists]; b=Length[twoLists[[1]]]; n=Length[twoLists[[1,1]]]; capN=a b n; sumOfVals=Sum[twoLists[[i,j,l]],{i,1,a},{j,1,b},{l,1,n} ]; sumOfValsSq=Sum[twoLists[[i,j,l]]^2,{i,1,a},{j,1,b},{l, 1,n}]; sumOverA=Table[Sum[twoLists[[i,j,l]],{j,1,b},{l,1,n}],{ i,1,a}]; sqValsA=Table[sumOverA[[i]]^2,{i,1,Length[sumOverA]}]; factorASumOfSq=Apply[Plus,sqValsA]/(b n)-c;

٥١٤

sumOverB=Table[Sum[twoLists[[i,j,l]],{i,1,a},{l,1,n}],{ j,1,b}]; sqValsB=Table[sumOverB[[j]]^2,{j,1,Length[sumOverB]}]; factorBSumOfSq=Apply[Plus,sqValsB]/(a n)-c; c=sumOfVals^2/capN; totalSS=sumOfValsSq-c; totalDF=a b-1; sumOverColumnAndSquare[v_]:=Sum[v[[k]],{k,1,Length[v]}] ^2; flatList=Flatten[twoLists,1]; num=Map[sumOverColumnAndSquare,flatList]; cellsSS=Apply[Plus,num]/n-c; cellsDF=a b-1; withinCellsSS=totalSS-cellsSS; withinCellsDF=(a-1)(b-1); factorADF=a-1; factorBDF=b-1; ACrossBinteractionDF=factorADF factorBDF; factorAMS=factorASumOfSq/factorADF; factorBMS=factorBSumOfSq/factorBDF; SSAB=totalSS-factorASumOfSq-factorBSumOfSq; MSAB=SSAB/ACrossBinteractionDF; withinCellsMS=withinCellsSS/withinCellsDF; fA=factorAMS/MSAB; pA=1CDF[FRatioDistribution[factorADF,ACrossBinteractionDF ],fA]; fB=factorBMS/MSAB; pB=1CDF[FRatioDistribution[factorBDF,withinCellsDF ],fB]; cd=N[(totalSS-withinCellsSS)/totalSS]; Print["Analysis of Variance Table"]; lineOne={"Factor A",factorASumOfSq//N,factorADF,factorAMS//N,fA//N,pA//N}; lineTwo={"Factor B",factorBSumOfSq//N,factorBDF,factorBMS//N,fB//N,pB//N}; lineFour={"Error",SSAB//N,withinCellsDF,MSAB//N," "," "}; lastLine={"Total",totalSS//N,totalDF," "," "," "}; Print[TableForm[{lineOne,lineTwo,lineFour,lastLine},Tab leHeadings->{None,{"Source","Sum of Squares","DF","Mean Squares","Fratio","P-value"}}]]; ]

٥١٥

diskDrive={{{60.,55,52,38,31},{58.,53,50,35,30},{37.,43,57,6 0,66}},{{44.,37,54,57,65},{63.,59,54,38,38},{59,51,53,62,71. }},{{46.,51,63,66,74},{63.,44,46,66,71.},{51.,80,68,71,55}}} ; twoWayAnova[diskDrive] Analysis of Variance Table

Source Factor A Factor B A B Interaction Error Total

Sum of Squares 1213.33 496.933 11.7333 4740. 6462.

DF 2 2 4 36 44

Mean Squares 606.667 248.467 2.93333 131.667

Fratio 4.60759 1.88709 0.0222785

Pvalue 0.016532 0.166197 0.998986

twoWayAnova[diskDrive,model->random] Analysis of Variance Table

Source Factor A Factor B A B Interaction Error Total

Sum of Squares 1213.33 496.933 11.7333 4740. 6462.

DF 2 2 4 36 44

Mean Squares 606.667 248.467 2.93333 131.667

Fratio 206.818 84.7045 0.0222785

Pvalue 0.0000917325 0.000532079 0.998986

twoWayAnova[diskDrive,model->mixed] Analysis of Variance Table

Source Factor A Factor B A B Interaction Error Total

Sum of Squares 1213.33 496.933 11.7333 4740. 6462.

DF 2 2 4 36 44

Mean Squares 606.667 248.467 2.93333 131.667

Fratio 206.818 1.88709 0.0222785

Pvalue 0.0000917325 0.166197 0.998986

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ : ‫اوﻻ اﻟﻣدﺧﻼت‬

‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ diskDrive : ‫وھﻰ‬ diskDrive={{{60.,55,52,38,31},{58.,53,50,35,30},{37.,43,57,6 0,66}},{{44.,37,54,57,65},{63.,59,54,38,38},{59,51,53,62,71. }},{{46.,51,63,66,74},{63.,44,46,66,71.},{51.,80,68,71,55}}} ; ٥١٦

‫وﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﻴﺎﻧﺎت اﻟﺨﺎﺻﺔ ﻓﻰ ﻣﺜﺎل )‪.(١٢-٦‬‬ ‫ﺛﺎﻧﯾﺎ ‪ :‬اﻟﻣﺧرﺟﺎت‬

‫ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﯾﺗم اﻟﺣﺻول ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ ‫]‪twoWayAnova[diskDrive‬‬ ‫اﻟﻣﺧرج ھو‬

‫‪Analysis of Variance Table‬‬

‫‪Pvalue‬‬ ‫‪0.016532‬‬ ‫‪0.166197‬‬ ‫‪0.998986‬‬

‫‪Fratio‬‬ ‫‪4.60759‬‬ ‫‪1.88709‬‬ ‫‪0.0222785‬‬

‫‪DF‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪36‬‬ ‫‪44‬‬

‫‪Mean Squares‬‬ ‫‪606.667‬‬ ‫‪248.467‬‬ ‫‪2.93333‬‬ ‫‪131.667‬‬

‫‪Sum of Squares‬‬ ‫‪1213.33‬‬ ‫‪496.933‬‬ ‫‪11.7333‬‬ ‫‪4740.‬‬ ‫‪6462.‬‬

‫‪Source‬‬ ‫‪Factor A‬‬ ‫‪Factor B‬‬ ‫‪A B Interaction‬‬ ‫‪Error‬‬ ‫‪Total‬‬

‫)‪ (٥-٦‬ﺑﻌض ﺗﺻﺎﻣﯾم اﻟﺗﺟﺎرب اﻟﺑﺳﯾطﺔ ‪:‬‬ ‫طرﯾﻘﺔ ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن اﻟﺗﻲ ﺗﻧﺎوﻟﻧﺎھﺎ ﻓﻲ اﻟﺑﻧود اﻟﺳﺎﺑﻘﺔ ﺗطﺑق ﺑﻌد اﻟﺣﺻول ﻋﻠﻰ اﻟﺑﯾﺎﻧﺎت اﻟﺧﺎﺻﺔ‬ ‫ﺑﺗﺟرﺑﺔ ﺗم اﺟراﺋﮭﺎ‪ .‬ﻓﻲ ﺑﻌض اﻻﺣﯾﺎن ‪ ،‬ﻟﻠﺣﺻول ﻋﻠﻰ أﻓﺿل اﻟﻣﻌﻠوﻣﺎت اﻟﻣﻣﻛﻧﺔ ‪ ،‬ﻓﺈن اﻟﺗﺟرﺑﺔ ﻻﺑد‬ ‫أن ﯾﺧطط ﻟﮭﺎ ﻗﺑل اﺟراﺋﮭﺎ ‪ .‬وھذا ھو ﻣﺎ ﯾﻌرف ﺑﺗﺻﻣﯾم اﻟﺗﺟﺎرب ‪ .‬ﻓﻲ اﻟﺑﻧد اﻟﺗﺎﻟﻲ ﺳوف ﻧﻧﺎﻗش‬ ‫ﺑﻌض اﻻﻣﺛﻠﺔ اﻟﻣﮭﻣﺔ ﻓﻲ ﺗﺻﻣﯾم اﻟﺗﺟﺎرب‪.‬‬

‫)‪ (١-٥-٦‬ﺗﺻﻣﯾم و ﺗﺣﻠﯾل اﻟﺗﺟﺎرب ذات اﻟﻌﺎﻣل اﻟواﺣد ‪:‬اﻟﺗﺻﻣﯾم اﻟﺗﺎم ﻟﻠﺗﻌﺷﯾﺔ ‪:‬‬ ‫‪Design and Analysis of Single Factor Experiments:‬‬ ‫‪Completely Randomized Design‬‬ ‫ﯾﻌﺗﺑر اﻟﺗﺻﻣﯾم اﻟﺗﺎم ﻟﻠﺗﻌﺷﯾﺔ )اﻟﺗﺻﻣﯾم اﻟﻛﺎﻣل اﻟﻌﺷ واﺋﻲ( ﻣ ن اﺑﺳ ط اﻟﺗﺻ ﺎﻣﯾم اﻟﺗﺟرﯾﺑﯾ ﺔ ﻣ ن ﺣﯾ ث‬ ‫ﺗﻌﯾﯾن اﻟوﺣدات اﻟﺗﺟرﯾﺑﯾﺔ ﻋﻠﻰ اﻟﻣﻌﺎﻟﺟﺎت‪ ،‬ﻣﺳﺗوﯾﺎت اﻟﻌﺎﻣ ل ﻣوﺿ ﻊ اﻟدراﺳ ﺔ‪ ،‬وﺗﺣﻠﯾ ل اﻟﺑﯾﺎﻧ ﺎت ﺣﯾ ث‬ ‫ﯾﻌﺗﺑر اﻷﺳﺎس ﻟﺑﻧﺎء ﺗﺻﺎﻣﯾم ﺗﺟرﯾﺑﯾﺔ أﻛﺛر ﺗﻌﻘﯾدا‪ .‬إن اﻟﮭدف اﻷﺳﺎﺳﻲ ﻣن ھ ذا اﻟﺗﺻ ﻣﯾم ھ و اﺧﺗﺑ ﺎر ﻣ ﺎ‬ ‫إذا ﻛﺎﻧت ﻣﺗوﺳطﺎت ﻣﺟﺗﻣﻌﺎت اﻟﻣﻌﺎﻟﺟ ﺎت ﻣﺗﺳ ﺎوﯾﺔ أم ﻻ؟ وذﻟ ك ﻟﺗﺟرﺑ ﺔ ذات ﻋﺎﻣ ل واﺣ د‪ ،‬وﻟﮭ ﺎ ﻋ دة‬ ‫ﻣﺳﺗوﯾﺎت‪ .‬ﯾﺷﺗرط ﻓﻲ ھ ذا اﻟﺗﺻ ﻣﯾم أن ﺗﻛ ون اﻟوﺣ دات ﻣﺗﺟﺎﻧﺳ ﺔ ﺗﻣﺎﻣ ﺎ‪ ،‬وﯾﻧطﺑ ق ذﻟ ك ﻋﻠ ﻰ اﻟﺗﺟ ﺎرب‬ ‫اﻟﻣﻌﻣﻠﯾﺔ ﻛﺗﺟﺎرب اﻟطﺑﯾﻌﺔ‪ ،‬واﻟﻛﯾﻣﯾﺎء ﺣﯾث ﺗﻘﺳم ﻛﻣﯾﺔ ﻣن ﻣ ﺎدة اﻟﺗﺟرﺑ ﺔ ﺑﻌ د ﺧﻠطﮭ ﺎ ﺟﯾ دا إﻟ ﻰ ﻋﯾﻧ ﺎت‬ ‫ﺻﻐﯾرة ﺗﺟرب ﻋﻠﯾﮭ ﺎ اﻟﻣﻌﺎﻟﺟ ﺎت اﻟﻣﺧﺗﻠﻔ ﺔ‪ .‬أﻣ ﺎ ﻓ ﻲ اﻟﺗﺟ ﺎرب اﻟﺗ ﻲ ﺗﻧﺗﻣ ﻲ إﻟ ﻰ اﻟﻧ واﺣﻲ اﻻﻗﺗﺻ ﺎدﯾﺔ أو‬ ‫اﻟزراﻋﯾ ﺔ أو اﻟﺗﺟﺎرﯾ ﺔ ﻓ ﻼ ﯾﻼﺋﻣﮭ ﺎ ھ ذا اﻟﺗﺻ ﻣﯾم ﺣﯾ ث ﺗﻛ ون اﻟوﺣ دات اﻟﺗﺟرﯾﺑﯾ ﺔ ﻏﯾ ر ﻣﺗﺟﺎﻧﺳ ﺔ‬ ‫)اﻟﺷﺧص أو اﻟﻌﺎﺋﻠﺔ أو اﻟﻣﺧزون… اﻟﺦ (‪ ،‬وﺗﺳﺗﺧدم ﺗﺻﺎﻣﯾم أﺧرى‪.‬‬ ‫وﻓﯾﻣﺎ ﯾﻠﻲ أﻣﺛﻠﺔ ﻟﺗﺟﺎرب اﺳﺗﺧدم ﻓﯾﮭﺎ اﻟﺗﺻﻣﯾم اﻟﺗﺎم ﻟﻠﺗﻌﺷﯾﺔ‪:‬‬ ‫)أ( ﺗﺟرﺑﺔ ﻟدراﺳﺔ ﺗﺄﺛﯾر ﺧﻣس أﻧواع ﻣن اﻟﺑﻧزﯾن ﻋﻠﻰ ﻛﻔﺎءة ﻋﻣل ﺳﯾﺎرة )‪.(mpg‬‬ ‫‪٥١٧‬‬

‫)ب( ﺗﺟرﺑ ﺔ ﻟدراﺳ ﺔ ﺗ ﺄﺛﯾر أرﺑﻌ ﺔ أﻧ واع ﻣﺧﺗﻠﻔ ﺔ ﻣ ن اﻟﻣﺣﺎﻟﯾ ل اﻟﺳ ﻛرﯾﺔ )ﺟﻠوﻛ وز‪ ،‬ﻓرﻛﺗ وز‪،‬‬ ‫ﺳﻛروز‪ ،‬ﺧﻠﯾط ﻣن اﻟﺛﻼﺛﺔ ( ﻋﻠﻰ ﻧﻣو اﻟﺑﻛﺗﯾرﯾﺎ‪.‬‬ ‫)ج( ﺗﺟرﺑﺔ ﻟدراﺳﺔ اﻟﻛﻣﯾﺎت اﻟﻣﺧﺗﻠﻔﺔ ﻣن دواء ﻣﮭدئ ﻋﻠﻰ ﺷﻔﺎء ﻣرﯾض ﯾﻌﺎﻧﻲ ﻣرض ﻧﻔﺳﻲ‪.‬‬ ‫)د( ﺗﺟرﺑﺔ ﻟدراﺳﺔ ﺗ ﺄﺛﯾر أرﺑﻌ ﺔ أﻧ واع ﻣ ن اﻟﺗﻐطﯾ ﺔ ﻋﻠ ﻰ اﻟﻘ وة اﻟﺗوﺻ ﯾﻠﯾﺔ ﻟﺻ ﻣﺎﻣﺎت اﻟﺗﻠﯾﻔزﯾ ون‪.‬ﻓ ﻲ‬ ‫اﻟﺗﺟرﺑ ﺔ )أ( اﻟﻌﺎﻣ ل ﻣوﺿ ﻊ اﻟدراﺳ ﺔ ھ و اﻟﺑﻧ زﯾن‪ ،‬وﻟ ﮫ ﺧﻣﺳ ﺔ ﻣﺳ ﺗوﯾﺎت ﻣﺧﺗﻠﻔ ﺔ‪ .‬ﻓ ﻲ اﻟﺗﺟرﺑ ﺔ‬ ‫)ب( اﻟﻌﺎﻣل ھو اﻟﺳﻛر وﻟﮫ أرﺑ ﻊ ﻣﺳ ﺗوﯾﺎت ) أو ﺧﻣﺳ ﺔ وذﻟ ك ﻋﻧ د اﺳ ﺗﻌﻣﺎل ﻣﺣﻠ ول ﻣراﻗﺑ ﺔ ﻻ‬ ‫ﯾﺣﺗ وي ﻋﻠ ﻰ أي ﺳ ﻛرﯾﺎت (‪ .‬ﻓ ﻲ اﻟﺗﺟرﺑ ﺔ )ج( ‪ ،‬ﻣ ﺛﻼ ‪ ،‬اﻟﻌﺎﻣ ل ﻛﻣ ﻰ واﻟﻣﺳ ﺗوﯾﺎت ﺗﺗﻣﺛ ل ﻓ ﻲ‬ ‫اﻟﺗﺻ ﻧﯾﻔﺎت اﻟﻣﻣﻛﻧ ﺔ ﻣ ن اﻟﻌﺎﻣ ل‪ .‬ﻓ ﻲ اﻟﺗﺟرﺑ ﺔ )د( اﻟﻌﺎﻣ ل ﯾﻣﺛ ل اﻷﻧ واع اﻟﻣﺧﺗﻠﻔ ﺔ ﻣ ن اﻷﻏطﯾ ﺔ‬ ‫وھو ﻋﺎﻣل وﺻﻔﻰ‪.‬‬ ‫ﻣزاﯾﺎ اﻟﺗﺻﻣﯾم‪:‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬

‫ﯾﻣﻛن اﺳﺗﺧدام أي ﻋدد ﻣن اﻟﻣﻌﺎﻟﺟﺎت‪.‬‬ ‫ﯾﻣﻛن اﺧﺗﻼف ﺣﺟم اﻟﻌﯾﻧﺔ ﻣن ﻣﻌﺎﻟﺟﺔ إﻟﻰ أﺧرى‪.‬‬ ‫ﺳﮭوﻟﺔ اﻟﺗﺣﻠﯾل اﻹﺣﺻﺎﺋﻲ‪.‬‬ ‫ﻻ ﯾﺳﺑب ﻓﻘدان أي ﻣﺷﺎھدة ﻣﺷﺎﻛل ﻓﻲ اﻟﺗﺣﻠﯾل اﻹﺣﺻﺎﺋﻲ‪.‬‬ ‫ﯾﻌﺗﻣد ھذا اﻟﺗﺻﻣﯾم ﻋﻠﻰ ﻋدد ﻗﻠﯾل ﻣن اﻟﻔروض ﺑﺧﻼف اﻟﺗﺻﺎﻣﯾم اﻟﺗﺟرﯾﺑﯾﺔ اﻷﺧرى‪.‬‬

‫ﻣﺛﺎل)‪(١٣-٦‬‬ ‫وﺿﻊ اﻣﺗﺣﺎن ﻟﻣﺟﻣوﻋﺎت ﻣن ‪ 26‬طﺎﻟب ﻻﺧﺗﺑﺎر اﻟذﻛﺎء واﻟﻘدرة ﻋﻠﻰ اﻟﺗرﻛﯾز ‪ .‬واﻟﻧﺗﺎﺋﺞ ﻣﻌطﺎة ﻓﻲ‬ ‫اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫)أ( أوﺟد ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﻟﮭذه اﻟﺗﺟرﺑﺔ‪.‬‬ ‫)ب( ھل ﺗوﺟد ﻓروق ﺑﯾن ﻣﺗوﺳطﺎت اﻟﻣﻌﺎﻟﺟﺎت ﻋﻧد ‪.   0.01‬‬ ‫اﻟﻣﺟﻣوع‬

‫‪D‬‬ ‫‪120‬‬ ‫‪109‬‬ ‫‪112‬‬

‫‪C‬‬ ‫‪119‬‬ ‫‪97‬‬ ‫‪128‬‬ ‫‪116‬‬ ‫‪130‬‬ ‫‪150‬‬ ‫‪114‬‬

‫‪B‬‬ ‫‪109‬‬ ‫‪121‬‬ ‫‪113‬‬ ‫‪112‬‬ ‫‪130‬‬

‫‪A‬‬ ‫‪127‬‬ ‫‪119‬‬ ‫‪138‬‬

‫‪26‬‬ ‫‪2934‬‬ ‫‪337596‬‬

‫‪E‬‬ ‫‪94‬‬ ‫‪96‬‬ ‫‪101‬‬ ‫‪72‬‬ ‫‪103‬‬ ‫‪105‬‬ ‫‪103‬‬ ‫‪96‬‬ ‫‪8‬‬ ‫‪770‬‬ ‫‪74896‬‬

‫‪7‬‬ ‫‪854‬‬ ‫‪105806‬‬

‫‪5‬‬ ‫‪585‬‬ ‫‪68735‬‬

‫‪3‬‬ ‫‪384‬‬ ‫‪49334‬‬

‫‪3‬‬ ‫‪341‬‬ ‫‪38825‬‬

‫‪104188‬‬

‫‪68445‬‬

‫‪49152‬‬

‫‪334657.83‬‬

‫‪74112.5‬‬

‫‪38760.33‬‬

‫‪ni‬‬ ‫‪Ti.‬‬ ‫‪2‬‬ ‫‪ xij‬‬ ‫‪j‬‬

‫‪٥١٨‬‬

‫‪Ti.2 / ni‬‬

‫اﻟﺣــل‪:‬‬ ‫)ا( ﻟﻠﺣﺻول ﻋﻠﻰ ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﻧﺗﺑﻊ اﻵﺗﻲ‪:‬‬ ‫‪= (2934)² / 26 = 331090.62‬‬

‫‪  T..2 / N‬‬

‫)‪(1‬‬

‫‪(2)    x ij2 = 337596‬‬ ‫‪(3)   (Ti.2 / ni ) = 334657.83.‬‬ ‫ﺗﺣﺳب ﻣﺟﺎﻣﯾﻊ اﻟﻣرﺑﻌﺎت ‪ SSTO, SSTC, SSE‬ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬

‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت اﻟﻛﻠﻲ ﺳﯾﻛون‪:‬‬ ‫‪SSTO = (2) - (1) = 6505.39 ,‬‬ ‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻠﻣﻌﺎﻟﺟﺎت ﺳﯾﻛون‪:‬‬ ‫‪,‬‬

‫‪SSC = (3) - (1) = 3567.22‬‬

‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻠﺧطﺄ ﺳﯾﻛون‪:‬‬ ‫‪SSE = SSTO - SSC = 2938.17 .‬‬ ‫ﺗﻠﺧص اﻟﻧﺗﺎﺋﺞ ﻓﻲ ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن اﻟﻣﻌطﻰ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫) ‪f  (1 ,  2‬‬

‫‪ f‬اﻟﻣﺣﺳوﺑﺔ‬

‫)‪= 4.37 f 0.01 (4, 21‬‬

‫‪6.37‬‬

‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت‬

‫ﻣﺗوﺳط‬ ‫اﻟﻣرﺑﻌﺎت‬ ‫‪891.81‬‬

‫‪SSC = 3567.22‬‬

‫‪139.91‬‬

‫‪SSE = 2938.17‬‬ ‫‪SSTO = 6505.39‬‬

‫ﻣﺻدر‬ ‫اﻻﺧﺗﻼف‬ ‫اﻟﻣﻌﺎﻟﺟﺎت‬

‫درﺟﺎت‬ ‫اﻟﺣرﯾﺔ‬ ‫‪k-1 = 4‬‬

‫اﻟﺧطﺄ‬

‫‪N-k= 21‬‬

‫‪N-1 = 25‬‬

‫اﻟﻛﻠﻲ‬

‫)ب( ﻣ ن اﻟﺟ دول اﻟﺳ ﺎﺑق وﺑﻣ ﺎ أن ﻗﯾﻣ ﺔ ‪ F‬اﻟﻣﺣﺳ وﺑﺔ ﺗزﯾ د ﻋ ن اﻟﻘﯾﻣ ﺔ اﻟﺟدوﻟﯾ ﺔ ﻋﻧ د ﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﮫ‬ ‫‪   0.01‬ﻟذﻟك ﺗرﻓض ‪ H 0‬ﻋﻧد ‪   0.01‬وﻧﺳﺗﺧﻠص ﻣن ھ ذه اﻟﺗﺟرﺑ ﺔ أن ھﻧ ﺎك ﻓ روق ﻣﻌﻧوﯾ ﺔ‬ ‫ﺑﯾن ﻣﺗوﺳطﺎت اﻟذﻛﺎء واﻟﻘدرة ﻋﻠﻰ اﻟﺗرﻛﯾز ﺑﯾن اﻟﻣﺟﻣوﻋﺎت اﻟﻣﺧﺗﻠﻔﺔ‪.‬‬ ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﻣﻛﺗوب ﺑﻠﻐﺔ ‪ Mathematica‬وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬ ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫‪=0.01‬‬ ‫‪0.01‬‬ ‫‪a={{127,119,138.},{109,121,113.,112,130.},{119.,97,128,116‬‬ ‫}}‪,130,150,114},{120,109.,112},{94,96,101.,72,103,105,103,96‬‬ ‫‪{{127,119,138.},{109,121,113.,112,130.},{119.,97,128,116,130‬‬ ‫}}‪,150,114},{120,109.,112},{94,96,101.,72,103,105,103,96‬‬ ‫]‪f[x_]:=Apply[Plus,x‬‬ ‫‪٥١٩‬‬

h[x_]:=Length[x] k=h[a] 5 m=Table[h[a[[i]]],{i,1,k}] {3,5,7,3,8} n=f[m] 26 xy=Map[f,a] {384.,585.,854.,341.,770.} xp=xy/m {128.,117.,122.,113.667,96.25} f[xy] 2934. cf=((%)^2)/n 331091. x1=xy^2 {147456.,342225.,729316.,116281.,592900.} x2=x1/m {49152.,68445.,104188.,38760.3,74112.5} sbet=f[x2] 334658. sbet=sbet-cf 3567.22 xx=Map[f,a^2] {49334.,68735.,105806.,38825.,74896.} xx1=f[xx] 337596. ssto=xx1-cf 6505.38 sser=ssto-sbet 2938.17 k1=k-1 4 msb=sbet/k1 891.804 n1=n-1 25 sx=n1-k1 21 msse=sser/sx 139.913 f1=msb/msse 6.37401 rt2=List[" df "," ss "," mss { df , ss , mss , f } rt3=List[k1,sbet,msb,f1] ٥٢٠

","

{4,3567.22,891.804,6.37401} rt4=List[sx,sser,msse,"-"] {21,2938.17,139.913,-} rt5=List[n1,ssto,"-","-"] {25,6505.38,-,-} a11=TableHeadings->{{ S.V,bet,within,total},{ANOVA}} TableHeadings{{S.V,bet,within,total},{ANOVA}} uu1=TableForm[{rt2,rt3,rt4,rt5},a11]

S.V bet within total

ANOVA df 4 21 25

ss 3567.22 2938.17 6505.38

mss 891.804 139.913

 <<Statistics`ContinuousDistributions` ff1=Quantile[FRatioDistribution[k1,sx],1-] 4.36882 If[f1>ff1,Print["RjectHo"],Print["AccpetHo"]] RjectHo

f 6.37401  

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬ ‫ﻣﺴﺘﻮى اﻟﻤﻌﻨﻮﯾﺔ ﻣﻦ اﻻﻣﺮ‬ =0.05 ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬a (١٣- ٦) ‫ﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﯿﺎﻧﺎت اﻟﺨﺎﺻﺔ ﺑﺎﻟﻤﺜﺎل‬

‫ اﻟﻣﺧرﺟﺎت‬: ‫ﺛﺎﻧﯾﺎ‬

‫ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ uu1=TableForm[{rt2,rt3,rt4,rt5},a11]

: ‫ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم‬

H 0 : 1   2  3 :‫ﺿد اﻟﻔرض اﻟﺑدﯾل‬ H1 : ‫ ﯾﺧﺗﻠف ﻋن اﻟﺑﺎﻗﻲ‬i ‫واﺣد ﻋﻠﻲ اﻷﻗل ﻣن‬ ‫ اﻟﺟدوﻟﯾﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬f ff1=Quantile[FRatioDistribution[k1,sx],1-]

‫ اﻟﻣﺣﺳوﺑﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر‬f f1=msb/msse

‫اﻟﻘرار اﻟذى ﯾﺗﺧذ ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ If[f1>ff1,Print["RjectHo"],Print["AccpetHo"]]

‫واﻟﻣﺧرج ھو‬ Reject H0

‫اى رﻓض ﻓرض اﻟﻌدم‬ ٥٢١

‫)‪ (٢-٥-٦‬ﺗﺻﻣﯾم اﻟﻘطﺎﻋﺎت اﻟﻛﺎﻣﻠﺔ اﻟﻌﺷواﺋﯾﺔ )ﺗﺟﺎرب ذات اﻟﻌﺎﻣل اﻟواﺣد( ‪:‬‬ ‫‪Randomized complete blocks design: Single – factor experiments.‬‬ ‫إﺣ دى اﻟط رق اﻟ ﺗﺣﻛم اﻟﻣﺑﺎﺷ ر ﻓ ﻲ اﻟﺧط ﺄ ‪ ،‬أو ﺗﻘﻠﯾ ل اﻟﺧط ﺄ اﻟﺗﺟرﯾﺑ ﻲ ‪ ،‬وھ و ﺗﺟﻣﯾ ﻊ وﺣ دات‬ ‫اﻟﺗﺟرﺑﺔ ﻓﻲ ﻣﺟﻣوﻋﺎت أو ﻣﺎﯾﺳﻣﻰ اﻟﻘطﺎﻋ ﺎت وھ ذه اﻟﻘطﺎﻋ ﺎت ﺗﺗﺻف ﺑﺗﺟ ﺎﻧس ذاﺗﯾ ﺎ ً أي أن اﻟوﺣ دات‬ ‫اﻟﺗﺟرﯾﺑﯾ ﺔ اﻟﺗ ﻲ ﺗﺷ ﻛل اﻟﻘط ﺎع ﺗﻛ ون وﺣ دات ﻣﺗﺟﺎﻧﺳ ﺔ أو ﻗرﯾﺑ ﮫ ﻣ ن اﻟﺗﺟ ﺎﻧس وﻻ ﯾﺷ ﺗرط أن ﺗﻛ ون‬ ‫اﻟﻘطﺎﻋﺎت ﻣﺗﺷﺎﺑﮭﮫ ﻓﯾﻣﺎ ﺑﯾﻧﮭﺎ ﻣن ﻧﺎﺣﯾﺔ اﻟﺗﺟﺎﻧس ﻓﻘد ﺗﻛون ﻣﺧﺗﻠﻔﺔ ‪.‬‬ ‫وﯾ ﺗم ﺗﻌ ﯾن اﻟﻣﻌﺎﻟﺟ ﺎت ﻋﻠ ﻰ اﻟﻘط ﺎع ﺑطرﯾﻘ ﮫ ﻋﺷ واﺋﯾﺔ ‪ .‬وإذا ﻛ ﺎن ﻋ دد اﻟوﺣ دات اﻟﺗﺟرﯾﺑﯾ ﺔ داﺧ ل ﻛ ل‬ ‫ﻗط ﺎع ﯾﺳ ﺎوي ﻋ دد اﻟﻣﻌﺎﻟﺟ ﺎت اﻟﻣﺳ ﺗﺧدﻣﺔ ﻓ ﺎن اﻟﺗﺻ ﻣﯾم ﯾﺳ ﻣﻰ ﺗﺻ ﻣﯾم ﻗطﺎﻋ ﺎت ﻛﺎﻣﻠ ﺔ اﻟﻌﺷ واﺋﯾﺔ‬ ‫‪ complete blocks design‬ﺣﯾ ث ﻋ دد اﻟﻘطﺎﻋ ﺎت ‪ ،‬و ﻟ ﯾﻛن ‪ ،n‬ﯾﺳ ﺎوي ﻋ دد اﻟﺗﻛ رارات اﻟﻣﻘ ررة‬ ‫ﻣن ﻗﺑل اﻟﻘﺎﺋم ﻋﻠﻰ اﻟﺗﺟرﺑﺔ ‪.‬‬ ‫وھﻧﺎك ﻣﺟﺎﻻت ﻛﺛﯾرة ﺗﺳﺗﺧدم ﺗﺻﻣﯾم اﻟﻘطﺎﻋﺎت اﻟﻌﺷواﺋﯾﺔ‪:‬‬ ‫ﻓﻔﻲ ﻣﺟ ﺎل ﻋﻠ م اﻟﺣﯾ وان ﻗ د ﯾﻛ ون اﻟﻘط ﺎع ﻋﺑ ﺎرة ﻋ ن ﻣﺟﻣوﻋ ﮫ ﻣ ن اﻟﺣﯾواﻧ ﺎت ﺣدﯾﺛ ﺔ اﻟ وﻻدة ﻣ ن‬ ‫وﻟ ده واﺣ ده ‪ .‬وﯾ راد دراﺳ ﺔ ﺗ ﺄﺛﯾر أﻧ واع ﻣﺧﺗﻠﻔ ﺔ ﻣ ن اﻟﺧﻠط ﺎت اﻟﻐذاﺋﯾ ﺔ ﻋﻠ ﻰ زﯾ ﺎدة اﻟ وزن ﻟﮭ ذه‬ ‫اﻟﺣﯾواﻧﺎت‪.‬‬ ‫وﻓﻲ اﻟﺗﺟﺎرب اﻟزراﻋﯾﺔ ﻗد ﯾﻛون اﻟﻘطﺎع ﻋﺑﺎرة ﻋن ﻗطﻌﺔ ارض ﻣﺗﺟﺎﻧﺳﺔ ﻣ ن ﻧﺎﺣﯾ ﺔ درﺟ ﺔ اﻟﺧﺻ وﺑﺔ‬ ‫وﻣواﺻﻔﺎت اﻟﺗرﺑﺔ ‪.‬‬ ‫وﻓﻲ ﻣﺟﺎل اﻟﺻﻧﺎﻋﺔ ﻗد ﯾﻛون اﻟﻘطﺎع ﺳﯾﺎرة ﯾوزع ﻋﻠﻰ ﻋﺟﻼﺗﮭﺎ أرﺑﻌﺔ أﻧ واع ﻣ ن اﻻﯾط ﺎرات ﺑطرﯾﻘ ﺔ‬ ‫ﻋﺷواﺋﯾﺔ وذﻟك ﻟﻠوﺻول إﻟﻰ أﻓﺿل ﻣﻘﺎرﻧﮫ ﻣن اﻻﯾطﺎرات وذﻟك ﻟﯾﺗﺧذ اﻟﻘﺎﺋم ﻋﻠﻰ اﻟﺗﺟرﺑ ﺔ ﻗ رار ﺑﺷ ﺎن‬ ‫أي اﻷﻧواع ﯾﻛون ﺟزﺋﮭﺎ اﻟﻣﻼﻣس ﻟﻸرض اﻗل اﺳﺗﮭﻼك ﻣ ن ﻏﯾرھ ﺎ ﺑﻌ د ﻗط ﻊ ﻋ دد ﻣﻌ ﯾن ﻣ ن اﻷﻣﯾ ﺎل‬ ‫ﺣﯾث اﻟﺗﻐﯾ ر اﻟﻣﻘ ﺎس ھ و اﻟﻔ رق ﻓ ﻲ ﺳ ﻣك اﻻﯾط ﺎر ﺑ ﯾن وﻗ ت اﺳ ﺗﺧداﻣﮫ ﻓ ﻲ ﻋﺟﻠ ﺔ اﻟﺳ ﯾﺎرة وﺑﯾﻧ ﮫ ﺑﻌ د‬ ‫إﻛﻣﺎل اﻟﻣﺳﺎﻓﺔ اﻟﻣﻘررة ‪ .‬وﯾﻛون اﻟﻌﺎﻣل اﻟوﺣﯾد اﻟذي ﯾﺟب اﻻھﺗﻣﺎم ﺑﮫ ھو ﻧوع اﻻﯾطﺎر ‪.‬‬ ‫ﻣزاﯾﺎ اﻟﺗﺻﻣﯾم ‪:‬‬ ‫ﺗؤدي ﻋﻣﻠﯾﺔ اﻟﺗﺟﻣﯾﻊ إﻟﻰ ﻗطﺎﻋﺎت ﻟﺗﺣﺳﯾن اﻟﻧﺗﺎﺋﺞ ﺑﺧﻠف اﻟﺗﺻﻣﯾم اﻟﺗﺎم ﻟﺗﻌﯾﺷﮫ‪.‬‬ ‫ﯾﻣﻛن اﺳﺗﺧدام أي ﻋدد ﻣن اﻟﻣﻌﺎﻟﺟﺎت وأي ﻋدد ﻣن اﻟﺗﻛرارات ‪.‬‬ ‫ﺳﮭوﻟﺔ اﻟﺗﺣﻠﯾل اﻹﺣﺻﺎﺋﻲ ﻧﺳﺑﯾﺎ ً‪.‬‬ ‫ﯾﻣﻛن إﻏﻔﺎل ﻣﻌﺎﻟﺟﺔ أو ﻗطﺎع ﻷي ﺳﺑب ﻣن اﻷﺳﺑﺎب ﻣ ن اﻟﺗﺣﻠﯾ ل اﻹﺣﺻ ﺎﺋﻲ ﻣ ﻊ ﻋ دم ﺗﻌﻘﯾ د اﻟﻌﻣﻠﯾ ﺎت‬ ‫اﻟﺣﺳﺎﺑﯾﺔ ‪.‬‬ ‫ﻋﯾوب اﻟﺗﺻﻣﯾم ‪:‬‬ ‫‪ ‬إن ﻓﻘدان أي ﻣﺷﺎھدة داﺧل اﻟﻘطﺎع ﺗؤدي إﻟﻰ ﺣﺳﺎﺑﺎت إﺿﺎﻓﯾﺔ ‪.‬‬ ‫‪ ‬اﻟﺣﺎﺟﺔ إﻟﻰ اﻟﻌدﯾد ﻣن ﻓروض اﻟﻧﻣوذج اﻟرﯾﺎﺿﻲ ‪.‬‬

‫ﻣﺛﺎل)‪(١٤-٦‬‬ ‫ﻓﻲ دراﺳﺔ ﻻﺧﺗﺑﺎر ﺗﺄﺛﯾر دواﺋﯾﯾن ‪ A , B‬ﻋﻠﻰ ﻋدد ‪ lymphocyte‬ﻓﻲ دم اﻟﻔﺋران وذﻟك ﺑﻣﻘﺎرﻧﺔ‬ ‫‪ A,B‬ﻣﻊ ﻣﻌﺎﻟﺟﺔ اﻟﻣراﻗﺑﺔ ‪ .P‬وﺑﺈﺟراء اﻟﺗﺟرﺑﺔ ﺑﺎﺳﺗﺧدام ﺗﺻﻣﯾم اﻟﻘطﺎﻋﺎت اﻟﻛﺎﻣﻠﺔ اﻟﻌﺷواﺋﯾﺔ ﺣﯾث‬ ‫‪ n=7, k=3‬وﻗد ﺗم ﺗوزﯾﻊ اﻟﻣﻌﺎﻟﺟﺎت ﻋﺷواﺋﯾﺎ ً ﻋﻠﻰ ﻛل ﻗطﺎع ﻛﻣﺎ ﺗم ﺣﺳﺎب ﻋدد ‪lymphocyte‬‬ ‫ﺑﺎﻷﻟف ﻟﻛل ﻣﻠﻠﯾﻣﺗر ﻣﻛﻌب ﻣن اﻟدم وﻧﺗﺎﺋﺞ اﻟﺗﺟرﺑﺔ ﻣﻌطﺎة ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫‪٥٢٢‬‬

‫اﻟﻘطﺎﻋﺎت‬

‫‪Yi‬‬ ‫‪38.8‬‬ ‫‪45.4‬‬ ‫‪37.4‬‬ ‫‪T..  121.6‬‬

‫‪7‬‬ ‫‪5.5‬‬ ‫‪6.8‬‬ ‫‪5.4‬‬ ‫‪17.7‬‬

‫‪5‬‬ ‫‪3.5‬‬ ‫‪5.5‬‬ ‫‪3.9‬‬ ‫‪12.9‬‬

‫‪6‬‬ ‫‪7.6‬‬ ‫‪9.0‬‬ ‫‪7.0‬‬ ‫‪23.6‬‬

‫‪4‬‬ ‫‪5.8‬‬ ‫‪6.4‬‬ ‫‪5.6‬‬ ‫‪17.8‬‬

‫اﻟﻣﻌﺎﻟﺟﺎت‬ ‫‪3‬‬ ‫‪7.0‬‬ ‫‪6.9‬‬ ‫‪6.5‬‬ ‫‪20.4‬‬

‫‪2‬‬ ‫‪4.0‬‬ ‫‪4.8‬‬ ‫‪3.9‬‬ ‫‪12.7‬‬

‫‪1‬‬ ‫‪5.4‬‬ ‫‪6.0‬‬ ‫‪5.1‬‬ ‫‪16.5‬‬

‫‪P‬‬ ‫‪A‬‬ ‫‪B‬‬

‫‪x. j‬‬

‫اﻟﺣــل‪:‬‬ ‫ﻹﯾﺟﺎد ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﻧﺗﺑﻊ اﻵﺗﻲ‪:‬‬

‫‪T..2 (121.6)2‬‬ ‫‪‬‬ ‫‪ 704.12‬‬ ‫‪nk‬‬ ‫‪21‬‬

‫‪(1) ‬‬

‫‪(2)    y ij2  741.36‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫)‪ Ti. (38.8)  (45.4)  (37.4‬‬ ‫‪‬‬ ‫‪ 709.33‬‬ ‫‪n‬‬ ‫‪7‬‬ ‫‪2‬‬ ‫‪ T. j (16.5)2  (12.7)2    (23.6)2  (17.7) 2‬‬ ‫‪(4) ‬‬ ‫‪‬‬ ‫‪k‬‬ ‫‪3‬‬ ‫‪ 734.4.‬‬

‫‪(3) ‬‬

‫وﻋﻠﻰ ذﻟك ﻓﺈن ‪ SST , SSTr , SSBL, SSE‬ﯾﺗم ﺣﺳﺎﺑﮭم ﻛﺎﻟﺗﺎﻟﻲ‪:‬‬ ‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت اﻟﻛﻠﻲ ﺳﯾﻛون‪:‬‬ ‫‪SSTO = (2) - (1) = 37.2 ,‬‬ ‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻠﻣﻌﺎﻟﺟﺎت )اﻟﺻﻔوف( ﺳﯾﻛون‪:‬‬ ‫‪SSTR = (3) - (1) = 5.21 ,‬‬ ‫‪SSBL= (4) - (1) = 30.28 ,‬‬ ‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻠﻘطﺎﻋﺎت )اﻻﻋﻣدة( ﺳﯾﻛون‪:‬‬ ‫‪SSBC= (4) - (1) = 30.28 ,‬‬ ‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت ﻟﻠﺧطﺄ ﺳﯾﻛون‪:‬‬ ‫‪SSE = SSTO – [ SSTR + SSBC] = 1.71.‬‬ ‫ﺗﻠﺧص اﻟﻧﺗﺎﺋﺞ ﻓﻲ ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن اﻟﻣﻌطﻰ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫) ‪f  (1 ,  2‬‬

‫‪ f‬اﻟﻣﺣﺳوﺑﺔ‬

‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت‬

‫ﻣﺗوﺳط‬ ‫‪٥٢٣‬‬

‫درﺟﺎت‬

‫ﻣﺻدر‬

‫)‪= 6.93 f 0.01 (2,12‬‬

‫**‪18.57‬‬

‫اﻟﻣرﺑﻌﺎت‬ ‫‪2.60‬‬

‫‪5.21‬‬ ‫‪30.28‬‬ ‫‪1.71‬‬

‫‪0.14‬‬

‫اﻻﺧﺗﻼف‬ ‫اﻟﻣﻌﺎﻟﺟﺎت‬ ‫اﻟﻘطﺎﻋﺎت‬

‫اﻟﺣرﯾﺔ‬ ‫‪2‬‬ ‫‪6‬‬

‫اﻟﺧطﺄ‬

‫‪12‬‬

‫‪20‬‬ ‫‪37.2‬‬ ‫ﻣ ن اﻟﺟ دول اﻟﺳ ﺎﺑق وﺑﻣ ﺎ أن ﻗﯾﻣ ﺔ ‪ F‬اﻟﻣﺣﺳ وﺑﺔ ﺗزﯾ د ﻋ ن اﻟﻘﯾﻣ ﺔ اﻟﺟدوﻟﯾ ﺔ ﻋﻧ د ﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﮫ‬ ‫‪   0.01‬ﻟذﻟك ﺗرﻓض ‪ H 0‬ﻋﻧد ‪   0.01‬وﻧﺳﺗﺧﻠص ﻣن ھذه اﻟﺗﺟرﺑﺔ أن ھﻧﺎك‬ ‫ﻓروق ﻣﻌﻧوﯾﺔ ﺑﯾن ﻣﺗوﺳطﺎت ﻣﺟﺗﻣﻌﺎت‪.‬‬ ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﻣﻛﺗوب ﺑﻠﻐﺔ ‪ Mathematica‬وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬ ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫‪=0.01‬‬ ‫‪0.01‬‬ ‫{‪a={{5.4,4,7.,5.8,3.5,7.6,5.5},{6,4.8,6.9,6.4,5.5,9,6.8},‬‬ ‫}}‪5.1,3.9,6.5,5.6,3.9,7,5.4‬‬ ‫‪{{5.4,4,7.,5.8,3.5,7.6,5.5},{6,4.8,6.9,6.4,5.5,9,6.8},{5.1,3‬‬ ‫}}‪.9,6.5,5.6,3.9,7,5.4‬‬ ‫]‪Dimensions[a‬‬ ‫}‪{3,7‬‬ ‫]]‪b=N[Transpose[a‬‬ ‫‪{{5.4,6.,5.1},{4.,4.8,3.9},{7.,6.9,6.5},{5.8,6.4,5.6},{3.5,5‬‬ ‫}}‪.5,3.9},{7.6,9.,7.},{5.5,6.8,5.4‬‬ ‫]‪f[x_]:=Apply[Plus,x‬‬ ‫]‪g[x_]:=Length[x‬‬ ‫]]]‪na=g[a[[1‬‬ ‫‪7‬‬ ‫]]]‪nb=g[b[[2‬‬ ‫‪3‬‬ ‫‪k1=na-1‬‬ ‫‪6‬‬ ‫‪k2=nb-1‬‬ ‫‪2‬‬ ‫‪N1=na*nb‬‬ ‫‪21‬‬ ‫]‪xxx=Flatten[a‬‬ ‫‪{5.4,4,7.,5.8,3.5,7.6,5.5,6,4.8,6.9,6.4,5.5,9,6.8,5.1,3.9,6.‬‬ ‫}‪5,5.6,3.9,7,5.4‬‬ ‫]‪xx2=f[xxx‬‬ ‫‪121.6‬‬ ‫]‪sxx22=f[xxx^2‬‬ ‫‪٥٢٤‬‬

‫اﻟﻛﻠﻲ‬

741.36 cf=(xx2^2)/N1 704.122 x1=Map[f,a] {38.8,45.4,37.4} aaaa1=f[x1^2]/na 709.337 aaaa=aaaa1-cf 5.21524 ssa=aaaa/k2 2.60762 xx2=Map[f,b] {16.5,12.7,20.4,17.8,12.9,23.6,17.7} bbb1=f[xx2^2]/nb 734.4 bbbb=bbb1-cf 30.2781 tot=sxx22-cf 37.2381 sser=tot-aaaa-bbbb 1.74476 v1=N1-1-k1-k2 12 msser=sser/v1 0.145397 v1=N1-1-k1-k2 12 ff1=ssa/msser 17.9345 rt2=List[" df "," ss "," mss "," f "] { df , ss , mss , f } rt1=List[k2,aaaa,ssa,ff1] {2,5.21524,2.60762,17.9345} rt3=List[k1,bbbb,"_","_"] {6,30.2781,_,_} rt9=List[v1,sser,msser,"-"] {12,1.74476,0.145397,-} rr9=List[N1-1,tot,"_","_"] {20,37.2381,_,_} a11=TableHeadings->{{ S.V,A,B,error,total},{ANOVA}} TableHeadings{{S.V,A,B,error,total},{ANOVA}} uu1=TableForm[{rt2,rt1,rt3,rt9,rr9},a11]

٥٢٥

‫‪f‬‬ ‫‪17.9345‬‬ ‫_‬ ‫‪‬‬

‫_‬

‫‪mss‬‬ ‫‪2.60762‬‬ ‫_‬ ‫‪0.145397‬‬ ‫_‬

‫‪ss‬‬ ‫‪5.21524‬‬ ‫‪30.2781‬‬ ‫‪1.74476‬‬ ‫‪37.2381‬‬

‫‪ANOVA‬‬ ‫‪df‬‬ ‫‪2‬‬ ‫‪6‬‬ ‫‪12‬‬ ‫‪20‬‬

‫‪S.V‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪error‬‬ ‫‪total‬‬

‫`‪<<Statistics`ContinuousDistributions‬‬ ‫;‪=0.01‬‬ ‫]‪f1=Quantile[FRatioDistribution[k2,v1],1-‬‬ ‫‪6.92661‬‬ ‫]]"‪If[ff1>f1,Print["RjectHo"],Print["AccpetHo‬‬ ‫‪RjectHo‬‬

‫اوﻻ‪ :‬اﻟﻣدﺧﻼت‬ ‫ﻣﺴﺘﻮى اﻟﻤﻌﻨﻮﯾﺔ ﻣﻦ اﻻﻣﺮ‬ ‫‪=0.01‬‬ ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ ‫‪ a‬وھﻰ‬ ‫{‪a={{5.4,4,7.,5.8,3.5,7.6,5.5},{6,4.8,6.9,6.4,5.5,9,6.8},‬‬ ‫}}‪5.1,3.9,6.5,5.6,3.9,7,5.4‬‬ ‫وﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﯿﺎﻧﺎت اﻟﺨﺎﺻﺔ ﺑﺎﻟﻤﺜﺎل )‪(١٤-٦‬‬

‫ﺛﺎﻧﯾﺎ ‪ :‬اﻟﻣﺧرﺟﺎت‬

‫ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﯾﺗم اﻟﺣﺻول ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ ‫]‪uu1=TableForm[{rt2,rt1,rt3,rt9,rr9},a11‬‬

‫ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪:‬‬

‫‪H 0 : 1   2   3‬‬ ‫ﺿد اﻟﻔرض اﻟﺑدﯾل‪:‬‬ ‫واﺣد ﻋﻠﻲ اﻷﻗل ﻣن ‪  i‬ﯾﺧﺗﻠف ﻋن اﻟﺑﺎﻗﻲ ‪H1 :‬‬ ‫‪ f‬اﻟﺟدوﻟﯾﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪f1=Quantile[FRatioDistribution[k2,v1],1-‬‬

‫‪ f‬اﻟﻣﺣﺳوﺑﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر‬ ‫‪ff1=ssa/msser‬‬

‫اﻟﻘرار اﻟذى ﯾﺗﺧذ ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ ‫]]"‪If[ff1>f1,Print["RjectHo"],Print["AccpetHo‬‬

‫‪Reject H0‬‬

‫اى رﻓض ﻓرض اﻟﻌدم‬

‫)‪ (٣-٥-٦‬ﺗﺻﻣﯾم اﻟﻣرﺑﻊ اﻟﻼﺗﯾﻧﻲ‪ :‬ﺗﺟﺎرب ذات ﻋﺎﻣل واﺣد‬ ‫‪٥٢٦‬‬

‫إن ﺗﺻ ﻣﯾم اﻟﻘطﺎﻋ ﺎت اﻟﻛﺎﻣﻠ ﺔ اﻟﻌﺷ واﺋﯾﺔ ﯾﮭ دف إﻟ ﻰ ﺗﺻ ﻐﯾر ﺧط ﺄ اﻟﺗﺟرﺑ ﺔ وذﻟ ك ﺑﺈزاﻟ ﺔ واﺣ د ﻣ ن‬ ‫ﻣﺻ ﺎدر اﻻﺧ ﺗﻼف )اﻟﻘطﺎﻋ ﺎت(‪٠‬ھﻧ ﺎك ﺗﺻ ﻣﯾم أﺧ ر ﻟ ﮫ أھﻣﯾ ﺔ ﻓ ﻲ اﻟ ﺗﺣﻛم ﻓ ﻲ ﻣﺻ درﯾن ﻟﻼﺧ ﺗﻼف‬ ‫واﻟﻣﺳ ﻣﻰ ﺑ ﺎﻟﻣرﺑﻊ اﻟﻼﺗﯾﻧ ﻲ‪٠‬ﻓ ﻲ ھ ذا اﻟﺗﺻ ﻣﯾم ﯾ ﺗم ﺗﺟﻣﯾ ﻊ اﻟوﺣ دات اﻟﺗﺟرﯾﺑﯾ ﺔ اﻟﻐﯾ ر ﻣﺗﺟﺎﻧﺳ ﺔ ﻓ ﻲ‬ ‫ﻣﺟﻣوﻋ ﺎت ﺗﺿ م وﺣ دات ﻣﺗﺟﺎﻧﺳ ﺔ أو ﻗرﯾﺑ ﺔ ﻣ ن اﻟﺗﺟ ﺎﻧس وھ ذا اﻟﺗﺟﻣﯾ ﻊ ﯾﻛ ون ﻓ ﻲ اﺗﺟ ﺎھﯾن وﯾﺳ ﻣﻰ‬ ‫اﺣدھﻣﺎ اﺗﺟﺎه اﻟﺻﻔوف وﯾﺳﻣﻰ اﻷﺧر اﺗﺟﺎه اﻷﻋﻣدة وﻣﻌﻧﻰ ذﻟك إن ﻛل ﺻف وﻛ ل ﻋﻣ ود ﻣ ﺎ ھ و إﻟ ﻰ‬ ‫ﻣﺟﻣوﻋ ﺔ أو ﻗط ﺎع أو ﻣﻛ رر ﻛﺎﻣ ل‪٠‬ﻓ ﻲ ھ ذا اﻟﺗﺻ ﻣﯾم ﺗ وزع اﻟﻣﻌﺎﻟﺟ ﺎت اﻟﻣ راد دراﺳ ﺔ ﺗﺄﺛﯾرھ ﺎ ﻓ ﻲ‬ ‫اﻟﺗﺟرﺑﺔ ﻋﻠﻰ اﻟوﺣدات اﻟﺗﺟرﯾﺑﯾﺔ ﺗﺑﻌﺎ ﻟﻠﺷرطﯾن اﻟﺗﺎﻟﯾﯾن‪:‬‬ ‫أ ‪ -‬ﻋدد اﻟﻣﻌﺎﻟﺟﺎت = ﻋدد اﻟﺻﻔوف = ﻋدد اﻻﻋﻣده‪٠‬‬ ‫ب‪ -‬ﻛل ﻣﻌﺎﻟﺟﮫ ﺗظﮭر ﻣرة واﺣدة ﻓﻲ اﻟﺻف أو اﻟﻌﻣود‪٠‬‬ ‫ﻋﺎدة ﯾﺷﺎر إﻟﻰ ﺗﺻﻣﯾم اﻟﻣرﺑ ﻊ اﻟﻼﺗﯾﻧ ﻲ وﻓﻘ ﺎ ﻟرﺗﺑﺗ ﮫ ﻓﺣﯾﻧﻣ ﺎ ﻧﻘ ول ﺗﺻ ﻣﯾم ﻣرﺑ ﻊ ﻻﺗﯾﻧ ﻲ ‪ 4×4‬ﻓﮭ ذا‬ ‫ﯾﻌﻧ ﻲ إن ﻋ دد اﻟﻣﻌﺎﻟﺟ ﺎت اﻟداﺧﻠ ﺔ ﻓ ﻲ اﻟﺗﺟرﺑ ﺔ ھ ﻲ أرﺑﻌ ﺔ ﻣﻌﺎﻟﺟ ﺎت واﻟﻣﺳ ﺎوﯾﺔ ﻟﻌ دد اﻟﺻ ﻔوف وﻋ دد‬ ‫اﻷﻋﻣ دة ‪ .‬وﺣﯾﻧﻣ ﺎ ﻧﻘ ول ﻣرﺑ ﻊ ﻻﺗﯾﻧ ﻲ‪ 5×5‬ﻓﮭ ذا ﯾﻌﻧ ﻲ إن ﻋ دد اﻟﻣﻌﺎﻟﺟ ﺎت ﺧﻣﺳ ﺔ واﻟﻣﺳ ﺎوﯾﺔ ﻟﻌ دد‬ ‫اﻟﺻ ﻔوف وﻋ دد اﻷﻋﻣ دة ‪ .‬ﻓﻌﻠ ﻰ ﺳ ﺑﯾل اﻟﻣﺛ ﺎل ﺑﻔ رض إﻧﻧ ﺎ ﻧﮭ ﺗم ﺑﺈﻧﺗﺎﺟﯾ ﺔ أرﺑﻌ ﺔ أﻧ واع ﻣ ن اﻟﻘﻣ ﺢ‬ ‫‪A,B,C,D‬وذﻟك ﺑﺎﺳﺗﺧدام أرﺑﻌﺔ أﻧواع ﻣن اﻷﺳﻣدة ‪) 1,2,3,4‬اﻟﺻﻔوف( وذﻟك ﺧ ﻼل أرﺑﻌ ﺔ ﻓﺗ رات‬ ‫زﻣﻧﯾﺔ ‪) 1,2,3,4‬اﻷﻋﻣدة(‪ ٠‬اھﺗﻣﺎﻣﻧﺎ ھﻧﺎ ﺳوف ﯾﻛ ون ﻓ ﻲ دراﺳ ﺔ ﻋﺎﻣ ل واﺣ د ﻓﻘ ط وھ و ﻧ وع اﻟﻘﻣ ﺢ‪.‬‬ ‫ﻧﻼﺣظ ﻣن اﻟﺟدول اﻟﺗﺎﻟﻰ ﻣرﺑﻊ ﻻﺗﯾﻧﻲ ‪ 4×4‬ﺣﯾ ث ﻛ ل ﻣﻌﺎﻟﺟ ﺔ ظﮭ رت ﻣ رة واﺣ دة ﺑﺎﻟﺿ ﺑط ﻓ ﻲ ﻛ ل‬ ‫ﺻ ف وﻛ ل ﻋﻣ ود‪ ٠‬ﺑﺎﺳ ﺗﺧدام ﻣﺛ ل ھ ذه اﻟﺗرﺗﯾﺑ ﺎت اﻟﻣﺗوازﻧ ﺔ وﺑﺎﺳ ﺗﺧدام ﺗﺣﻠﯾ ل اﻟﺗﺑ ﺎﯾن ﯾﻣﻛﻧﻧ ﺎ ﻓﺻ ل‬ ‫اﻻﺧ ﺗﻼف اﻟ ذي ﯾرﺟ ﻊ إﻟ ﻰ اﺧ ﺗﻼف اﻷﺳ ﻣدة أو اﺧ ﺗﻼف اﻟﺳ ﻧوات ﻣ ن ﻣﺟﻣ وع اﻟﻣرﺑﻌ ﺎت اﻟﻛﻠ ﻰ‬ ‫وﺑﺎﻟﺗﺎﻟﻲ اﻟﺣﺻول ﻋﻠﻰ اﺧﺗﺑﺎر أﻛﺛ ر دﻗ ﺔ ﻟﻼﺧ ﺗﻼف ﻓ ﻲ اﻹﻧﺗﺎﺟﯾ ﺔ ﻟﻸﻧ واع اﻷرﺑﻌ ﺔ ﻣ ن اﻟﻘﻣ ﺢ‪ ٠‬ﻋﻧ دﻣﺎ‬ ‫ﯾﻛون ھﻧﺎك ﺗﻔﺎﻋل ﺑﯾن أي ﻣن ﻣﺻﺎدر اﻻﺧﺗﻼف ﻓﺎن ﻗﯾﻣﺔ ‪ F‬ﻓﻲ ﺗﺣﻠﯾ ل اﻟﺗﺑ ﺎﯾن ﺗﻛ ون ﻏﯾ ر ﺻ ﺣﯾﺣﺔ ‪،‬‬ ‫وﻓﻲ ھذه اﻟﺣﺎﻟﺔ ﻓﻠن ﯾﻛون ﺗﺻﻣﯾم اﻟﻣرﺑﻊ اﻟﻼﺗﯾﻧﻲ ھو اﻟﺗﺻﻣﯾم اﻟﻣﻧﺎﺳب‪.‬‬ ‫‪4‬‬

‫‪3‬‬

‫‪2‬‬

‫‪1‬‬

‫‪D‬‬ ‫‪C‬‬ ‫‪B‬‬ ‫‪A‬‬

‫‪C‬‬ ‫‪B‬‬ ‫‪A‬‬ ‫‪D‬‬

‫‪B‬‬ ‫‪A‬‬ ‫‪D‬‬ ‫‪C‬‬

‫‪A‬‬ ‫‪D‬‬ ‫‪C‬‬ ‫‪B‬‬

‫اﻷﻋﻣدة‬ ‫اﻟﺻﻔوف‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬

‫ﻓﯾﻣﺎ ﯾﻠﻲ ﻋدة إﺷﻛﺎل ﻟﻣرﺑﻌﺎت ﻻﺗﯾﻧﯾﺔ‪.‬‬ ‫‪C‬‬ ‫‪B‬‬ ‫‪A‬‬

‫ﻣرﺑﻊ ﻻﺗﯾﻧﻲ ‪3×3‬‬ ‫‪B‬‬ ‫‪A‬‬ ‫‪C‬‬

‫ﻣرﺑﻊ ﻻﺗﯾﻧﻲ ‪4×4‬‬ ‫‪٥٢٧‬‬

‫‪A‬‬ ‫‪C‬‬ ‫‪B‬‬

‫‪A‬‬ ‫‪B‬‬ ‫‪D‬‬ ‫‪C‬‬

‫‪B‬‬ ‫‪E‬‬ ‫‪A‬‬ ‫‪D‬‬ ‫‪C‬‬

‫‪B‬‬ ‫‪A‬‬ ‫‪C‬‬ ‫‪D‬‬

‫‪C‬‬ ‫‪D‬‬ ‫‪B‬‬ ‫‪A‬‬

‫ﻣرﺑﻊ ﻻﺗﯾﻧﻲ ‪5×5‬‬ ‫‪E‬‬ ‫‪C‬‬ ‫‪C‬‬ ‫‪D‬‬ ‫‪B‬‬ ‫‪E‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪D‬‬ ‫‪A‬‬

‫‪D‬‬ ‫‪C‬‬ ‫‪A‬‬ ‫‪B‬‬

‫‪A‬‬ ‫‪B‬‬ ‫‪D‬‬ ‫‪C‬‬ ‫‪E‬‬

‫‪D‬‬ ‫‪A‬‬ ‫‪C‬‬ ‫‪E‬‬ ‫‪B‬‬

‫ﻣن اﻟطﺑﯾﻌﻲ إن ﺗﻛون اﻟﻣرﺑﻌﺎت اﻟﻼﺗﯾﻧﯾﺔ اﻟﻛﺑﯾ رة اﻟرﺗﺑ ﺔ ﻗﻠﯾﻠ ﺔ اﻻﺳ ﺗﺧدام ﺣﺗ ﻰ إن اﻟﻣرﺑﻌ ﺎت ﻣ ن رﺗﺑ ﺔ‬ ‫اﻛﺑ ر ﻣ ن ‪ 12 ×12‬ﺗﻛ ون ﻏﯾ ر ﻋﻣﻠﯾ ﺔ وﻏﯾ ر ﻣﺳ ﺗﺧدﻣﺔ‪ .‬ﺑﯾﻧﻣ ﺎ اﻷﻛﺛ ر اﺳ ﺗﺧداﻣﺎ ھ ﻲ اﻟﻣرﺑﻌ ﺎت اﻟﺗ ﻲ‬ ‫رﺗﺑﺗﮭﺎ ﺗﻘﻊ ﻓﻲ اﻟﻣدى‪ 5×5‬و ‪. 8×8‬‬ ‫ھذا اﻟﺗﺻﻣﯾم ﺷﺎﺋﻊ اﻻﺳﺗﺧدام ﻓﻲ ‪:‬‬ ‫‪ -١‬اﻟﺗﺟﺎرب اﻟزراﻋﯾﺔ ﺣﯾث ﯾوﺟد ﻣﺻدرﯾن ﻟﻼﺧﺗﻼف ﻣﺛ ل اﻟﺧﺻ وﺑﺔ واﻟﻣﯾ ل أو ﺧﺻ وﺑﺗﻲ ﻣﺧﺗﻠﻔﺗ ﯾن‬ ‫‪٠‬وﻗد ﺗﻣﺛل اﻟﺻﻔوف إﺷراف ﻋدد ﻣن اﻷﺷﺧﺎص ذو ﺧﺑرات ﻣﺗﺑﺎﯾﻧ ﺔ ﻓ ﻲ اﻟزراﻋ ﺔ واﻷﻋﻣ دة ﺗﻣﺛ ل‬ ‫أﺻﻧﺎﻓﺎ ﻣﺧﺗﻠﻔﺔ ﻓﻲ اﻟﺣﻘول‪.‬‬ ‫‪ -٢‬اﻟﺗﺟﺎرب اﻟﻛﯾﻣﺎﺋﯾﺔ ﻓﻘد ﯾﺗوﻗﻊ اﻟﺑﺎﺣث ﻓﻲ ﺑﻌ ض اﻷﺣﯾ ﺎن اﻻﺧ ﺗﻼف ﻓ ﻲ اﻻﺳ ﺗﺟﺎﺑﺔ واﻟﺗ ﻲ ﺗﻧ ﺗﺞ ﻣ ن‬ ‫اﺧﺗﻼف ﻓﻲ أﯾﺎم اﻟﺗﺣﻠﯾل واﻟﺗرﺗﯾب ﻓﻲ اﻟﺗﺣﻠﯾل ﺧﻼل اﻟﯾوم اﻟواﺣد‪.‬‬ ‫‪ -٣‬ﻓﻲ ﺗﺟ ﺎرب اﺧﺗﺑ ﺎر أﻧ واع ﻣﺧﺗﻠﻔ ﺔ ﻣ ن اﻹط ﺎرات)أرﺑﻌ ﺔ أﻧ واع ﻋﻠ ﻰ ﺳ ﺑﯾل اﻟﻣﺛ ﺎل( ﻓ ﺎن اﺳ ﺗﮭﻼك‬ ‫اﻹطﺎرات اﻷﻣﺎﻣﯾﺔ ﻟﺳﯾﺎرة ﻣﻌﯾﻧﺔ ﯾﺧﺗﻠف ﻋن اﺳﺗﮭﻼك اﻹطﺎرات اﻟﺧﻠﻔﯾ ﺔ وﻛ ذﻟك ﯾﺧﺗﻠ ف اﺳ ﺗﮭﻼك‬ ‫اﻹط ﺎر ﻣ ن ﺟﺎﻧ ب إﻟ ﻰ أﺧ ر‪ ٠‬ﺑﺎﺳ ﺗﺧدام ﺗﺻ ﻣﯾم اﻟﻣرﺑ ﻊ أﻻﺗﯾﻧ ﻲ ﯾﻣﻛ ن إن ﯾﺳ ﺗﺧدم ﻛ ل ﻧ وع ﻣ ن‬ ‫اﻹطﺎرات ﻣرة واﺣدة ﻓﻲ ﻛل ﺳﯾﺎرة )اﻟﺻﻔوف( وﻣرة واﺣ دة ﻓ ﻲ ﻛ ل ﻣوﻗ ﻊ ﻣ ن اﻟﻣواﻗ ﻊ اﻟﻣﺧﺗﻠﻔ ﺔ‬ ‫إﻣﺎﻣﻲ أﯾﺳر‪،‬ﺧﻠﻔﻲ أﯾﺳر‪ ،‬أﻣﺎﻣﻲ أﯾﻣن ‪ ،‬ﺧﻠﻔﻲ أﯾﻣن)اﻷﻋﻣدة(‪٠‬‬ ‫‪ -٤‬ﻓﻲ اﻟدراﺳ ﺎت اﻟﺗﺳ وﯾﻘﯾﺔ‪-‬ﻣ ﺛﻼ‪-‬إذا ﻛ ﺎن ﻣﺗوﻗﻌ ﺎ إن اﻻﺧ ﺗﻼف ﻓ ﻲ ﺣﺟ م اﻟﻣﺑﯾﻌ ﺎت ﺑﺣﺳ ب اﻷﯾ ﺎم أو‬ ‫اﻟﻔروع أو أﻗﺳﺎم اﻟﻣؤﺳﺳﺔ وأﺳﻠوب اﻹﻋﻼن أو طرﯾﻘ ﺔ ﻋ رض اﻟﺳ ﻠﻌﺔ‪٠‬وﻋﻠ ﻰ ذﻟ ك ﯾﻣﻛ ن اﻋﺗﺑ ﺎر‬ ‫اﻷﯾﺎم ﺗﻣﺛل اﻟﺻﻔوف وﻓروع اﻟﻣؤﺳﺳﺔ ﺗﻣﺛل اﻻﻋﻣدة‪٠‬‬ ‫‪ -٥‬ﻓ ﻲ اﻟﻣﺟ ﺎﻻت اﻟطﺑﯾ ﺔ ﺣﯾ ث اﻟﺻ ﻔوف ﺗﻣﺛ ل اﻟﻣﺳﺗﺷ ﻔﯾﺎت واﻷﻋﻣ دة ﺗﻣﺛ ل اﻷدوﯾ ﺔ واﻟﺣ روف ﺗﻣﺛ ل‬ ‫اﻹﻣراض اﻟﻣﺧﺗﻠﻔﺔ )اﻟﻣﻌﺎﻟﺟﺎت اﻟﻣﺧﺗﻠﻔﺔ(‪.‬‬ ‫‪ -٦‬وﻓﻲ اﻟﻣﺟﺎﻻت اﻟﻧﻔﺳﯾﺔ ﺣﯾث اﻟﺻﻔوف ﺗﻣﺛل ﻣﺟﺎﻣﯾﻊ ﻣﺧﺗﻠﻔﺔ ﻣ ن اﻷﺷ ﺧﺎص واﻷﻋﻣ دة ھ ﻲ ﺗرﺗﯾ ب‬ ‫اﻟﻛﺷف ﻋﻠﯾﮭم إﻣﺎ اﻟﺣروف ﻓﺗﻣﺛل اﻟﻣﻌﺎﻟﺟﺎت اﻟﻣﺧﺗﻠﻔﺔ‪.‬‬ ‫‪ -٧‬ﻓﻲ اﻟﺗﺟﺎرب اﻟﺗﻲ ﺗﺟرى ﻋﻠﻰ اﻟﺣﯾواﻧﺎت ﻓ ﺈن اﻟﺻ ﻔوف ﺗﻣﺛ ل اﻟواﻟ دات إﻣ ﺎ اﻷﻋﻣ دة ﻓﮭ ﻲ إﺣﺟ ﺎم‬ ‫اﻟﺣﯾواﻧﺎت وﺗﻣﺛل اﻟﺣروف اﻟﻣﻌﺎﻟﺟﺎت اﻟﻣﺧﺗﻠﻔﺔ‪.‬‬ ‫وﻣن أھم ﻋﯾوب ھذا اﻟﺗﺻﻣﯾم‪:‬‬ ‫‪٥٢٨‬‬

‫ﺗﺳﺎوى ﻋدد ﻛل ﻣن اﻟﺻﻔوف واﻷﻋﻣدة واﻟﻣﻌﺎﻟﺟﺎت ﯾﻘﻠ ل ﻣ ن درﺟ ﺎت اﻟﺣرﯾ ﺔ ﻟﻠﺧط ﺄ اﻟﺗﺟرﯾﺑ ﻲ‬ ‫ﻋﻧ دﻣﺎ ﯾﻛ ون ﻟ دﯾﻧﺎ ﻋ دد ﻗﻠﯾل ﻣ ن اﻟﻣﻌﺎﻟﺟ ﺎت واﻟﻌﻛ س ﺻ ﺣﯾﺢ ﻋﻧ دﻣﺎ ﯾ زداد ﻋ دد اﻟﻣﻌﺎﻟﺟ ﺎت ﻓ ﺈن ذﻟ ك‬ ‫ﯾ ؤدي إﻟ ﻰ أن درﺟ ﺎت اﻟﺣرﯾ ﺔ ﻟﻠﺧط ﺄ ﺗﻛ ون اﻛﺑ ر ﻣ ن اﻟ ﻼزم‪ .‬ﻓﻌﻠ ﻲ ﺳ ﺑﯾل اﻟﻣﺛ ﺎل ﻻ ﺗوﺟ د درﺟ ﺎت‬ ‫ﺣرﯾﺔ ﻟﺗﻘدﯾر اﻟﺧطﺄ ﻓﻲ اﻟﻣرﺑﻊ اﻟﻼﺗﯾﻧﻲ ‪ 2  2‬ﺑﯾﻧﻣﺎ ﯾﻌط ﻲ اﻟرﺗﺑ ﺔ ‪ 3 x 3‬درﺟﺗ ﻲ ﺣرﯾ ﺔ ﻟﺗﻘ دﯾر اﻟﺧط ﺄ‬ ‫واﻟﻣرﺑﻊ ‪ 4  4‬ﯾﻌطﻲ ﺳﺗﺔ درﺟﺎت ﺣرﯾﺔ اﻟﺧطﺄ‪.‬‬ ‫ﻋﻣﻠﯾﮫ أﻟﺗﻌﯾﺷﮫ أﻛﺛر ﺗﻌﻘﯾدا ﻣﻣﺎ أﺗﺑﻊ ﻓﻲ ﺗﺻﻣﯾم اﻟﻘطﺎﻋﺎت اﻟﻛﺎﻣﻠﺔ اﻟﻌﺷواﺋﯾﺔ‪.‬‬ ‫ﻓﻘدان اﺣد اﻟﻣﺷﺎھدات ﯾؤدى إﻟﻰ ﺣﺳﺎﺑﺎت إﺿﺎﻓﯾﺔ‪.‬‬

‫ﻣﺛﺎل)‪(١٥-٦‬‬ ‫اﺟري إﺣدى اﻟﻣﺣﻼت اﻟﺗﺟﺎرﯾﺔ دراﺳﺔ اﺳﺗطﻼﻋﯾﮫ ﻋﻠﻲ ﺗﺄﺛﯾر اﻷﺳﻌﺎر ﻋﻠﻲ اﻟﻣﺑﯾﻌﺎت ﻷﺣد‬ ‫اﻟﻣﻧﺗﺟﺎت اﻟﺻﻧﺎﻋﯾﺔ وﻟﻣﺎ ﻛﺎن ﺗﻐﯾﯾر اﻟﺳﻌر ﻋدة ﻣرات داﺧل ﻓرع واﺣد ﻗد ﯾﻛون ﻟﮫ ﺗﺄﺛﯾر ﺳﻠﺑﻲ ﻋﻠﻲ‬ ‫اﻟﻣﺳﺗﮭﻠﻛﯾن ﻓﻘد رؤى اﺳﺗﺧدام ﺳﻌر واﺣد داﺧل اﻟﻔرع اﻟواﺣد وان ﻛﺎن اﻟﺳﻌر ﻗد ﯾﺧﺗﻠف ﻣن ﻓرع‬ ‫ﻷﺧر‪ .‬ھذا وﻗد ﺗﻘرر إﺟراء اﻟدراﺳﺔ ﻟﻣدة ﺳﺗﺔ ﺷﮭور واﺷﺗرك ﻓﯾﮭﺎ ‪ 16‬ﻓرﻋﺎ‪ .‬وﻋﻣﻼ ﻋﻠﻲ ﺗﺧﻔﯾض‬ ‫اﻟﺧطﺄ اﻟﺗﺟرﯾﺑﻲ ﻓﻘد ﺗم اﺧﺗﯾﺎر اﻟﻔروع ﺑﺈﺣﺟﺎم ﻣﺑﯾﻌﺎت ﻣﺧﺗﻠﻔﺔ وﻓﻲ ﻣواﻗﻊ ﺟﻐراﻓﯾﮫ ﻣﺧﺗﻠﻔﺔ ھذا وﻗد ﺗم‬ ‫ﺗﺣدﯾد ﻣﺳﺗوﯾﺎت اﻷﺳﻌﺎر ﻋﻠﻲ اﻟﻧﺣو اﻟﺗﺎﻟﻲ‪:‬‬ ‫‪A :1.79$ , B :1.69$ , C :1.59$ , D :1.49$‬‬ ‫ﯾﻌطﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ اﻟﻣﺑﯾﻌﺎت ﺑﻣﺋﺎت اﻟدوﻻرات ﺧﻼل ﻓﺗرة اﻟﺳﺗﺔ ﺷﮭور ‪.‬‬ ‫اﻟﻣطﻠـوب ‪) :‬أ( أوﺟد ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﻟﮭذه اﻟﺗﺟرﺑﺔ‪.‬‬ ‫)ب( اﺧﺗﺑر ﻣﺎ إذا ﻛﺎﻧت ھﻧﺎك ﻓروق ﺑﯾن اﻟﻣﻌﺎﻟﺟﺎت )اﻟﻣﺑﯾﻌﺎت(‪.‬‬ ‫اﻟﻣوﻗﻊ اﻟﺟﻐراﻓـﻲ‬

‫‪Ti..‬‬

‫اﻟﺷﻣﺎل اﻟﺷرﻗﻲ‬ ‫‪5.3‬‬ ‫‪6.4‬‬ ‫‪9.8‬‬ ‫‪11.5‬‬ ‫‪33‬‬

‫‪B: 1.1‬‬ ‫‪A: 1.4‬‬ ‫‪C: 2.8‬‬ ‫‪D: 3.4‬‬ ‫‪8.7‬‬

‫اﻟﺷﻣﺎل‬ ‫اﻟﻐرﺑﻲ‬ ‫‪C: 1.5‬‬ ‫‪D: 1.9‬‬ ‫‪B: 2.2‬‬ ‫‪A: 2.5‬‬ ‫‪8.1‬‬

‫ﺣﺟم‬ ‫اﻟﻣﺑﯾﻌﺎت‬

‫اﻟﺟﻧوب‬ ‫اﻟﺷرﻗﻲ‬ ‫‪A: 1.0‬‬ ‫‪B: 1.6‬‬ ‫‪D: 2.7‬‬ ‫‪C: 2.9‬‬ ‫‪8.2‬‬

‫اﻟﺟﻧوب‬ ‫اﻟﻐرﺑﻲ‬ ‫‪D: 1.7‬‬ ‫‪C: 1.5‬‬ ‫‪A: 2.1‬‬ ‫‪B: 2.7‬‬ ‫‪8‬‬

‫اﻟﺣــل‪:‬‬ ‫ﻣن اﻟﺟدول اﻟﺳﺎﺑق ﯾﻣﻛن اﻟﺣﺻول ﻋﻠﻰ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫‪D‬‬ ‫‪1.7‬‬ ‫‪1.9‬‬ ‫‪2.7‬‬

‫‪B‬‬ ‫‪1.1‬‬ ‫‪1.6‬‬ ‫‪2.2‬‬

‫‪C‬‬ ‫‪1.5‬‬ ‫‪1.5‬‬ ‫‪2.8‬‬ ‫‪٥٢٩‬‬

‫‪A‬‬ ‫‪1.0‬‬ ‫‪1.4‬‬ ‫‪2.1‬‬

‫)‪ (١‬اﻷﻗل‬ ‫)‪(٢‬‬ ‫)‪(٣‬‬ ‫)‪ (٤‬اﻷﻛﺑر‬

‫‪T.j.‬‬

‫‪3.4‬‬ ‫‪9.7‬‬ ‫ﻓﯾﻣﺎ ﯾﻠﻰ ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ‪:‬‬ ‫‪ f‬اﻟﻣﺣﺳوﺑﺔ‬ ‫) ‪f  (1 ,  2‬‬

‫‪2.9‬‬ ‫‪8.7‬‬

‫ﻣﺗوﺳط‬ ‫اﻟﻣرﺑﻌﺎت‬

‫‪2.7‬‬ ‫‪7.6‬‬

‫ﻣﺟﻣوع اﻟﻣرﺑﻌﺎت‬ ‫‪6.28‬‬ ‫‪0.08‬‬ ‫‪1.08‬‬ ‫‪0.12‬‬ ‫‪7.56‬‬

‫‪f 0.01 (3,6)  9.78‬‬

‫‪18‬‬

‫‪0.36‬‬ ‫‪0.02‬‬

‫‪2.5‬‬ ‫‪Y..k 7.0‬‬

‫درﺟﺎت‬ ‫اﻟﺣرﯾﺔ‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪6‬‬ ‫‪15‬‬

‫ﻣﺻدر‬ ‫اﻻﺧﺗﻼف‬ ‫اﻟﺻﻔوف‬ ‫اﻻﻋﻣدة‬ ‫اﻟﻣﻌﺎﻟﺟﺎت‬ ‫اﻟﺧطﺄ‬ ‫اﻟﻛﻠﻰ‬

‫ﻣن اﻟﺟدول اﻟﺳﺎﺑق ﻓﺈن ﻗﯾﻣﺔ ‪ f‬اﻟﻣﺣﺳ وﺑﺔ أﻛﺑ ر ﻣ ن اﻟﻘﯾﻣ ﺔ اﻟﺟدوﻟﯾ ﺔ وھ ذا ﯾﻌﻧ ﻰ وﺟ ود ﻓ روق ﻣﻌﻧوﯾ ﺔ‬ ‫ﺑﯾن ﻣﺗوﺳطﺎت اﻟﻣﻌﺎﻟﺟﺎت ﻋﻧد ‪.   0.01‬‬ ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﻣﻛﺗوب ﺑﻠﻐﺔ ‪ Mathematica‬وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬ ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫‪=0.01‬‬ ‫‪0.01‬‬ ‫‪b={{1.7,1,1.5,1.1},{1.5,1.6,1.9,1.4},‬‬ ‫}}‪{2.1,2.7,2.2,2.8},{2.7,2.9,2.5,3.4‬‬ ‫‪{{1.7,1,1.5,1.1},{1.5,1.6,1.9,1.4},{2.1,2.7,2.2,2.8},{2.7,2.‬‬ ‫}}‪9,2.5,3.4‬‬ ‫]‪a=Transpose[b‬‬ ‫‪{{1.7,1.5,2.1,2.7},{1,1.6,2.7,2.9},{1.5,1.9,2.2,2.5},{1.1,1.‬‬ ‫}}‪4,2.8,3.4‬‬ ‫‪c={{1,1.4,2.1,2.5},{1.1,1.6,2.2,2.7},‬‬ ‫}}‪{1.5,1.5,2.8,2.9},{1.7,1.9,2.7,3.4‬‬ ‫‪{{1,1.4,2.1,2.5},{1.1,1.6,2.2,2.7},{1.5,1.5,2.8,2.9},{1.7,1.‬‬ ‫}}‪9,2.7,3.4‬‬ ‫]‪f[x_]:=Apply[Plus,x‬‬ ‫]‪g[x_]:=Length[x‬‬ ‫]‪x1=Map[f,a‬‬ ‫}‪{8.,8.2,8.1,8.7‬‬ ‫]‪xx1=Map[f,b‬‬ ‫}‪{5.3,6.4,9.8,11.5‬‬ ‫]‪xx2=Map[f,c‬‬ ‫}‪{7.,7.6,8.7,9.7‬‬ ‫]‪x2=Map[f,a^2‬‬ ‫}‪{16.84,19.26,16.95,22.57‬‬ ‫‪٥٣٠‬‬

n=g[a[[1]]] 4 w3=n*n 16 cf=(f[x1]^2)/w3 68.0625 st=f[x2]-cf 7.5575 f[x1^2]/n 68.135 rrw=%-cf 0.0725 f[xx1^2]/n 74.335 ccw=%-cf 6.2725 f[xx2^2]/n 69.135 sttw=%-cf 1.0725 se=st-rrw-ccw-sttw 0.14 w1=n-1 3 w3=w3-1 15 w4=w3-3w1 6 msw=sttw/w1 0.3575 msee=se/w4 0.0233333 f1=msw/msee 15.3214 rt2=List[" df "," ss "," mss { df , ss , mss , f } rt1=List[w1,sttw,msw,f1] {3,1.0725,0.3575,15.3214} rt3=List[w1,rrw,"-","-"] {3,0.0725,-,-} rt9=List[w1,ccw,"-","-"] {3,6.2725,-,-} rr9=List[w4,se,msee,"_"] {6,0.14,0.0233333,_} rr10=List[w3,st,"_","_"] {15,7.5575,_,_} ٥٣١

","

‫}}‪a11=TableHeadings->{{ S.V,treat,A,B,error,total},{ANOVA‬‬ ‫}}‪TableHeadings{{S.V,treat,A,B,error,total},{ANOVA‬‬ ‫]‪uu1=TableForm[{rt2,rt1,rt3,rt9,rr9,rr10},a11‬‬

‫‪f‬‬ ‫‪15.3214‬‬

‫‪mss‬‬ ‫‪0.3575‬‬

‫‪‬‬ ‫‪‬‬

‫_‬ ‫_‬

‫‪0.0233333‬‬ ‫_‬

‫‪ss‬‬ ‫‪1.0725‬‬ ‫‪0.0725‬‬ ‫‪6.2725‬‬ ‫‪0.14‬‬ ‫‪7.5575‬‬

‫‪ANOVA‬‬ ‫‪df‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪6‬‬ ‫‪15‬‬

‫‪S.V‬‬ ‫‪treat‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪error‬‬ ‫‪total‬‬

‫`‪<<Statistics`ContinuousDistributions‬‬ ‫]‪ff1=Quantile[FRatioDistribution[w1,w4],1-‬‬ ‫‪9.77954‬‬ ‫]]"‪If[f1>ff1,Print["RjectHo"],Print["AccpetHo‬‬ ‫‪RjectHo‬‬

‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ ‪:‬‬ ‫اوﻻ‪ :‬اﻟﻣدﺧﻼت‬ ‫ﻣﺴﺘﻮى اﻟﻤﻌﻨﻮﯾﺔ ﻣﻦ اﻻﻣﺮ‬ ‫‪=0.01‬‬ ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ ‫‪ b‬وھﻰ‬ ‫‪b={{1.7,1,1.5,1.1},{1.5,1.6,1.9,1.4},‬‬ ‫}}‪{2.1,2.7,2.2,2.8},{2.7,2.9,2.5,3.4‬‬ ‫وﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﯿﺎﻧﺎت اﻟﺨﺎﺻﺔ ﺑﻤﺜﺎل )‪.(١٥- ٦‬‬ ‫اﻟﻘﺎﺋﻤﺔ اﻟﻤﺴﻤﻰ‬ ‫‪ c‬وھﻰ‬ ‫‪c={{1,1.4,2.1,2.5},{1.1,1.6,2.2,2.7},‬‬ ‫}}‪{1.5,1.5,2.8,2.9},{1.7,1.9,2.7,3.4‬‬

‫وﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﯿﺎﻧﺎت اﻟﻤﻮﺟﻮدة ﻓﻰ اﻟﺠﺪول اﻟﺨﺎص ﺑﺎﻟﻤﻌﺎﻟﺠﺎت واﻟﻤﺸﺘﻖ ﻣﻦ اﻟﺠﺪول اﻻول‪.‬‬ ‫ﺛﺎﻧﯾﺎ ‪ :‬اﻟﻣﺧرﺟﺎت‬

‫ﺟدول ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﯾﺗم اﻟﺣﺻول ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ ‫]‪uu1=TableForm[{rt2,rt1,rt3,rt9,rr9,rr10},a11‬‬

‫ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪:‬‬

‫‪H 0 : 1   2  3‬‬ ‫ﺿد اﻟﻔرض اﻟﺑدﯾل‪:‬‬ ‫واﺣد ﻋﻠﻲ اﻷﻗل ﻣن ‪ i‬ﯾﺧﺗﻠف ﻋن اﻟﺑﺎﻗﻲ ‪H1 :‬‬ ‫‪ f‬اﻟﺟدوﻟﯾﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪ff1=Quantile[FRatioDistribution[w1,w4],1-‬‬ ‫‪٥٣٢‬‬

‫‪ f‬اﻟﻣﺣﺳوﺑﺔ ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر‬ ‫‪ff1=ssa/msser‬‬

‫واﻟﺗﻰ ﺗﺧﺗﻠف ﻗﻠﯾﻼ ﻋن اﻟﻣﺣﺳوﺑﺔ ﯾدوﯾﺎ وذﻟك ﻧﺗﯾﺟﺔ ﻟﻌﻣﻠﯾﺔ اﻟﺗﻘرﯾب‪.‬‬ ‫اﻟﻘرار اﻟذى ﯾﺗﺧذ ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر‬ ‫]]"‪If[f1>ff1,Print["RjectHo"],Print["AccpetHo‬‬

‫‪Reject H0‬‬

‫اى رﻓض ﻓرض اﻟﻌدم‬ ‫ﻣﻠﺣوظﺔ ‪ B :‬ھﻰ اﻟﺻﻔوف و ‪ A‬ھﻰ اﻻﻋﻣدة‪.‬‬

‫‪٥٣٣‬‬

‫اﻟﻔﺻل اﻟﺳﺎﺑﻊ‬ ‫اﻻﺧﺗﺑﺎرات اﻟﻼﻣﻌﻠﻣﯾﺔ‬

‫‪٥٣٤‬‬

‫)‪ (١-٧‬ﻣﻘدﻣــﺔ‬

‫‪Introduction‬‬

‫ﺗﻌﺗﻣد اﻟطرق اﻟﻣﺳﺗﺧدﻣﺔ ﻓﻲ اﺧﺗﺑﺎرات اﻟﻔروض وﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن وﺗﺣﻠﯾل اﻻﻧﺣدار وﺗﺣﻠﯾل‬ ‫اﻻرﺗﺑﺎط ) اﻟطرق اﻟﻣﻌﻠﻣﯾﺔ ‪ ،( parametric methods‬واﻟﺗﻲ ﺳﺑق ﻣﻧﺎﻗﺷﺗﮭﺎ ﻓﻲ اﻟﻔﺻول اﻟﺳﺎﺑﻘﺔ‪،‬‬ ‫ﻋﻠﻰ ﻋدد ﻣن اﻟﻔروض ‪ .‬ﻓﻌﻠﻰ ﺳﺑﯾل اﻟﻣﺛﺎل ﯾﺷﺗرط ﻓﻲ اﻻﺧﺗﺑﺎر اﻟذي ﯾﺧص ﻣﺗوﺳط ﻣﺟﺗﻣﻊ )‬ ‫اﺧﺗﺑﺎر ‪( t‬أن اﻟﻣﺟﺗﻣﻊ اﻟذي اﺧﺗﯾرت ﻣﻧﮫ اﻟﻌﯾﻧﺔ ﯾﺗﺑﻊ ﺗوزﯾﻌﺎ ً طﺑﯾﻌﯾﺎ ً وذﻟك ﻋﻧدﻣﺎ ﯾﻛون ﺗﺑﺎﯾن اﻟﻣﺟﺗﻣﻊ‬ ‫ﻏﯾر ﻣﻌروف وﺣﺟم اﻟﻌﯾﻧﺔ ﺻﻐﯾر ‪ .‬أﯾﺿﺎ ﻓﻲ ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﯾﺷﺗرط أن اﻟﻣﺟﺗﻣﻌﺎت اﻟﺗﻲ اﺧﺗﯾرت‬ ‫ﻣﻧﮭﺎ اﻟﻌﯾﻧﺎت ﺗﺗﺑﻊ ﺗوزﯾﻌﺎت طﺑﯾﻌﯾﺔ وﺗﺑﺎﯾﻧﺎﺗﮭﺎ ﻣﺗﺳﺎوﯾﺔ ‪ .‬اﻻﺧﺗﺑﺎرات اﻟﺳﺎﺑﻘﺔ ﺳوف ﺗﻛون ﻏﯾر ﻣﺟدﯾﺔ‬ ‫إذا ﻟم ﺗﺗﺣﻘق اﻟﺷروط اﻟﺧﺎﺻﺔ ﺑﮭﺎ ‪ ،‬وﻓﻲ ھذه اﻟﺣﺎﻟﺔ ﺗﻛون اﻻﺧﺗﺑﺎرات اﻟﻼﻣﻌﻠﻣﯾﺔ ھﻲ اﻟﺑدﯾل ‪ .‬ﻋﻣوﻣﺎ ً‬ ‫ﺗﺳﺗﺧدم اﻻﺧﺗﺑﺎرات اﻟﻼﻣﻌﻠﻣﯾﺔ ﻓﻲ اﻷﺣوال اﻵﺗﯾﺔ ‪:‬‬ ‫ﻋﻧدﻣﺎ ﺗﻛون اﻟﺷروط اﻟﻼزﻣﺔ ﻟﻼﺧﺗﺑﺎر اﻟﻣﻌﻠﻣﻲ ﻏﯾر ﻣﺳﺗوﻓﺎة ‪.‬‬ ‫)أ(‬ ‫)ب( ﻋﻧدﻣﺎ ﯾدور ﻓرض اﻟﻌدم واﻟﻔرض اﻟﺑدﯾل ﻋﻠﻰ أﺷﯾﺎء وﺻﻔﯾﺔ وﻟﯾﺳت ﻋﻠﻰ ﻣﻌﻠﻣﺔ ﻣﺟﺗﻣﻊ‬ ‫ﻛﻣﺎ ﻓﻲ اﺧﺗﺑﺎر ﺟودة اﻟﺗوﻓﯾق واﻟذي ﺳوف ﺗﺗﻧﺎوﻟﮫ ﻓﻲ اﻟﺑﻧد اﻟﺗﺎﻟﻲ ‪ .‬أﯾﺿﺎ ﻋﻧد اﻟرﻏﺑﺔ ﻓﻲ‬ ‫ﻋﻣل ﻣﻘﺎرﻧﺔ ﺑﯾن ﻣﺟﺗﻣﻌﯾن أو اﻛﺛر وذﻟك ﺑﺎﻻﻋﺗﻣﺎد ﻋﻠﻰ ﻋﯾﻧﺎت ﻋﺷواﺋﯾﺔ ﻣﺧﺗﺎرة ﻣن ھذه‬ ‫اﻟﻣﺟﺗﻣﻌﺎت دون اﻟﺗﻌرف ﻋﻠﻰ اﻟﺗوزﯾﻌﺎت اﻻﺣﺗﻣﺎﻟﯾﺔ أو اﻟﺗﻌرض ﻟﮭﺎ ‪.‬‬ ‫)ج( ﻋﻧدﻣﺎ ﯾﻛون ﻣﻘﯾﺎس اﻟﺑﯾﺎﻧﺎت وﺻﻔﻲ أو ﺗرﺗﯾﺑﻲ ‪.‬‬ ‫)د( ﻋﻧد اﻟﺣﺎﺟﺔ ﻟﻠوﺻول إﻟﻰ ﻗرار ﺳرﯾﻊ ﺑدون اﺳﺗﺧدام اﻵﻻت اﻟﺣﺎﺳﺑﺔ أو اﻟﺣﺎﺳﺑﺎت اﻹﻟﻛﺗروﻧﯾﺔ‬ ‫‪.‬‬ ‫ﺑﻌض ﻣﻣﯾزات اﻟطرق اﻟﻼﻣﻌﻠﻣﯾﺔ‬ ‫)أ( ﻓرص ﻋدم اﻟدﻗﺔ ﻗﻠﯾﻠﺔ ﻋﻧد ﺗطﺑﯾﻘﮭﺎ وذﻟك ﻻﻋﺗﻣﺎدھﺎ ﻋﻠﻰ اﻗل ﻗدر ﻣن اﻟﻔروض ‪.‬‬ ‫)ب( ﻟﺑﻌض اﻟطرق اﻟﻼﻣﻌﻠﻣﯾﺔ ‪ ،‬اﻟﻌﻣﻠﯾﺎت ﺗﺗم ﺑﺳرﻋﺔ ﻟذﻟك ﻓﺈﻧﮭﺎ ﺗوﻓر اﻟوﻗت وﺧﺻوﺻﺎ ً ﻋﻧد ﻋدم‬ ‫ﺗوﻓر اﻵﻻت اﻟﺣﺎﺳﺑﺔ ‪.‬‬ ‫)ج( ﺗﻼﺋم ﻛﺛﯾر ﻣن اﻟﺑﺎﺣﺛﯾن ﻓﻲ ﻣﺟﺎل ﻋﻠم اﻟﻧﻔس واﻻﺟﺗﻣﺎع ﻷن ﻣﻌظم اﻟﺑﯾﺎﻧﺎت اﻟذﯾن ﯾﺗﻌﺎﻣﻠون‬ ‫ﻣﻌﮭﺎ ﺗﻛون ﻣن اﻟﻧوع اﻟوﺻﻔﻲ أو اﻟﺗرﺗﯾﺑﻲ ‪.‬‬ ‫)د( ﺗﻼﺋم اﻟﺑﺎﺣﺛﯾن اﻟذﯾن ﻟدﯾﮭم أدﻧﻰ ﻣﻌﻠوﻣﺎت ﻓﻲ ﻣﺟﺎل اﻟرﯾﺎﺿﯾﺎت واﻹﺣﺻﺎء وذﻟك ﻟﺳﮭوﻟﺔ‬ ‫اﻟﻣﻔﺎھﯾم واﻟطرق اﻟﻼﻣﻌﻠﻣﯾﺔ ‪.‬‬ ‫ﺑﻌض ﻋﯾوب اﻟطرق اﻟﻼﻣﻌﻠﻣﯾﮫ‬ ‫)أ( وﻻن اﻟﻌﻣﻠﯾﺎت اﻟﻣﺳﺗﺧدﻣﺔ ﻓﻲ ﻣﻌظم اﻻﺧﺗﺑﺎرات اﻟﻼﻣﻌﻠﻣﯾﺔ ﺑﺳﯾطﺔ وﺳرﯾﻌﺔ ﻓﺈﻧﮭﺎ ﺗؤدى إﻟﻰ ﻓﻘد‬ ‫ﻓﻲ اﻟﻣﻌﻠوﻣﺎت اﻟﻣوﺟودة ﻓﻲ اﻟﺑﯾﺎﻧﺎت ﻛﻣﺎ ھو اﻟﺣﺎل ﻋﻧد ﺗﺣوﯾل اﻟﺑﯾﺎﻧﺎت إﻟﻰ رﺗب ‪ ،‬وھذا‬ ‫ﯾؤدى إﻟﻰ ﻓﻘد ﻛﺑﯾر ﻓﻲ اﻟدﻗﺔ ‪.‬‬ ‫)ب( ﺑﻌض اﻟطرق اﻟﻼﻣﻌﻠﻣﯾﺔ ﺗﻛون ﻣﻌﻘدة ‪.‬‬

‫)‪ (١-٧‬اﺧﺗﺑﺎر ﻣرﺑﻊ ﻛﺎى ﻟﻼﺳﺗﻘﻼل‬ ‫‪٥٣٥‬‬

‫‪The Chi-square Test of Independent‬‬ ‫ﻓﻲ ﻛﺛﯾر ﻣن اﻷﺣﯾﺎن ﯾرﻏب اﻟﺑﺎﺣث ﻓﻲ اﻟﺗﻌرف ﻋﻣﺎ إذا ﻛﺎﻧت ھﻧ ﺎك ﻋﻼﻗ ﺔ ﺑ ﯾن ﺻ ﻔﺗﯾن ﻣ ن‬ ‫ﺻﻔﺎت ﻣﺟﺗﻣﻊ ﻣﺎ‪ .‬ﻓﻌﻠﻲ ﺳﺑﯾل اﻟﻣﺛ ﺎل ﻗ د ﯾرﻏ ب ﻣﺳ ﺋول اﻟﺗﻐذﯾ ﺔ ﻓ ﻲ ﻣدرﺳ ﺔ ﻣ ﺎ ﻓ ﻲ اﻟﺗﻌ رف ﻋﻣ ﺎ إذا‬ ‫ﻛﺎﻧت اﻟﺣﺎﻟﺔ اﻟﻐذاﺋﯾﺔ ﻟﻠطﺎﻟب ﻟﮭﺎ ﻋﻼﻗﺔ ﺑﻛﻔﺎءﺗﮫ اﻟﺗﻌﻠﯾﻣﯾ ﺔ ‪ .‬أﯾﺿ ﺎ ﻗ د ﯾرﻏ ب ﺑﺎﺣ ث ﻓ ﻲ ﻣﺟ ﺎل اﻟوراﺛ ﺔ‬ ‫ﻓﻲ اﻟﺗﻌرف ﻋﻣﺎ إذا ﻛﺎﻧت ھﻧﺎك ﻋﻼﻗﺔ ﺑﯾن ﻟون اﻟﺷﻌر وﻟون اﻟﻌﯾﻧﯾن … اﻟﺦ ‪.‬‬ ‫ﻹﺟ راء اﻻﺧﺗﺑ ﺎر ﻧﺧﺗ ﺎر ﻋﯾﻧ ﺔ ﻋﺷ واﺋﯾﺔ ﻣ ن اﻟﺣﺟ م ‪ n‬ﻣ ن اﻟﻣﺟﺗﻣ ﻊ ﻣوﺿ ﻊ اﻟدراﺳ ﺔ ‪.‬ﺗﺻ ﻧف‬ ‫ﻣﺷﺎھدات ھ ذه اﻟﻌﯾﻧ ﺔ ﺣﺳ ب ﻣﺳ ﺗوﯾﺎت ﻛ ل ﻣ ن اﻟﺻ ﻔﺗﯾن ﻣوﺿ ﻊ اﻟدراﺳ ﺔ ﻓ ﻲ ﺟ دول ﻣ زدوج ﯾﺳ ﻣﻰ‬ ‫ﺟدول اﻟﺗواﻓ ق ‪ . contingency table‬ﺑﻔ رض أن ‪ A1, A 2 ,..., A k‬ﺗرﻣ ز ﻟﻣﺳ ﺗوﯾﺎت اﻟﺻ ﻔﺔ ‪ A‬و‬ ‫‪ B1 , B2 ,..., B k‬ﺗرﻣ ز ﻟﻣﺳ ﺗوﯾﺎت اﻟﺻ ﻔﺔ ‪ B‬ﻓ ﺈن ﺟ دول اﻟﺗواﻓ ق ﯾﻛ ون ﻋﻠ ﻰ اﻟﺷ ﻛل اﻟﻣوﺿ ﺢ ﻓ ﻲ‬ ‫اﻟﺟدول اﻟﺗﺎﻟﻰ ‪ ،‬ﺣﯾث أن ‪ Oij‬ﺗرﻣز ﻟﻌدد اﻟﻣﺷﺎھدات اﻟﺗﻲ ﯾﺗ وﻓر ﻓﯾﮭ ﺎ اﻟﻣﺳ ﺗوى ‪ A i‬ﻣ ن اﻟﺻ ﻔﺔ ‪ A‬و‬ ‫اﻟﻣﺳﺗوى ‪ B j‬ﻣن اﻟﺻﻔﺔ ‪ B‬ﺣﯾ ث ‪ i  1,2,...,r‬و ‪ . j  1,2,...,c‬أﯾﺿ ﺎ ‪ ni‬ﺗرﻣ ز ﻟﻌ دد اﻟﻣﺷ ﺎھدات‬ ‫‪c‬‬

‫اﻟﺗ ﻲ ﯾﺗ وﻓر ﻓﯾﮭ ﺎ اﻟﻣﺳ ﺗوى ‪ A i‬ﻣ ن اﻟﺻ ﻔﺔ ‪ A‬أي أن ‪ . n i.   Oij‬أﯾﺿ ﺎ ‪ n. j‬ﺗرﻣ ز ﻟﻌ دد‬ ‫‪j1‬‬

‫‪r‬‬

‫اﻟﻣﺷﺎھدات اﻟﺗﻲ ﯾﺗوﻓر ﻓﯾﮭﺎ اﻟﻣﺳﺗوى ‪ B j‬ﻣن اﻟﺻﻔﺔ ‪ B‬أي أن ‪ . n j.   Oij‬وﻋﻠﻰ ذﻟك ‪:‬‬ ‫‪i 1‬‬

‫‪r‬‬

‫‪c‬‬

‫‪i 1‬‬

‫‪j1‬‬

‫‪ n.j   ni.    Oij .‬‬ ‫‪j1 i 1‬‬

‫ﯾﺣﺗوي ﺟدول اﻟﺗواﻓق ﻋﻠﻰ ﺧﺎﻧﺎت ) ﺧﻼﯾﺎ ( ﻋددھﺎ ) ‪ (r x c‬ﺧﻠﯾﺔ‪.‬‬ ‫اﻟﻣﺟﻣوع‬ ‫‪n1.‬‬

‫‪n 2.‬‬ ‫‪‬‬ ‫‪n r.‬‬

‫‪… Bc‬‬ ‫‪... O1c‬‬

‫‪B2‬‬

‫‪B1‬‬

‫‪O11 O12‬‬ ‫‪O 21 O 22 ... O 2c‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪O r 2 ... O rc‬‬

‫‪‬‬ ‫‪O r1‬‬

‫‪n 2.‬‬

‫‪n1.‬‬

‫‪nc.‬‬

‫‪A1‬‬ ‫‪A2‬‬ ‫‪‬‬ ‫‪Ar‬‬

‫‪n‬‬ ‫ﻓرض اﻟﻌدم واﻟﻔرض اﻟﺑدﯾل ﺳوف ﯾﻛوﻧﺎن ﻋﻠﻰ اﻟﺷﻛل ‪:‬‬ ‫‪ : H 0‬اﻟﻣﺗﻐﯾرﯾن ﻣﺳﺗﻘﻠﯾن ‪.‬‬ ‫‪ : H1‬اﻟﻣﺗﻐﯾرﯾن ﻏﯾر ﻣﺳﺗﻘﻠﯾن ‪.‬‬ ‫ﯾﻌﺗﻣد اﺧﺗﺑﺎر ﻣرﺑ ﻊ ﻛ ﺎي ﻟﻼﺳ ﺗﻘﻼل ﻋﻠ ﻰ ﻣﻘﺎرﻧ ﺔ اﻟﺗﻛ رارات اﻟﻣﺷ ﺎھدة ﺑ ﺎﻟﺗﻛرارات اﻟﻣﺗوﻗﻌ ﺔ‬ ‫ﻓﻲ ﻛل ﺧﻠﯾﺔ ﻋﻧدﻣﺎ ‪ H 0‬ﺻﺣﯾﺢ ‪.‬‬ ‫وﯾﻣﻛن ﺣﺳﺎب اﻟﺗﻛرارات اﻟﻣﺗوﻗﻌﺔ ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬ ‫‪n i.‬‬ ‫‪n .j‬‬ ‫‪E ij ‬‬ ‫‪,i  1, 2,..., r, ; j  1,2,...,c.‬‬ ‫‪n‬‬ ‫‪٥٣٦‬‬

‫‪...‬‬

‫ﺑﺎﻓﺗراض أن ‪ H0‬ﺻﺣﯾﺢ ﻓﺈن ‪:‬‬

‫‪.‬‬

‫‪(Oij  Eij ) 2‬‬ ‫‪Eij‬‬

‫‪c‬‬

‫‪r‬‬

‫‪2‬‬

‫‪  ‬‬ ‫‪j1 i 1‬‬

‫ﻗﯾﻣ ﺔ ﻟﻣﺗﻐﯾ ر ﻋﺷ واﺋﻲ ‪ X2‬ﺗﻘرﯾﺑ ﺎ ﯾﺗﺑ ﻊ ﺗوزﯾ ﻊ ‪ 2‬ﺑ درﺟﺎت ﺣرﯾ ﺔ )‪ (r  1)(c  1‬ﺣﯾ ث ‪ r‬ﻋ دد‬

‫اﻟﺻﻔوف و ‪ c‬ﻋدد اﻷﻋﻣدة ﻓﻲ ﺟ دول اﻟﺗواﻓ ق ‪ .‬ﻟﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ ‪ ‬ﻓ ﺈن ﻣﻧطﻘ ﺔ اﻟ رﻓض ‪X2  2‬‬ ‫‪2‬‬ ‫ﺣﯾ ث ‪ ‬ﺗﺳ ﺗﺧرج ﻣ ن ﺟ دول ﺗوزﯾ ﻊ ‪ 2‬ﻓ ﻲ ﻣﻠﺣ ق )‪ (٣‬ﺑ درﺟﺎت ﺣرﯾ ﺔ )‪ . (r  1)(c  1‬إذا‬

‫وﻗﻌت ‪ 2‬ﻓﻲ ﻣﻧطﻘﺔ اﻟرﻓض ﻧرﻓض ‪. H0‬‬ ‫وﯾﺟب ان ﯾﻛون ﻋدد اﻟﻣﺷﺎھدات ﻓﻰ ﻛل ﺧﻠﯾﺔ ﻻ ﯾﻘل ﻋن ‪.5‬‬

‫ﻣﺛﺎل)‪(١-٧‬‬ ‫ﻟدراﺳﺔ اﻟﻌﻼﻗ ﺔ ﺑ ﯾن ﻟ ون ﺷ ﻌر اﻟ زوج واﻟزوﺟ ﺔ ﻗ ﺎم ﺑﺎﺣ ث ﺑﺈﺧﺗﯾ ﺎر ﻋﯾﻧ ﺔ ﻋﺷ واﺋﯾﺔ ﻣ ن اﻟﺣﺟ م‬ ‫‪) 500‬زوج وزوﺟﺔ( وﺗم ﺳؤاﻟﮭم واﻟﺑﯾﺎﻧﺎت ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫اﻟﻣﺟﻣوع‬ ‫‪50‬‬ ‫‪150‬‬ ‫‪150‬‬ ‫‪150‬‬ ‫‪500‬‬

‫اﻟزوج‬ ‫ﺑﻧﻰ‬ ‫‪20‬‬ ‫‪50‬‬ ‫‪52‬‬ ‫‪78‬‬ ‫‪200‬‬

‫أﺳود‬ ‫‪10‬‬ ‫‪50‬‬ ‫‪60‬‬ ‫‪30‬‬ ‫‪150‬‬

‫اﻟزوﺟﺔ‬ ‫أﺻﻔر‬ ‫‪10‬‬ ‫‪40‬‬ ‫‪25‬‬ ‫‪25‬‬ ‫‪100‬‬

‫أﺣﻣر‬ ‫‪10‬‬ ‫‪10‬‬ ‫‪13‬‬ ‫‪17‬‬ ‫‪50‬‬

‫أﺣﻣر‬ ‫اﺻﻔر‬ ‫اﺳود‬ ‫ﺑﻧﻰ‬ ‫اﻟﻣﺟﻣوع‬

‫اﻟﻣطﻠ وب اﺧﺗﺑ ﺎر ﻣ ﺎ إذا ﻛﺎﻧ ت ھﻧ ﺎك ﻋﻼﻗ ﺔ ﺑ ﯾن ﻟ ون ﺷ ﻌر اﻟ زوج واﻟزوﺟ ﺔ أم ﻻ ؟ وذﻟ ك ﻋﻧ د‬ ‫ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪.   0.05‬‬

‫اﻟﺣــل‪:‬‬ ‫‪ : H0‬ﻟون ﺷﻌر اﻟزوج وﻟون ﺷﻌر اﻟزوﺟﺔ ﻣﺳﺗﻘﻠﯾن‪.‬‬ ‫‪ : H1‬ﻟون ﺷﻌر اﻟزوج وﻟون ﺷﻌر اﻟزوﺟﺔ ﻏﯾر ﻣﺳﺗﻘﻠﯾن‪.‬‬ ‫اﻟﺗﻛرارات اﻟﻣﺗوﻗﻌﺔ ﻣﻌطﺎة ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬

‫‪٥٣٧‬‬

‫اﻟزوﺟﺔ‬ ‫أﺣﻣر‬ ‫اﺻﻔر‬ ‫اﺳود‬ ‫ﺑﻧﻰ‬ ‫اﻟﻣﺟﻣوع‬

‫اﻟزوج‬ ‫أﺣﻣر‬ 5 15 15 15 50

‫أﺻﻔر‬ 10 30 30 30 100 2

  

‫اﻟﻣﺟﻣوع‬ ‫أﺳود‬ 15 45 45 45 150

‫ﺑﻧﻰ‬ 20 50 60 150 60 150 60 150 200 500 : ‫ﻣن اﻟﺟدوﻟﯾن اﻟﺳﺎﺑﻘﯾن ﻓﺈن‬

(Oij  Eij )2

j1 i 1

E ij

 32.56. 2  16.919 ‫ ﻓﺈن‬  0.05 ‫ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ‬ ) ‫ ﻓﻲ ﻣﻠﺣ ق‬2 ‫ واﻟﻣﺳﺗﺧرﺟﺔ ﻣن ﺟدول ﺗوزﯾﻊ‬.05

‫ اﻟﻣﺣﺳ وﺑﺔ ﺗﻘ ﻊ ﻓ ﻲ‬ 2 ‫ وﺑﻣﺎ أن‬X2 > 16.919 ‫ ﻣﻧطﻘﺔ اﻟرﻓض‬. 3 × 3 = 9 ‫ ( ﺑدرﺟﺎت ﺣرﯾﺔ‬٣ . H0 ‫ﻣﻧطﻘﺔ اﻟرﻓض ﻓﺈﻧﻧﺎ ﻧرﻓض‬ ‫ وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬Mathematica ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ‬ . ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت‬ Off[General::spell1]; <<Statistics`DataManipulation` <<Statistics`NormalDistribution` Options[npmChiSquareTest]={mthd->chiSquare}; npmChiSquareTest[freqList_,opts___]:=Module[{c,r,rVals,cVals ,n,fHat,fHatTable,sqDiff,cs,pVal}, mtype=mthd/. {opts} /. Options[npmChiSquareTest]; c=Length[freqList[[1]]]; r=Length[freqList]; rowTotal[i_]:=Sum[freqList[[i,j]],{j,1,c}]; rVals=Table[rowTotal[i],{i,1,r}]; colTotal[j_]:=Sum[freqList[[i,j]],{i,1,r}]; cVals=Table[colTotal[j],{j,1,c}]; n=Apply[Plus,rVals]; fHat[i_,j_]:=rVals[[i]] cVals[[j]]/n; fHatTable=Table[fHat[i,j],{i,1,r},{j,1,c}]//N; sqDiff=Table[(freqList[[i,j]]fHatTable[[i,j]])^2/fHatTable[[i,j]],{i,1,r},{j,1,c}]; cs=Apply[Plus,Flatten[sqDiff]]; pVal=1-CDF[ChiSquareDistribution[(r-1)(c-1)],cs]; ٥٣٨

Print["Title: Chi Square Test"]; Print["Distribution: ChiSquare[",(r-1)(c-1),"]"]; Print["PValue: ",pVal]; If[mtype==ExpectedFrequencies,Print[TableForm[fHatTable ]]]; lessThanFiveQ[x_]:=If[x<5,True,False]; tfTable=Table[lessThanFiveQ[fHatTable[[i,j]]],{i,1,r},{ j,1,c}]; If[MemberQ[Flatten[tfTable],True]==True,Print[TableForm [fHatTable]]]; ] npmChiSquareTest[{{10,10,10,20},{10,40,50,50},{13,25,60,52}, {17,25,30,78}},mthd->ExpectedFrequencies] Title: Chi Square Test Distribution: ChiSquare[ 9 ] PValue: 0.000159554

5. 15. 15. 15.

10. 30. 30. 30.

15. 45. 45. 45.

20. 60. 60. 60.

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ﺗﺪﺧﻞ ﻣﺤﺘﻮﯾﺎت ﺟﺪول اﻟﺘﻮاﻓﻖ ﻣﻦ ﺧﻼل اﻻﻣﺮ اﻟﺘﺎﻟﻰ‬ npmChiSquareTest[{{10,10,10,20},{10,40,50,50},{13,25,60,52}, {17,25,30,78}},mthd->ExpectedFrequencies] ‫ﺣﯿﺚ ﺗﺪﺧﻞ ﺻﻒ ﺻﻒ واﻟﻤﺨﺮج ﻟﮭﺬا اﻻﻣﺮ ھﻮ‬ Title: Chi Square Test Distribution: ChiSquare[ 9 ] PValue: 0.000159554

5. 15. 15. 15.

10. 30. 30. 30.

15. 45. 45. 45.

20. 60. 60. 60.

9 ‫واﻟﺬى ﯾﻮﺿﺢ ان اﺣﺼﺎء ھﺬا اﻻﺧﺘﺒﺎر ﯾﺘﺒﻊ ﺗﻮزﯾﻊ ﻣﺮﺑﻊ ﻛﺎى ﺑدرﺟﺎت ﺣرﯾﺔ‬ ‫ﻛﻤﺎ ﯾﺤﺘﻮى اﻟﻤﺨﺮج ﻋﻠﻰ ﻗﯿﻤﺔ‬ ‫وھﻰ‬ p-value

PValue:

0.000159554

‫اﯾﺿ ﺎ ﻣ ن اﻻﻣ ر اﻟﺳ ﺎﺑق ﯾ ﺗم اﻟﺣﺻ ول ﻋﻠ ﻰ‬. ‫ ﻧ رﻓض ﻓ رض اﻟﻌ دم‬  0.05 ‫وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻗل ﻣن‬ ‫اﻟﺗﻛرارات اﻟﻣﺗوﻗﻌﺔ وھﻰ‬

٥٣٩

‫‪15.‬‬ ‫‪45.‬‬ ‫‪45.‬‬ ‫‪45.‬‬

‫‪20.‬‬ ‫‪60.‬‬ ‫‪60.‬‬ ‫‪60.‬‬

‫‪5.‬‬ ‫‪15.‬‬ ‫‪15.‬‬ ‫‪15.‬‬

‫‪10.‬‬ ‫‪30.‬‬ ‫‪30.‬‬ ‫‪30.‬‬

‫إذا ﻛﺎن ﻟﻛل ﻣن اﻟﺻﻔﺗﯾن ‪ A , B‬ﻣﺳﺗوﯾﺎن ﻓﻘط ﻓﺈن اﻟﺟدول اﻟﻧ ﺎﺗﺞ ﯾﺗﻛ ون ﻣ ن ﺻ ﻔﯾن وﻋﻣ ودﯾن )أي‬ ‫أرﺑ ﻊ ﺧﻼﯾ ﺎ (‪ .‬ﯾﺳ ﻣﻰ اﻟﺟ دول اﻟﻧ ﺎﺗﺞ ﺟ دول اﻻﻗﺗ ران)‪ .(2×2‬اﻟﺟ دول اﻟﺗ ﺎﻟﻰ ﯾﻣﺛ ل ﺟ دول اﻗﺗران‪.‬‬ ‫ﻋدد درﺟﺎت اﻟﺣرﯾﺔ اﻟﺗﻲ ﺗرﺗﺑط ﺑﺟدول اﻻﻗﺗران ﺳوف ﺗﺳﺎوى اﻟواﺣد اﻟﺻﺣﯾﺢ‪.‬‬

‫اﻟﺻﻔﺔ اﻷوﻟﻰ‬

‫اﻟﺻﻔﺔ اﻟﺛﺎﻧﯾﺔ‬ ‫‪B2‬‬

‫‪B1‬‬

‫‪ab‬‬

‫‪b‬‬

‫‪a‬‬

‫‪A1‬‬

‫‪cd‬‬

‫‪d‬‬

‫‪c‬‬

‫‪A2‬‬

‫‪n‬‬

‫‪bd‬‬

‫ﯾﻣﻛن اﺳﺗﺧدام ﺻﯾﻐﺔ ﺑﺳﯾطﺔ ﻟﺣﺳﺎب ﻗﯾﻣﺔ‬

‫‪a c‬‬ ‫‪2‬‬

‫‪ ‬ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬

‫‪n(ad  bc)2‬‬ ‫‪.‬‬ ‫‪ ‬‬ ‫)‪(a  c)(b  d)(c  d)(a  b‬‬ ‫‪2‬‬

‫ﻣﺛﺎل)‪(٢-٧‬‬ ‫ﻟدراﺳﺔ اﻟﻌﻼﻗﺔ ﺑﯾن اﻟﻧ وم ﻟ ﯾﻼ ً واﻟﺗ دﺧﯾن اﺧﺗﯾ رت ﻋﯾﻧ ﺔ ﻋﺷ واﺋﯾﺔ ﻣ ن ‪ 56‬ﺷﺧﺻ ﺎ ً واﻟﺑﯾﺎﻧ ﺎت‬ ‫ﻣﻌطﺎة ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫اﻟﺗدﺧﯾن‬ ‫اﻟﻧــوم‬ ‫اﻟﻣﺟﻣوع‬ ‫ﻻ‬

‫ﻧﻌم‬

‫‪36‬‬

‫‪16‬‬

‫‪20‬‬

‫‪20‬‬ ‫‪56‬‬

‫‪14‬‬ ‫‪30‬‬

‫‪6‬‬ ‫‪26‬‬

‫ﻧﻌم‬ ‫ﻻ‬ ‫اﻟﻣﺟﻣوع‬

‫اﻟﻣطﻠوب اﺧﺗﺑﺎر ﻣﺎ إذا ﻛﺎﻧت اﻟﻧوم ﻟﯾﻼ ً واﻟﺗدﺧﯾن ﻣﺳﺗﻘﻠﯾن ام ﻻ وذﻟك ﻋﻧ د ﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ‬ ‫‪.   0.05‬‬

‫‪٥٤٠‬‬

‫اﻟﺣــل‪:‬‬ ‫‪ : H 0‬اﻟﻣﺗﻐﯾرﯾن ﻣﺳﺗﻘﻠﯾن ‪.‬‬ ‫‪ : H1‬اﻟﻣﺗﻐﯾرﯾن ﻏﯾر ﻣﺳﺗﻘﻠﯾن ‪.‬‬ ‫ﻣن اﻟﺟدول اﻟﺳﺎﺑق ﻓﺈن ‪:‬‬ ‫‪n(ad  bc)2‬‬ ‫‪ ‬‬ ‫)‪(a  c)(b  d)(c  d)(a  b‬‬ ‫‪2‬‬

‫‪56[(20)(14)  (16)(6)]2‬‬ ‫‪‬‬ ‫‪ 3.376 .‬‬ ‫)‪(26)(30)(20)(36‬‬ ‫‪2‬‬ ‫ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪   0.05‬ﻓﺈن ‪ 3.843‬‬ ‫‪ .05‬واﻟﻣﺳﺗﺧرﺟﺔ ﻣن ﺟدول ﺗوزﯾﻊ ‪ 2‬ﻓ ﻲ ﻣﻠﺣ ق ) ‪٣‬‬

‫( ﺑدرﺟﺎت ﺣرﯾﺔ واﺣدة ‪ .‬ﻣﻧطﻘﺔ اﻟرﻓض ‪ . X2 > 3.843‬وﺑﻣﺎ أن ‪  2‬ﺗﻘﻊ ﻓ ﻲ ﻣﻧطﻘ ﺔ اﻟﻘﺑ ول ﻧﻘﺑ ل‬ ‫‪. H0‬‬ ‫ﺳﺑق أن ذﻛرﻧﺎ أن اﻟﺗﻛرارات اﻟﻣﺗوﻗﻌﺔ ﻓﻲ ﻛل ﺧﻠﯾﺔ ﯾﺟ ب أن ﻻ ﯾﻘ ل ﻋ ن ‪ 5‬وإذا ﺣ دث وﻛ ﺎن‬ ‫أﺣد اﻟﺗﻛرارات اﻟﻣﺗوﻗﻌﺔ أﻗل ﻣن ‪ 5‬ﻓﺈﻧﻧﺎ ﻧﻘوم ﺑدﻣﺞ اﻟﺗﻛ رارات ‪ .‬وﻋﻠ ﻰ أي ﺣ ﺎل ﻓ ﺈن ھ ذه اﻟطرﯾﻘ ﺔ ﻻ‬ ‫ﺗﺳﺗﺧدم ﻓﻲ ﺣﺎﻟﺔ ﺟدول اﻻﻗﺗران ‪ .‬وﻗد أﻗﺗرح )‪ Yates (1934‬ﺗﺻﺣﯾﺣﺎ ً ﯾﺳﺗﺧدم ﻓﻲ ﺣﺎﻟ ﺔ ﻣ ﺎ إذا ﻛ ﺎن‬ ‫أﺣد اﻟﺗﻛرارات اﻟﻣﺗوﻗﻌ ﺔ أﻗ ل ﻣ ن ‪ . 5‬وﺑﺎﺳ ﺗﺧدام اﻟﺗﺻ ﺣﯾﺢ ﯾﺻ ﺑﺢ ﻗﯾﻣ ﺔ اﻹﺣﺻ ﺎء اﻟ ذي ﯾﻌﺗﻣ د ﻋﻠﯾ ﮫ‬ ‫ﻗراراﻧﺎ ھو ‪:‬‬ ‫‪n‬‬ ‫‪n(| ad  bc |  )2‬‬ ‫‪2‬‬ ‫‪2 ‬‬ ‫)‪(a  c)(b  d)(c  d)(a  b‬‬ ‫ﺑﺗطﺑﯾق ﺗﺻﺣﯾﺢ ‪ Yates‬ﻋﻠﻰ اﻟﺑﯾﺎﻧﺎت ﻓﻲ ﺟدول اﻟﺳﺎﺑق ﻓﺈن ﻗﯾﻣﺔ اﻹﺣﺻﺎء ﺗﺻﺑﺢ ‪:‬‬ ‫‪56‬‬ ‫‪56[ (20)(14)  (16)(6)  ]2‬‬ ‫‪2‬‬ ‫‪2 ‬‬ ‫)‪(26)(30)(20)(36‬‬ ‫‪ 2.427.‬‬ ‫ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪   0.05‬ﻓﺈﻧﻧﺎ ﻧﺣﺻ ل إﻟ ﻰ ﻧﻔ س اﻻﺳ ﺗﻧﺗﺎج اﻟ ذي ﺣﺻ ﻠﻧﺎ ﻋﻠﯾ ﮫ ﺑ دون ﺗﺻ ﺣﯾﺢ‪ ،‬أي‬ ‫إﻧﻧﺎ ﻧﻘﺑل ‪. H 0‬‬ ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﻧﻔس اﻟﺑرﻧﺎﻣﺞ اﻟﺧﺎص ﺑﺎﻟﻣﺛﺎل )‪ (١-٧‬وﺳوف ﻧﻛﺗﻔﻰ ھﻧﺎ ﺑﺗوﺿﯾﺢ‬ ‫اﻟﻣدﺧﻼت واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫ﺗﺪﺧﻞ ﻣﺤﺘﻮﯾﺎت ﺟﺪول اﻻﻗﺘﺮان ﻣﻦ ﺧﻼل اﻻﻣﺮ اﻟﺘﺎﻟﻰ‬ ‫]‪npmChiSquareTest[{{20,16},{6,14}},mthd->ExpectedFrequencies‬‬ ‫ﺣﯿﺚ ﺗﺪﺧﻞ ﺻﻒ ﺻﻒ واﻟﻤﺨﺮج ﻟﮭﺬا اﻻﻣﺮ ھﻮ‬ ‫‪Title: Chi Square Test‬‬ ‫] ‪Distribution: ChiSquare[ 1‬‬ ‫‪PValue: 0.0661543‬‬

‫‪19.2857‬‬ ‫‪10.7143‬‬ ‫‪٥٤١‬‬

‫‪16.7143‬‬ ‫‪9.28571‬‬

‫واﻟﺬى ﯾﻮﺿﺢ ان اﺣﺼﺎء ھﺬا اﻻﺧﺘﺒﺎر ﯾﺘﺒﻊ ﺗﻮزﯾﻊ ﻣﺮﺑﻊ ﻛﺎى‬ ‫ﻛﻤﺎ ﯾﺤﺘﻮى اﻟﻤﺨﺮج ﻋﻠﻰ ﻗﯿﻤﺔ‬ ‫وھﻰ‬ ‫‪p-value‬‬

‫‪0.0661543‬‬

‫‪PValue:‬‬

‫وﺑﻣ ﺎ ان اﻟﻘﯾﻣ ﺔ اﻛﺑ ر ﻣ ن ‪   0.05‬ﻧﻘﺑ ل ﻓ رض اﻟﻌ دم ‪.‬اﯾﺿ ﺎ ﻣ ن اﻻﻣ ر اﻟﺳ ﺎﺑق ﯾ ﺗم اﻟﺣﺻ ول ﻋﻠ ﻰ‬ ‫اﻟﺗﻛرارات اﻟﻣﺗوﻗﻌﺔ وھﻰ‬ ‫‪19.2857‬‬ ‫‪10.7143‬‬

‫‪16.7143‬‬ ‫‪9.28571‬‬

‫)‪ (٢- ٧‬اﺧﺗﺑﺎر ﻣرﺑﻊ ﻛﺎي ﻟﻠﺗﺟﺎﻧس‬ ‫‪The Chi-square Test of Homogeneity‬‬ ‫ﺑﻔرض أن ﻟدﯾﻧﺎ ﻣﺟﺗﻣﻌﺎت ﻋددھﺎ ‪ r‬وﺟﻣﯾﻌﮭﺎ ﻣﺗﻣﺎﺛﻠﺔ ﻣن ﺣﯾث اﻟﺗﺻﻧﯾف وﺑﻔرض أن ‪ c‬ھﻲ‬ ‫ﻋدد ﻓﺋﺎت اﻟﺗﺻﻧﯾف ﻓﻲ ﻛل ﻣﺟﺗﻣﻊ ‪ .‬ﺑﻔرض أن ‪ Pj|i‬ﯾرﻣز ﻟﻧﺳﺑﺔ ﻣﺷﺎھدات اﻟﻣﺟﺗﻣﻊ رﻗم ‪ i‬اﻟﺗﻲ ﺗﻘﻊ‬ ‫ﻓﻲ اﻟﻔﺋﺔ رﻗم ‪ . j‬ﯾﻣﻛن ﺗﻣﺛﯾل ھذه اﻟﻣﺟﺗﻣﻌﺎت ﺑﺎﻟﺟدول اﻟﺗﺎﻟﻰ ‪.‬‬

‫ﻓﺋﺎت اﻟﺗﺻﻧﯾف‬ ‫‪2 … j … c‬‬ ‫‪P2|1 ... Pj|1 ... Pc|1‬‬

‫‪1‬‬ ‫‪P1|1‬‬

‫‪1‬‬

‫‪P2|2 ... Pj|2 ... Pc|2‬‬

‫‪P1|2‬‬

‫‪2‬‬

‫‪‬‬ ‫‪1‬‬

‫‪P2|i ... Pj|i‬‬

‫‪P1|i‬‬

‫‪P2|r ... Pj|r ... Pc|r‬‬

‫‪P1|r‬‬

‫‪1‬‬

‫‪P2  Pj  Pc‬‬

‫‪1‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪1‬‬

‫‪Pc|i‬‬

‫‪...‬‬

‫اﻟﻣﺟﺗﻣﻊ‬

‫‪‬‬ ‫‪i‬‬ ‫‪‬‬ ‫‪r‬‬

‫‪P1‬‬

‫ﻋﻧدﻣﺎ ﺗﻛون اﻟﻧﺳب ‪ Pj|i‬ﻣﺟﮭوﻟﺔ ﻓﺈﻧﻧﺎ ﻧرﻏب ﻓﻲ ﻣﻌرﻓﺔ ﻣﺎ إذا ﻛﺎﻧت اﻟﻣﺟﺗﻣﻌﺎت اﻟﺗﻲ ﻋددھﺎ ‪r‬‬ ‫ﻣﺗﺟﺎﻧﺳﺔ أي إﻧﻧﺎ ﻧرﻏب ﻓﻲ اﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪:‬‬ ‫‪H 0 : Pj|1  Pj|2  ....  Pj|r  Pj‬‬ ‫‪; j  1, 2,....c.‬‬

‫‪٥٤٢‬‬

‫ﻹﺟراء اﻻﺧﺗﺑﺎر ﻓﺈﻧﻧﺎ ﻧﺧﺗﺎر ﻋﯾﻧ ﺎت ﻋﺷ واﺋﯾﺔ ﻋ ددھﺎ ‪ r‬واﺣ دة ﻣ ن ﻛ ل ﻣﺟﺗﻣ ﻊ وأﺣﺟﺎﻣﮭ ﺎ ھ ﻲ‬ ‫‪ n1 ,n 2 ,..., n r‬ﻋﻠﻰ أن ﺗﻛون اﻟﻌﯾﻧﺎت اﻟﻌﺷواﺋﯾﺔ ﻣﺳﺗﻘﻠﺔ ﻋن ﺑﻌﺿﮭﺎ اﻟﺑﻌض‪ .‬ﺑﻔﺣ ص ﻣﺷ ﺎھدات ھ ذه‬ ‫اﻟﻌﯾﻧﺎت ووﺿﻊ ﻛل ﻣﺷﺎھدة ﺣﺳب ﺗﺻﻧﯾﻔﮭﺎ ﻧﺣﺻل ﻋﻠﻰ اﻟﺟدول اﻟﺗﺎﻟﻰ‪:‬‬

‫‪… c‬‬

‫‪… j‬‬

‫‪1‬‬

‫‪2‬‬

‫‪O11 O12 ... O1 j ... O1c‬‬

‫‪n1‬‬ ‫‪n2‬‬

‫‪O 21 O 22 ... O 2 j ... O 2 c‬‬

‫‪‬‬ ‫‪ni‬‬

‫‪Oi1 O i 2 ... Oij‬‬

‫‪O ic‬‬

‫‪‬‬ ‫‪nr‬‬

‫‪O r1 O r 2 ... O rj ... O rc‬‬

‫‪N‬‬

‫‪… n. j … n.c‬‬

‫‪c‬‬

‫ﺣﯾث ‪ n i   Oij‬و‬ ‫‪j1‬‬

‫‪n .2‬‬

‫‪n .1‬‬

‫‪r‬‬

‫‪c‬‬

‫‪r‬‬

‫‪i 1‬‬

‫‪j1‬‬

‫‪i 1‬‬

‫اﻟﻣﺟﺗﻣﻊ‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪i‬‬ ‫‪‬‬ ‫‪r‬‬

‫اﻟﻣﺟﻣوع‬

‫‪ n.j   Oij‬و ‪. n   n i   n.j‬‬

‫إذا ﻛﺎن ﻓرض اﻟﻌدم ﺻﺣﯾﺢ ﻓﺈن ‪:‬‬ ‫‪2‬‬

‫‪.‬‬

‫) ‪(Oij  Eij‬‬ ‫‪Eij‬‬

‫‪r‬‬

‫‪c‬‬

‫‪2   ‬‬ ‫‪j1 i 1‬‬

‫ھ ﻲ ﻗﯾﻣ ﺔ ﻟﻠﻣﺗﻐﯾ ر اﻟﻌﺷ واﺋﻲ ‪ X2‬اﻟ ذي ﺗﻘرﯾﺑ ﺎ ﯾﺗﺑ ﻊ ﺗوزﯾ ﻊ ‪ 2‬ﺑ درﺟﺎت ﺣرﯾ ﺔ )‪. (r  1)(c  1‬‬ ‫ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪ ‬ﻧﺗﺑﻊ اﻟﺧط وات اﻟﺗ ﻲ اﺳ ﺗﺧدﻣﻧﺎھﺎ ﻓ ﻲ اﺧﺗﺑ ﺎر ﻣرﺑ ﻊ ﻛ ﺎي‬ ‫ﻟﻼﺳﺗﻘﻼل ‪.‬‬ ‫‪٥٤٣‬‬

‫ﻣﺛﺎل)‪(٣-٧‬‬ ‫ﻗﺎﻣت ﺷرﻛﺔ ﻟﻠﻣﯾﺎه اﻟﻐﺎزﯾﺔ ﺑدراﺳﺔ ﻟﻣﻌرﻓﺔ ﻣ ﺎ إذا ﻛ ﺎن ھﻧ ﺎك اﺧ ﺗﻼف ﺑ ﯾن ﺷ راﺋﺢ ﻣﺧﺗﻠﻔ ﺔ ﻣ ن‬ ‫اﻟﻣﺟﺗﻣ ﻊ ﻣ ن ﻧﺎﺣﯾ ﺔ اﻟﺗﻔﺿ ﯾل ﻟﺛﻼﺛ ﺔ أﻧ واع ﻣ ن اﻟﻣﺷ روﺑﺎت ‪ .‬اﺳ ﺗﺧدﻣت ﻟﮭ ذه اﻟدراﺳ ﺔ أرﺑ ﻊ‬ ‫ﻋﯾﻧ ﺎت ﻣﺳ ﺗﻘﻠﺔ واﻟﻧﺗ ﺎﺋﺞ ﻣﻌط ﺎة ﻓ ﻲ اﻟﺟ دول اﻟﺗ ﺎﻟﻰ ‪ .‬اﺳ ﺗﺧدم اﺧﺗﺑ ﺎر ﻣرﺑ ﻊ ﻛ ﺎى ﻟﻠﺗﺟ ﺎﻧس‬ ‫ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪ : H 0 :‬اﻟﻣﺟﺗﻣﻌﺎت اﻷرﺑﻌﺔ ﻣﺗﺳﺎوﯾﯾن ﻓﻲ ﺗﻔﺿﯾل اﻟﻣﺷروب‪.‬‬ ‫ﺿد اﻟﻔرض اﻟﺑدﯾل ‪ : H1 :‬اﻟﻣﺟﺗﻣﻌﺎت اﻷرﺑﻌﺔ ﻏﯾر ﻣﺗﺳﺎوﯾﯾن ﻓﻲ ﺗﻔﺿﯾل اﻟﻣﺷروب‪.‬‬ ‫وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪.   0.05‬‬ ‫اﻟﻌﯾﻧﺎت اﻷرﺑﻌﺔ‬ ‫ﻧوع اﻟﻣﺷروب‬ ‫اﻟﻣﺟﻣوع‬ ‫‪100‬‬ ‫‪200‬‬ ‫‪47‬‬ ‫‪300‬‬ ‫‪647‬‬

‫‪C‬‬

‫‪B‬‬

‫‪A‬‬

‫‪5‬‬ ‫‪20‬‬ ‫‪17‬‬ ‫‪100‬‬ ‫‪142‬‬

‫‪20‬‬ ‫‪130‬‬ ‫‪25‬‬ ‫‪100‬‬ ‫‪275‬‬

‫‪75‬‬ ‫‪50‬‬ ‫‪5‬‬ ‫‪100‬‬ ‫‪230‬‬

‫رﺑﺎت اﻟﺑﯾوت‬ ‫رﺟﺎل اﻷﻋﻣﺎل‬ ‫ﻋﻣﺎل‬ ‫طﻠﺑﺔ‬ ‫اﻟﻣﺟﻣوع‬

‫اﻟﺗﻛرارات اﻟﻣﺗوﻗﻌﺔ ﺗم ﺣﺳﺎﺑﮭﺎ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬

‫اﻟﻣﺟﻣوع‬ ‫‪100‬‬ ‫‪200‬‬ ‫‪47.01‬‬ ‫‪300‬‬ ‫‪647.01‬‬

‫‪C‬‬ ‫‪21.95‬‬ ‫‪43.89‬‬ ‫‪10.32‬‬ ‫‪65.84‬‬ ‫‪142‬‬

‫ﻧوع اﻟﻣﺷروب‬ ‫‪B‬‬ ‫‪42.50‬‬ ‫‪85.01‬‬ ‫‪19.98‬‬ ‫‪127.51‬‬ ‫‪275‬‬

‫ﻧﺣﺳب ﻗﯾﻣﺔ اﻹﺣﺻﺎء ﻣن اﻟﺻﯾﻐﺔ اﻟﺗﺎﻟﯾﺔ ‪:‬‬

‫‪٥٤٤‬‬

‫اﻟﻌﯾﻧﺎت اﻷرﺑﻌﺔ‬ ‫‪A‬‬ ‫‪35.55‬‬ ‫‪71.10‬‬ ‫‪16.71‬‬ ‫‪106.65‬‬ ‫‪230.01‬‬

‫رﺑﺎت اﻟﺑﯾوت‬ ‫رﺟﺎل اﻷﻋﻣﺎل‬ ‫ﻋﻣﺎل‬ ‫طﻠﺑﺔ‬ ‫اﻟﻣﺟﻣوع‬

‫‪(Oij  Eij )2‬‬ ‫‪E ij‬‬

‫‪r‬‬

‫‪c‬‬

‫‪2‬‬

‫‪  ‬‬ ‫‪j1 i 1‬‬

‫‪ 149.72.‬‬ ‫‪2‬‬ ‫ﻟﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ ‪   0.05‬ﻓ ﺈن ‪ 12.592‬‬ ‫‪ .05‬واﻟﻣﺳ ﺗﺧرﺟﺔ ﻣ ن ﺟ دول ﺗوزﯾ ﻊ ‪ 2‬ﻓ ﻲ ﻣﻠﺣ ق‬

‫)‪ (٣‬ﻋﻧد درﺟﺎت ﺣرﯾﺔ ‪ . r  c  3  2  6‬ﻣﻧطﻘﺔ اﻟرﻓض ‪ . X 2  12.592‬وﺑﻣﺎ أن ‪ 2‬ﺗﻘ ﻊ ﻓ ﻲ‬ ‫ﻣﻧطﻘﺔ اﻟرﻓض ﻧرﻓض ‪. H 0‬‬ ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﻧﻔس اﻟﺑرﻧﺎﻣﺞ اﻟﺧﺎص ﺑﺎﻟﻣﺛﺎل )‪ (١-٧‬وﺳوف ﻧﻛﺗﻔﻰ ھﻧﺎ ﺑﺗوﺿﯾﺢ‬ ‫اﻟﻣدﺧﻼت واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫ﺗﺪﺧﻞ ﻣﺤﺘﻮﯾﺎت اﻟﺠﺪول ﻣﻦ ﺧﻼل اﻻﻣﺮ اﻟﺘﺎﻟﻰ‬ ‫‪npmChiSquareTest[{{75,20,5},{50,130,20},{5,25,17},{100,100,1‬‬ ‫]‪00}},mthd->ExpectedFrequencies‬‬ ‫ﺣﯿﺚ ﺗﺪﺧﻞ ﺻﻒ ﺻﻒ واﻟﻤﺨﺮج ﻟﮭﺬا اﻻﻣﺮ ھﻮ‬ ‫‪Title: Chi Square Test‬‬ ‫] ‪Distribution: ChiSquare[ 6‬‬ ‫‪PValue: 0.‬‬

‫‪21.9474‬‬ ‫‪43.8949‬‬ ‫‪10.3153‬‬ ‫‪65.8423‬‬

‫‪42.5039‬‬ ‫‪85.0077‬‬ ‫‪19.9768‬‬ ‫‪127.512‬‬

‫‪35.5487‬‬ ‫‪71.0974‬‬ ‫‪16.7079‬‬ ‫‪106.646‬‬

‫واﻟﺬى ﯾﻮﺿﺢ ان اﺣﺼﺎء ھﺬا اﻻﺧﺘﺒﺎر ﯾﺘﺒﻊ ﺗﻮزﯾﻊ ﻣﺮﺑﻊ ﻛﺎى ﺑدرﺟﺎت ﺣرﯾﺔ ‪6‬‬ ‫ﻛﻤﺎ ﯾﺤﺘﻮى اﻟﻤﺨﺮج ﻋﻠﻰ ﻗﯿﻤﺔ‬ ‫وھﻰ‬ ‫‪p-value‬‬

‫‪0.‬‬

‫‪PValue:‬‬

‫وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻗل ﻣن ‪   0.05‬ﻧ رﻓض ﻓ رض اﻟﻌ دم ‪.‬اﯾﺿ ﺎ ﻣ ن اﻻﻣ ر اﻟﺳ ﺎﺑق ﯾ ﺗم اﻟﺣﺻ ول ﻋﻠ ﻰ‬ ‫اﻟﺗﻛرارات اﻟﻣﺗوﻗﻌﺔ وھﻰ ‪:‬‬ ‫‪21.9474‬‬ ‫‪43.8949‬‬ ‫‪10.3153‬‬ ‫‪65.8423‬‬

‫‪42.5039‬‬ ‫‪85.0077‬‬ ‫‪19.9768‬‬ ‫‪127.512‬‬

‫‪35.5487‬‬ ‫‪71.0974‬‬ ‫‪16.7079‬‬ ‫‪106.646‬‬

‫ﻣﺛﺎل)‪(٤-٧‬‬ ‫‪٥٤٥‬‬

‫اﺧﺗﯾرت ﻋﯾﻧﺔ ﻋﺷواﺋﯾﺔ ﻣن ‪ 30‬ﻓردا ً ﻓﻰ ﺟﺎﻣﻌﺔ ﻣﺎ وﺗم ﺗﺻﻧﯾﻔﮭم ﺗﺑﻌﺎ ً ﻟﻠﺟ ﻧس وﻋ دد ﺳ ﺎﻋﺎت‬ ‫ﻣﺷﺎھدة اﻟﺗﻠﯾﻔزﯾون ﻓﻰ ﺧ ﻼل أﺳ ﺑوع‪ .‬اﻟﺑﯾﺎﻧ ﺎت اﻟﺗ ﻰ ﺗ م اﻟﺣﺻ ول ﻋﻠﯾﮭ ﺎ ﻣﻌط ﺎه ﻓ ﻲ اﻟﺟ دول‬ ‫اﻟﺗﺎﻟﻰ ‪:‬‬ ‫اﻟﻣﺷﺎھدة‬ ‫اﻟﺟﻧس‬ ‫اﻧﺛﻰ‬

‫ذﻛر‬

‫‪9‬‬ ‫‪7‬‬

‫‪5‬‬ ‫‪9‬‬

‫اﻛﺛر ﻣن ‪ 25‬ﺳﺎﻋﮫ‬ ‫أﻗل ﻣن ‪ 25‬ﺳﺎﻋﮫ‬

‫اﺳ ﺗﺧدم اﺧﺗﺑ ﺎر ﻣرﺑ ﻊ ﻛ ﺎى ﻟﻠﺗﺟ ﺎﻧس ﻻﺧﺗﺑ ﺎر ﻓ رض اﻟﻌ دم ‪ : H 0 :‬اﻟﻣﺟﺗﻣﻌ ﯾن ﻣﺗﺳ ﺎوﯾﯾن ﻓ ﻲ‬ ‫اﻟﻣﺷﺎھدة‬ ‫ﺿد اﻟﻔرض اﻟﺑدﯾل ‪ : H1 :‬اﻟﻣﺟﺗﻣﻌﯾن ﻏﯾر ﻣﺗﺳﺎوﯾﯾن ﻓﻲ اﻟﻣﺷﺎھدة‬ ‫وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪.   0.05‬‬ ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ ‪ Mathematica‬وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬ ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫;]‪Off[General::spell1‬‬ ‫`‪<<Statistics`NormalDistribution‬‬ ‫;}‪Options[npmChiSquare2x2Test]={mthd->uncorrected‬‬ ‫‪npmChiSquare2x2Test[fqlist_,opts___]:=Module[{r,c,n,prod,p1H‬‬ ‫‪at,p2Hat,fhat,det,fHatTab,minfHat,p,f},‬‬ ‫‪mtype=mthd/. {opts} /.‬‬ ‫;]‪Options[npmChiSquare2x2Test‬‬ ‫;]]‪f11=fqlist[[1,1‬‬ ‫;]]‪f12=fqlist[[1,2‬‬ ‫;]]‪f21=fqlist[[2,1‬‬ ‫;]]‪f22=fqlist[[2,2‬‬ ‫;‪r[1]=f11+f12‬‬ ‫;‪r[2]=f21+f22‬‬ ‫;‪c[1]=f11+f21‬‬ ‫;‪c[2]=f12+f22‬‬ ‫;‪n=f11+f12+f21+f22‬‬ ‫;]‪prod=r[1]*r[2]*c[1]*c[2‬‬ ‫;‪p1Hat=f11/c[1]//N‬‬ ‫;‪p2Hat=f12/c[2]//N‬‬ ‫;‪fhat[i_,j_]:=r[i]*c[j]/n‬‬ ‫;‪det=f11*f22-f12*f21‬‬ ‫‪٥٤٦‬‬

fHatTab=Table[fhat[i,j]//N,{i,1,2},{j,1,2}]//Flatten; minfHat=Min[fHatTab]; findMult[f_,minfHat_]:=Module[{}, k=1; While[0.5*k<Abs[f-minfHat],k=k+1]; 0.5*(k-1)]; subloop[x_,y_,z_,w_,minfHat_]:=Module[{p,d}, p=Position[fHatTab,minfHat]; f={x,y,z,w}[[p[[1,1]]]]; If[f<=2*minfHat,d=findMult[f,minfHat],d=Abs[fminfHat]-0.5]; n^3*d^2/prod]; If[mtype==uncorrected,cs=n*(det^2)/prod]; If[mtype==yates,cs=n*(Abs[det]-n/2)^2/prod]; If[mtype==haber,cs=subloop[f11,f12,f21,f22,minfHat]]; twotail=1-CDF[ChiSquareDistribution[1],cs]//N; If[mtype==uncorrected,corr=None]; If[mtype==yates,corr=Yates]; If[mtype==haber,corr=Haber]; Print["Title: Chi Square Test"]; Print["Distribution: Chi Square"]; Print["Sample Proportions: ",p1Hat,", ", p2Hat]; Print["Correction: ",corr]; Print["Two-Sided P-Value: ",twotail]; If[mtype==ExpectedFrequencies,Print["Expected Frequencies: ",Table[fhat[i,j]//N,{i,1,2},{j,1,2}]//TableForm]]; If[minfHat<5,Print["Expected Frequencies: ",Table[fhat[i,j]//N,{i,1,2},{j,1,2}]//TableForm]]] npmChiSquare2x2Test[{{5,9},{9,7}}] Title: Chi Square Test Distribution: Chi Square Sample Proportions: 0.357143 , 0.5625 Correction: None Two-Sided P-Value:

0.260679

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ﺗﺪﺧﻞ ﻣﺤﺘﻮﯾﺎت اﻟﺠﺪول ﻣﻦ ﺧﻼل اﻻﻣﺮ اﻟﺘﺎﻟﻰ‬ npmChiSquare2x2Test[{{5,9},{9,7}}] ‫ﺣﯿﺚ ﺗﺪﺧﻞ ﺻﻒ ﺻﻒ واﻟﻤﺨﺮج ﻟﮭﺬا اﻻﻣﺮ ھﻮ‬ Title: Chi Square Test Distribution: Chi Square Sample Proportions: 0.357143 , Correction: None Two-Sided P-Value: 0.260679 ٥٤٧

0.5625

‫‪0.260679‬‬

‫‪Two-Sided P-Value:‬‬

‫وذﻟك ﻻﺧﺗﺑ ﺎر ﻣ ن ﺟ ﺎﻧﺑﯾن وﺑﻣ ﺎ ان اﻟﻘﯾﻣ ﺔ اﻛﺑ ر ﻣ ن ‪   0.05‬ﻧﻘﺑ ل ﻓ رض اﻟﻌ دم ‪ .‬اﻳﻀـﺎ ﻳﺤﺘـﻮى‬

‫اﻟﻤﺨﺮج ﻋﻠﻰ ﻧﺴﺒﺔ اﻟﺬﻛﻮر اﻟـﺬﻳﻦ ﻳﺸـﺎﻫﺪون اﻟﺘﻠﻔﺰﻳـﻮن اﻛﺛ ر ﻣ ن ‪ 25‬ﺳ ﺎﻋﮫ واﻳﻀـﺎ ﻧﺴـﺒﺔ اﻻﻧـﺎث اﻟـﺬﻳﻦ ﻳﺸـﺎﻫﺪون‬

‫اﻟﺘﻠﻔﺰﻳﻮن اﻛﺛر ﻣن ‪ 25‬ﺳﺎﻋﮫ و ﻫﻤﺎ ﻋﻠﻰ اﻟﺘﻮاﻟﻰ‬

‫‪0.5625‬‬

‫‪0.357143 ,‬‬

‫‪Sample Proportions:‬‬

‫)‪ (٣ -٧‬اﺧﺗﺑﺎر اﻹﺷﺎرة ﻟﻌﯾﻧﺔ واﺣدة‬ ‫‪The One – sample Sign Test‬‬ ‫ﯾﺳﺗﺧدم اﺧﺗﺑﺎر اﻹﺷﺎرة ﻛﺑدﯾل ﻻﺧﺗﺑﺎر ‪ t‬اﻟﺧﺎص ﺑﻣﺗوﺳط ﻣﺟﺗﻣﻊ وذﻟك ﻋﻧد ﻋدم اﻟﺗﺣﻘ ق ﻣ ن‬ ‫أن اﻟﻣﺟﺗﻣﻊ اﻟذي اﺧﺗﺑرت ﻣﻧﺔ اﻟﻌﯾﻧﺔ ﯾﺗﺑﻊ ﺗوزﯾﻌﺎ ً طﺑﯾﻌﯾﺎ ً وﺣﺟم اﻟﻌﯾﻧﺔ أﻗل ﻣن ‪. 30‬‬ ‫ﺗﺗﻛ ون اﻟﺑﯾﺎﻧ ﺎت اﻟﻼزﻣ ﺔ ﻟﻠﺗﺣﻠﯾ ل ﻣ ن ﻋﯾﻧ ﺔ ﻋﺷ واﺋﯾﺔ ﻣ ن اﻟﺣﺟ م ‪ n‬ﻣ ن اﻟﻣﺷ ﺎھدات‬ ‫‪ x1, x 2 ,..., x n‬واﻟﻣﺧﺗﺎرة ﻣن ﻣﺟﺗﻣﻊ ﻣﺗﺻل وﺳ ﯾطﮫ ﻣﺟﮭ ول ‪ .‬ﻓ ﻲ اﻟﺣﻘﯾﻘ ﺔ إذا ﻛ ﺎن اﻟﺗوزﯾ ﻊ ﻣﺗﻣﺎﺛ ل‬ ‫ﻓﺈن اﻟوﺳﯾط ﯾﺳﺎوى اﻟوﺳط اﻟﺣﺳﺎﺑﻲ ﻟﻠﻣﺟﺗﻣﻊ ‪ ‬وﯾﻣﻛن اﺳﺗﺧدام اﺧﺗﺑﺎر اﻹﺷﺎرة ﻛﺎﺧﺗﺑ ﺎر ﻟﻠﻣﺗوﺳ ط ‪.‬‬ ‫ﻓرض اﻟﻌدم واﻟﻔرض اﻟﺑدﯾل ﺳوف ﯾﻛوﻧﺎن ﻋﻠﻰ اﻟﺷﻛل ‪:‬‬ ‫‪H0 : M  M0 ,‬‬ ‫‪H0 : M  M0.‬‬ ‫ﻹﺟ راء اﻻﺧﺗﺑ ﺎر ﻧﺣﺳ ب اﻟﻘﯾﻣ ﺔ ‪ k‬واﻟﺗ ﻲ ﺗﻣﺛ ل ﻋ دد اﻹﺷ ﺎرات اﻟﺳ ﺎﻟﺑﺔ ﻟﻠﻔ روق‬ ‫‪ (x i  M 0 ),i  1, 2,..., n‬وإذا وﺟ دت ﻣﺷ ﺎھدة ﺗﺳ ﺎوى اﻟوﺳ ﯾط ﻧﮭﻣﻠﮭ ﺎ وﻻ ﻧﺄﺧ ذھﺎ ﻓ ﻲ اﻻﻋﺗﺑ ﺎر ‪.‬‬ ‫ﺑﻔ رض أن ‪ H 0‬ﺻ ﺣﯾﺢ ﻓ ﺈن ‪ k‬ﺗﻣﺛ ل ﻗﯾﻣ ﺔ ﻟﻣﺗﻐﯾ ر ﻋﺷ واﺋﻲ ) إﺣﺻ ﺎء ( ‪ K‬ﻟ ﮫ ﺗوزﯾ ﻊ ذي اﻟﺣ دﯾن‬ ‫ﺑﻣﻌﺎﻟم‪ p  0.5 n,‬ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪ ‬ﻧرﻓض ‪ H 0‬إذا ﻛﺎن ‪:‬‬

‫‪P(K  k n,0.5)  . .‬‬ ‫ﻟﻠﻔرض اﻟﺑدﯾل ‪ H 0 :M  M 0‬ﻧرﻓض ‪ H 0‬إذا ﻛﺎن ‪:‬‬ ‫‪P(K  k n,0.5)  . .‬‬ ‫ﺣﯾث ‪ k‬ﺗﻣﺛل ﻋدد اﻹﺷﺎرات اﻟﻣوﺟﺑﺔ‪.‬‬ ‫ﻟﻠﻔرض اﻟﺑدﯾل ‪ H 0 :M  M 0‬ﻧرﻓض ‪ H 0‬إذا ﻛﺎن ‪:‬‬ ‫‪‬‬ ‫‪.‬‬ ‫‪2‬‬ ‫ﺣﯾث ‪ k‬ﺗﻣﺛل ﻋدد اﻹﺷﺎرات اﻟﻣوﺟﺑ ﺔ أو اﻟﺳ ﺎﻟﺑﺔ أﯾﮭﻣ ﺎ اﻗ ل ﻟﻠﻔ روق ‪(x i  M 0 ) , i  1, 2,..., n‬‬ ‫‪.‬‬ ‫‪P(K  k n,0.5) ‬‬

‫ﻣﺛﺎل)‪(٥-٧‬‬ ‫‪٥٤٨‬‬

‫ﯾﻌط ﻰ اﻟﺟ دول اﻟﺗ ﺎﻟﻰ اﻟ دﺧول اﻟﺳ ﻧوﯾﺔ ) ﺑ ﺎﻵﻻف اﻟ دوﻻرات ( ﻟﻌﯾﻧ ﺔ ﻋﺷ واﺋﯾﺔ ﻣ ن ‪21‬‬ ‫ﻋﺿ و ھﯾﺋ ﺔ ﺗ درﯾس ﺑﺈﺣ دى اﻟﺟﺎﻣﻌ ﺎت واﻟﻣطﻠ وب اﺧﺗﺑ ﺎر ﻓ رض اﻟﻌ دم‬ ‫‪ H 0 : M  25.1‬ﺿ د اﻟﻔ رض اﻟﺑ دﯾل ‪ H1 : M  25.1 :‬ﺑﺎﺳ ﺗﺧدام ﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ‬ ‫‪.   0.05‬‬ ‫‪25.1‬‬

‫‪26.7‬‬

‫‪25.2‬‬

‫‪24.1‬‬

‫‪23.6‬‬

‫‪22.2‬‬

‫‪19.9‬‬

‫‪19.4‬‬

‫‪14.0‬‬

‫‪12.9‬‬

‫‪11.1‬‬

‫‪50.6‬‬

‫‪50.5‬‬

‫‪40.1‬‬

‫‪38.9‬‬

‫‪34.8‬‬

‫‪33.9‬‬

‫‪32.2‬‬

‫‪30.7‬‬

‫‪29.6‬‬

‫‪28.1‬‬

‫اﻟﺣــل‪:‬‬ ‫ﻟﺣﺳﺎب ﻋدد اﻹﺷﺎرات اﻟﺳﺎﻟﺑﺔ ‪ k‬ﻧﺣدد إﺷﺎرات اﻟﻔروق ﺑ ﯾن ﻣﺷ ﺎھدات اﻟﻌﯾﻧ ﺔ واﻟوﺳ ﯾط ﺣﯾ ث وﺟ دت‬ ‫ﻛﺎﻵﺗﻲ ‪:‬‬ ‫‪--------++0++++++++++‬‬ ‫وﺣﯾث أن ﻟدﯾﻧﺎ ﻣﺷﺎھدة ﺗﺳﺎوى ﺻﻔر وﺑﻌد اﺳﺗﺑﻌﺎدھﺎ ﯾﺻﺑﺢ ﺣﺟم اﻟﻌﯾﻧﺔ ھو ‪ 20‬ﻣﺷﺎھدة ‪ .‬أي أن‬ ‫ﻋدد اﻹﺷﺎرات اﻟﺳﺎﻟﺑﺔ ‪ . k  8‬ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪   0.05‬ﻧرﻓض ‪ H 0‬إذا ﻛﺎﻧت ‪:‬‬

‫‪P(K  k n,0.5)   .‬‬ ‫ﻧﺣﺳب ‪:‬‬ ‫‪8‬‬

‫‪P(K  8 20,0.5)   b(x;20,0.5)  0.252 ,‬‬ ‫‪x 0‬‬

‫ﺣﯾث ‪ 0.252‬ﺗﺳﺗﺧرج ﻣن ﺟدول ذي اﻟﺣدﯾن ﻓﻲ ﻣﻠﺣق )‪ (٧‬ﻋﻧد ‪ n  20‬و ‪ . p  0.5‬ﺑﻣﺎ أن‬ ‫‪ 0.252‬أﻛﺑر ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧﻘﺑل ‪. H 0‬‬ ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ ‪ Mathematica‬وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬ ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫;]‪Off[General::spell1‬‬ ‫`‪<<Statistics`DiscreteDistributions‬‬ ‫‪f[y_,m0_]:=Module[{},‬‬ ‫]]]‪If[y<m0,-1,If[y>m0,1,0‬‬ ‫;}‪Options[npmSignTest]={sided->2‬‬ ‫‪npmSignTest[data_,m0_,opts___]:=Module[{m,n,signs,s,equalNum‬‬ ‫‪ber,tail,med,pval},‬‬ ‫;]‪m=Length[data‬‬ ‫;]‪signs=Map[f[#,m0]&,data‬‬ ‫;]‪s=Count[signs,1‬‬ ‫;]‪equalNumber=Count[signs,0‬‬ ‫;‪n=m-equalNumber‬‬ ‫;]‪tail=sided/. {opts} /. Options[npmSignTest‬‬ ‫‪If[s<=n/2,pval=N[CDF[BinomialDistribution[n,1/2],s]],pv‬‬ ‫;]]]‪al=1-N[CDF[BinomialDistribution[n,1/2],s‬‬ ‫;]‪If[tail==2,pval=2*pval‬‬ ‫‪٥٤٩‬‬

med=Median[data]//N; Print["Title: Sign Test"]; Print["Estimate: Sample Median -> ",med]; Print["Test Statistic: Number of Pluses is ",s]; Print["Distribution: BinomialDistribution[",n,",1/2]"]; Print[tail," - sided p-value -> ",pval]];

grades={11.1,12.9,14,19.4,19.9,22.2,23.6,24.1,25.2,26.7,25.1 ,28.1,29.6,30.7,32.2,33.9,34.8,38.9,40.1,50.5,50.6}; npmSignTest[grades,25.1] Title: Sign Test Estimate: Sample Median -> 26.7 Test Statistic: Number of Pluses is 12 Distribution: BinomialDistribution[ 20 ,1/2] 2 - sided p-value -> 0.263176 npmSignTest[grades,25.1,sided->1] Title: Sign Test Estimate: Sample Median -> 26.7 Test Statistic: Number of Pluses is 12 Distribution: BinomialDistribution[ 20 ,1/2] 1

- sided p-value ->

0.131588

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬: ‫اوﻻ‬

‫ وھﻰ‬grades ‫اﻟﻘﺎﺋﻣﺔ اﻟﻣﺳﻣﺎه‬ grades={11.1,12.9,14,19.4,19.9,22.2,23.6,24.1,25.2,26.7,25.1 ,28.1,29.6,30.7,32.2,33.9,34.8,38.9,40.1,50.5,50.6} (٥-٧) ‫وﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﯿﺎﻧﺎت اﻟﺨﺎﺻﺔ ﺑﺎﻟﻤﺜﺎل‬ : ‫ﺛﺎﻧﯾﺎ اﻟﻣﺧرﺟﺎت‬ ‫ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ npmSignTest[grades,25.1]

‫واﻟﻤﺨﺮج ھﻮ‬ Title: Sign Test Estimate: Sample Median -> 26.7 Test Statistic: Number of Pluses is 12 Distribution: BinomialDistribution[ 20 ,1/2] 2 - sided p-value -> 0.263176

‫ﺣﯿﺚ اﻟوﺳﯾط ﻟﻠﻌﯾﻧﺔ ھو‬ Estimate: Sample Median ->

26.7

‫وﻋدد اﻻﺷﺎرات اﻟﻣوﺟﺑﺔ ھو‬ Test Statistic: Number of Pluses is ٥٥٠

‫واﻟﺗﻰ ﺗﻌﻧﻰ ان ﻋدد اﻻﺷﺎرات اﻟﺳﺎﻟﺑﺔ ھو ‪ 8‬واﻟﺗﻰ ﺣﺻﻠن ﻋﻠﯾﮭﺎ ﻋﻧد ﺣل اﻟﻣﺛﺎل ﯾدوﯾﺎ‬ ‫وذﻟك ﺑﻌد اﺳﺗﺑﻌﺎد ﻣﺷﺎھدة واﺻﺑﺢ ﺣﺟم اﻟﻌﯾﻧﺔ ھو ‪ 20‬ﻣﺷﺎھدة‪.‬‬ ‫وﻗﯾﻣﺔ ‪ p-value‬ھﻰ‬ ‫‪0.263176‬‬

‫>‪- sided p-value -‬‬

‫‪2‬‬

‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻛﺑر ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧﻘﺑل ‪. H 0‬اﻣﺎ إذا ﻛﺎن اﻻﺧﺗﺑﺎر‬ ‫اﻟﻌدم ‪ H 0 : M  25.1‬ﺿد اﻟﻔرض اﻟﺑدﯾل‬ ‫ﻣن ﺟﺎﻧب واﺣد اى اﻟﻣطﻠوب اﺧﺗﺑﺎر ﻓرض‬ ‫‪H1 : M  25.1 :‬‬ ‫ﻓﺈﻧﻧﺎ ﻧﺳﺗﺧدم اﻻﻣر‬ ‫]‪npmSignTest[grades,25.1,sided->1‬‬

‫واﻟﻤﺨﺮج ھﻮ‬ ‫‪Title: Sign Test‬‬ ‫‪Estimate: Sample Median -> 26.7‬‬ ‫‪Test Statistic: Number of Pluses is 12‬‬ ‫]‪Distribution: BinomialDistribution[ 20 ,1/2‬‬ ‫‪0.131588‬‬

‫>‪- sided p-value -‬‬

‫‪1‬‬

‫وﻗﯾﻣﺔ ‪ p-value‬ھﻰ‬ ‫‪0.131588‬‬

‫>‪- sided p-value -‬‬

‫‪1‬‬

‫وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻛﺑر ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧﻘﺑل ‪ . H 0‬وﻣﻣﺎ ﯾﺟد اﻻﺷﺎرة إﻟﯾﮫ اﻧﮫ إذا ﻛﺎن اﻟﻣطﻠوب‬ ‫اﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪ H 0 : M  25.1‬ﺿد اﻟﻔرض اﻟﺑدﯾل ‪ H1 : M  25.1 :‬ﻓﺈن ﻗﯾﻣﺔ ‪p‬‬ ‫‪ -value‬ﺳوف ﺗﻛون ‪ 1- p‬وذﻟك ﻻن اﻟوﺳﯾط ﻟﻠﻌﯾﻧﺔ اﻛﺑر ﻣن ‪. 25.1‬‬

‫)‪ (٤ -٧‬اﺧﺗﺑﺎر إﺷﺎرة اﻟرﺗب‬ ‫‪The Signed – ranks Test‬‬ ‫ﯾﻌﺗﻣد اﺧﺗﺑﺎر اﻹﺷﺎرة ﻟﻌﯾﻧﺔ واﺣ دة ﻋﻠ ﻰ اﻟﻔ رق ﺑ ﯾن ﻗ ﯾم ﻣﺷ ﺎھدات اﻟﻌﯾﻧ ﺔ واﻟوﺳ ﯾط اﻟﻣﻔﺗ رض‬ ‫ﻣ ﻊ إﻏﻔ ﺎل ﻗﯾﻣ ﺔ اﻟﻔ روق واﻟ ذي ﯾ ؤدى إﻟ ﻰ أﺿ ﻌﺎف اﻻﺧﺗﺑ ﺎر ‪ .‬ﻟ ذﻟك أﻗﺗ رح اﻟﻌ ﺎﻟم ‪Wilcoxon‬‬ ‫ﺗﺑﺎرا ً ﻻﻣﻌﻠﻣﯾﺎ ً آﺧ ر أطﻠ ق ﻋﻠﯾ ﮫ أﺳ ﻣﮫ ﯾﻌﺗﻣ د ﻋﻠ ﻰ إﺷ ﺎرة اﻟﻔ رق وﻗﯾﻣ ﺔ اﻟﻔ رق ﺣﯾ ث ﯾﻌط ﻰ وزﻧ ﺎ ً‬ ‫إﺧ‬ ‫أﻛﺑر ﻟﻺﺷﺎرة اﻟﺗﻲ ﺗﺻﺎﺣب ﻓرﻗﺎ ً ﻛﺑﯾرا ً واﻟﻌﻛس ﺻﺣﯾﺢ ‪ .‬ﯾﺷﺗرك ھذا اﻻﺧﺗﺑﺎر ﻣﻊ اﺧﺗﺑ ﺎر اﻹﺷ ﺎرة ﻓ ﻲ‬ ‫أﻧﮫ ﯾﻣﻛن أن ﯾﺳﺗﺧدم ﻛﺎﺧﺗﺑﺎر ﻟﻠﻣﺗوﺳط ﻋﻧدﻣﺎ ﯾﻛون اﻟﻣﺟﺗﻣﻊ ﻣوﺿﻊ اﻟدراﺳﺔ ﻣﺗﻣﺎﺛل ‪.‬‬ ‫ﺗﺗﻛون اﻟﺑﯾﺎﻧﺎت اﻟﻼزﻣ ﺔ ﻟﻠﺗﺣﻠﯾ ل ﻣ ن ﻋﯾﻧ ﺔ ﻋﺷ واﺋﯾﺔ ﻣ ن اﻟﺣﺟ م ‪ n‬ﻣ ن اﻟﻣﺷ ﺎھدات اﻟﻣﺳ ﺗﻘﻠﺔ‬ ‫‪ x1, x 2 ,..., x n‬واﻟﻣﺧﺗ ﺎرة ﻣ ن ﻣﺟﺗﻣ ﻊ ﻣﺗﺻ ل‪ .‬ﻓ رض اﻟﻌ دم واﻟﻔ رض اﻟﺑ دﯾل ﺳ وف ﯾﻛوﻧ ﺎن ﻋﻠ ﻰ‬ ‫اﻟﺷﻛل ‪:‬‬ ‫‪H0 : M  M0 ,‬‬ ‫‪H1 : M  M 0 .‬‬ ‫ﻟﺣﺳﺎب ﻗﯾﻣﺔ اﻹﺣﺻﺎء اﻟذي ﯾﻌﺗﻣد ﻋﻠﯾﮫ ﻗرارﻧﺎ ‪ ،‬ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪ ، ‬ﻧﺗﺑﻊ اﻟﺧطوات اﻟﺗﺎﻟﯾﺔ ‪:‬‬ ‫‪٥٥١‬‬

‫)أ(‬

‫ﻧﺣدد إﺷﺎرة وﻗﯾﻣﺔ اﻟﻔروق ‪. Di  (x i  M 0 ) ; i  1, 2,..., n‬‬

‫)ب(‬

‫إذا ﻛﺎن اﻟﻔرق ﻣﺳﺎوﯾﺎ ً ﻟﻠﺻﻔر ﺗﺳ ﺗﺑﻌد اﻟﻣﺷ ﺎھدة ﻣ ن اﻟﺗﺣﻠﯾ ل وﯾﻌ دل ﺣﺟ م اﻟﻌﯾﻧ ﺔ ﺑط رح ﻋ دد‬ ‫ﯾﺳﺎوى ﻋدد اﻟﻣﺷﺎھدات اﻟﺗﻲ ﺗﺳﺎوي اﻟوﺳﯾط ‪.‬‬

‫)ج(‬

‫ﻧﮭﻣل إﺷﺎرة اﻟﻔروق ﻣؤﻗﺗﺎ ً وﻧرﺗب اﻟﻔروق ﺗﺻﺎﻋدﯾﺎ ً ﺑﻣﻌﻧﻰ آﺧ ر ﻧﻌط ﻰ رﺗ ب ﻟﻠﻘ ﯾم ‪ Di‬أي‬ ‫اﻟﻘ ﯾم اﻟﻣطﻠﻘ ﺔ ﻟﻠﻔ روق وإذا ﻛ ﺎن ھﻧ ﺎك ﻓ روق ﻣﺗﺳ ﺎوﯾﺔ ‪ ،‬أي ﺗ داﺧﻼت ‪ ، ties‬ﻓﺈﻧﻧ ﺎ ﻧﻌطﯾﮭ ﺎ‬ ‫ﻣﺗوﺳط اﻟرﺗب اﻟﺗﻲ ﻛﺎﻧت ﺳﺗﺄﺧذھﺎ ﻟو أﻧﮭﺎ ﻛﺎﻧت ﻣﺧﺗﻠﻔﺔ ‪.‬‬

‫)د(‬

‫ﺗﻌ ﺎد اﻹﺷ ﺎرات إﻟ ﻰ اﻟرﺗ ب اﻟﻣﻧ ﺎظرة وﻧوﺟ د ﻣﺟﻣ وع رﺗ ب اﻹﺷ ﺎرات اﻟﺳ ﺎﻟﺑﺔ وﻧرﻣ ز ﻟ ﮫ‬ ‫ﺑﺎﻟرﻣز ‪ t ‬وﻧوﺟ د ﻣﺟﻣ وع رﺗ ب اﻹﺷ ﺎرات اﻟﻣوﺟﺑ ﺔ وﻧرﻣ ز ﻟ ﮫ ﺑ ﺎﻟرﻣز ‪ t ‬وﯾﻣﻛ ن إﯾﺟ ﺎد ﻛ ل‬ ‫ﻣﻧﮭﻣﺎ ﺑدﻻﻟﺔ اﻵﺧر ﻣن اﻟﻌﻼﻗﺔ ‪:‬‬

‫)‪n(n  1‬‬ ‫‪ t.‬‬ ‫‪2‬‬

‫‪t ‬‬

‫ﻧﺣﺳب ﻣﺟﻣوع اﻟرﺗب اﻟﺳﺎﻟﺑﺔ أو ﻣﺟﻣوع اﻟرﺗب اﻟﻣوﺟﺑﺔ أﯾﮭﻣﺎ اﻗل وﻧرﻣز ﻟﮫ ﺑﺎﻟرﻣز ‪ t‬أي أن‪:‬‬ ‫‪t  min(t  , t  ) .‬‬

‫واﻟﺗ ﻲ ﺗﻣﺛ ل ﻗﯾﻣ ﺔ ﻹﺣﺻ ﺎء ‪ .‬ﯾﺳ ﺗﺧدم اﻟﺟ دول ﻓ ﻲ ﻣﻠﺣ ق )‪ (٨‬ﻻﺳ ﺗﺧراج اﻟﻘ ﯾم اﻟﺣرﺟ ﺔ ﻟﮭ ذا‬ ‫اﻹﺣﺻﺎء ﻟﻌﯾﻧﺎت ﻣن اﻟﺣﺟم ‪ 3‬وﺣﺗﻰ اﻟﺣﺟم ‪ 25‬وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧب أو ﺟ ﺎﻧﺑﯾن ‪ .‬ﺳ ﻧرﻣز ﻟﻠﻘ ﯾم‬ ‫اﻟﺟدوﻟﯾﺔ ﺑﺎﻟرﻣز )‪ d(n, ) , d(n, ‬ﻋﻧدﻣﺎ ﯾﻛون اﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧ ب أو ﺟ ﺎﻧﺑﯾن ﻋﻠ ﻰ اﻟﺗ واﻟﻲ‪.‬‬ ‫ﻟﻠﺑ دﯾل ‪ H 0 : M  M 0‬و ﻟﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ ‪ ‬ﻧ رﻓض ‪ H 0‬إذا ﻛ ﺎن )‪ . t  d(n, ‬ﻟﻠﻔ رض‬ ‫اﻟﺑدﯾل ‪ H 0 : M  M 0‬اھﺗﻣﺎﻣﻧ ﺎ ﺳ وف ﯾﻛ ون ﻓ ﻲ ﻣﺟﻣ وع اﻟرﺗ ب اﻟﺳ ﺎﻟﺑﺔ ‪ . t ‬ﻟﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ‬ ‫‪ ‬ﻧ رﻓض ‪ H0‬إذا ﻛ ﺎن )‪ . t   d(n, ‬ﻟﻠﻔ رض اﻟﺑ دﯾل ‪ H 0 : M  M 0‬اھﺗﻣﺎﻣﻧ ﺎ ﺳ وف‬ ‫ﯾﻛون ﻓﻲ ﻣﺟﻣوع اﻟرﺗب اﻟﻣوﺟﺑﺔ ‪ . t ‬ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪ ‬ﻧرﻓض ‪ H 0‬إذا ﻛ ﺎن )‪t   d(n, ‬‬ ‫‪.‬‬ ‫‪٥٥٢‬‬

‫ﻣﺜﺎل)‪(٦-٧‬‬

‫اﺧﺗﯾرت ﻋﯾﻧﺔ ﻋﺷواﺋﯾﺔ ﻣن ‪15‬ﺷﺧﺻﺎ ً ﻣﻣن ﯾﻣﺗﻠﻛون ﻣﻧزﻻ ً ﻓ ﻲ ﻣدﯾﻧ ﺔ ﻣ ﺎ ‪ .‬وﻗ د ﺗ م ﺳ ؤاﻟﮭم‬ ‫ﻋن ﻣﻘدار اﻟزﯾﺎدة ﻓﻲ ﻓﺎﺗورة اﻟﺿراﺋب اﻟﺳ ﻧوﯾﺔ ﺑﺎﻟ دوﻻر‪ .‬اﻟﺑﯾﺎﻧ ﺎت اﻟﺗ ﻲ ﺗ م اﻟﺣﺻ ول ﻋﻠﯾﮭ ﺎ‬ ‫ﻣﻌط ﺎة ﻓ ﻲ اﻟﺟ دول اﻟﺗ ﺎﻟﻰ اﻟﻣطﻠ وب اﺧﺗﺑ ﺎر ﻓ رض اﻟﻌ دم ‪ H1 : M  500‬ﺿ د اﻟﻔ رض‬ ‫اﻟﺑدﯾل ‪ H1 : M  500‬ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪.   0.05‬‬ ‫‪ Di‬ﻣﻀﺮوﺑﺎً ﰲ إﺷﺎرة ‪Di‬‬

‫رﺗﺐ ‪Di‬‬

‫)‪Di  ( xi  500‬‬

‫ﻣﻘﺪار اﻟﺰﻳﺎدة‬

‫‪-8‬‬ ‫‪+13‬‬ ‫‪-14‬‬ ‫‪-11‬‬ ‫‪+4‬‬ ‫‪-‬‬‫‪-6‬‬ ‫‪-3‬‬ ‫‪+2‬‬ ‫‪-5‬‬ ‫‪-12‬‬ ‫‪+7‬‬ ‫‪-1‬‬ ‫‪-10‬‬ ‫‪+9‬‬

‫‪8‬‬ ‫‪13‬‬ ‫‪14‬‬ ‫‪11‬‬ ‫‪4‬‬ ‫ﻳﺴﺘﺒﻌﺪ‬ ‫‪6‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪5‬‬ ‫‪12‬‬ ‫‪7‬‬ ‫‪1‬‬ ‫‪10‬‬ ‫‪9‬‬

‫‪-5.6‬‬ ‫‪+10.8‬‬ ‫‪-12.5‬‬ ‫‪-6.8‬‬ ‫‪+2.6‬‬ ‫‪0‬‬ ‫‪-4.1‬‬ ‫‪-1.8‬‬ ‫‪+1.6‬‬ ‫‪-2.7‬‬ ‫‪-8.0‬‬ ‫‪+4.3‬‬ ‫‪-0.8‬‬ ‫‪-6.5‬‬ ‫‪+5.8‬‬

‫‪494.4‬‬ ‫‪510.8‬‬ ‫‪487.5‬‬ ‫‪493.2‬‬ ‫‪502.6‬‬ ‫‪500.0‬‬ ‫‪495.9‬‬ ‫‪498.2‬‬ ‫‪501.6‬‬ ‫‪497.3‬‬ ‫‪492.0‬‬ ‫‪504.3‬‬ ‫‪499.2‬‬ ‫‪493.5‬‬ ‫‪505.8‬‬

‫اﻟﺣــل‪:‬‬ ‫ﻣن ا ﻟﺟدول اﻟﺳﺎﺑق وﻣن اﻟﻌﻣود اﻷﺧﯾر ) ﻋﻠﻰ اﻟﯾﻣﯾن ( ﻧﺣﺳب ﻣﺟﻣوع اﻟرﺗب اﻟﺳﺎﻟﺑﺔ‬

‫ﻧﺣﺻل ﻋﻠﻰ اﻟﻘﯾم اﻟﺳﺎﻟﺑﺔ واﻟﻣوﺟﺑﺔ ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬ ‫‪t   8  14  11  6  3  5  12  1  10  70,‬‬ ‫‪t   13  4  2  7  9  35.‬‬ ‫‪٥٥٣‬‬

. n  14 ‫وﺣﯾث أن ھﻧﺎك ﻓرق ﻣﺳﺎوﯾﺎ ً ﻟﻠﺻﻔر ﻓﺈﻧﻧﺎ ﻧﮭﻣﻠﮫ وﻧﻌدل ﺣﺟم اﻟﻌﯾﻧﺔ ﺗﺑﻌﺎ ً ﻟذﻟك أي أن‬ t  min(t  , t  )  35 . ‫ ﻣ ن‬n  14 ‫ وﻋﻧ د‬  0.049 ‫ ﺣﯾ ث‬d(14,0.049)  22 ‫ ﻓ ﺈن‬  0.05 ‫ﻟﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ‬ . H 0 ‫ ﻧﻘﺑل‬22 ‫ أﻛﺑر ﻣن‬t   35 ‫ ( وﺑﻣﺎ أن‬٨ ) ‫اﻟﺟدول ﻓﻲ ﻣﻠﺣق‬ ‫( ﻟﺗﻘ دﯾر اﻟﻣﻌﻧوﯾ ﺔ‬٨) ‫ ﻓﺈﻧﻧ ﺎ ﻻ ﻧﺳ ﺗطﯾﻊ اﺳ ﺗﺧدام اﻟﺟ دول ﻓ ﻲ ﻣﻠﺣ ق‬، 25 ‫ أﻛﺑ ر ﻣ ن‬n ‫ﻋﻧ دﻣﺎ ﺗﻛ ون‬ .‫ﻟﻠﻘﯾﻣﺔ اﻟﻣﺣﺳوﺑﺔ ﻟﻺﺣﺻﺎء‬ : ‫ﻟﻠﻌﯾﻧﺎت اﻟﻛﺑﯾرة ﻓﺈن‬ n(n  1) t  4 t*  . n(n  1)(2n  1) 24 ‫ ﻟﻼﺧﺗﺑ ﺎرات ﻣ ن ﺟﺎﻧ ب واﺣ د‬. ‫ اﻟذي ﺗﻘرﯾﺑﺎ ﯾﺗﺑﻊ اﻟﺗوزﯾ ﻊ اﻟطﺑﯾﻌ ﻲ اﻟﻘﯾﺎﺳ ﻲ‬T* ‫ﺗﻣﺛل ﻗﯾﻣﺔ ﻟﻺﺣﺻﺎء‬ .‫ ﺣﺳب اﻟﻔرض اﻟﻣﺳﺗﺧدم‬t  ‫ أو‬t  ‫ ﺑﺎﻟﻘﯾﻣﺔ‬t * ‫ ﻓﻲ ﺻﯾﻐﺔ‬t ‫ﻓﺈﻧﮫ ﯾﻣﻛن اﺳﺗﺑدال‬ ‫ وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬Mathematica ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ‬ . ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت‬ Off[General::spell1]; <<Statistics`DataManipulation` <<Statistics`NormalDistribution` f[y_,m0_]:=Module[{}, If[y<m0,-1,If[y>m0,1,0]]] d[y_,m0_]:=y-m0 Clear[findTieNumber] findTieNumber[x_,list3_]:=Module[{m,list2}, m=1; list2=Frequencies[list3]; While[!MatchQ[x,list2[[m,2]]],m=m+1]; list2[[m,1]]] Clear[rank] rank[j_,xlist_]:=Module[{k,flag,xsort,num2}, k=1; flag=0; xsort=Sort[xlist]; While[xlist[[j]]!=xsort[[k]],k=k+1]; num2=findTieNumber[xlist[[j]],xlist]; Sum[val,{val,k,k+num2-1}]/num2//N] tsig[list1_]:=Module[{tsum,fcount,n}, n=Length[list1]; fcount=Frequencies[list1]; Sum[fcount[[j,1]]^3fcount[[j,1]],{j,1,Length[fcount]}]] removeZeros[x_]:=If[x!=0,AppendTo[newlist,x]] greaterQ[x_]:=If[x>0,AppendTo[bigger,x]] ٥٥٤

largeSampleZ[n_,t_,ts_]:=Module[{numer,one}, numer=t-n(n+1)/4; one=n(n+1)(2n+1)/24; numer/ Sqrt[one-ts/48]//N] largeSamplePValue[n_,t_,ts_]:=Module[{z}, z=largeSampleZ[n,t,ts]; CDF[NormalDistribution[0,1],z]//N] t-probability Clear[n,p] n[k_,n1_]:=1 /; k===0 && n1===0 n[k_,n1_]:=0 /; k!=0 && n1===0 n[k_,n1_]:=0 /; k<0 n[k_,n1_]:=0 /; k>n1(n1+1)/2 n[k_,n1_]:=n[k,n1]=n[k,n1-1]+n[k-n1,n1-1] p[t_,n1_]:=Sum[n[k,n1]/(2^n1),{k,0,t}]//N Body of Program Options[npmSignedRanksTest]={sided->2};

٥٥٥

npmSignedRanksTestdata_, m0_, opts___ : Moduledlist, addNonzeros, total, nonZeroList, absList, n1, absRank, plusRanks, tPlus, tsum, pval, tail, med, dlist  Mapd#, m0 &, data; newlist  0; addNonzeros  MapremoveZeros# &, dlist; total  LengthaddNonzeros; nonZeroList  DropaddNonzerostotal, 1; n1  LengthnonZeroList; absList  Map Abs, nonZeroList; absRank  Tablerankk, absList, k, 1, n1; signedRanks  TableSignnonZeroListj  absRankj, j, 1, n1; bigger  0; MapgreaterQ, signedRanks; plusRanks  Dropbigger, 1; tPlus  ApplyPlus, plusRanks; tNeg  n1n1 1  2  tPlus; t  MintPlus, tNeg; intGreater  Ceilingt; intLess  Floort; tsum  tsigabsList; Ifn1  30, pval  largeSamplePValuen1, t, tsum, pval  pintLess, n1  pintGreater, n1  2; tail  sided . opts . OptionsnpmSignedRanksTest; Iftail  2, pval  2  pval; med  Mediandata  N; Print ٥٥٦ Test"; "Title: Wilcoxon SignedRanks Print"Sample Median  ", med; Print"Test Statistics: T  ", tPlus, " T  ", tNeg, " T  ", t; Printtail, "  sided PValue  ", pval

‫‪grades={494.4,510.8,487.5,493.2,502.6,500,495.9,498.2,501.6,‬‬ ‫;}‪497.3,492,504.3,499.2,493.5,505.8‬‬ ‫]‪npmSignedRanksTest[grades,500‬‬ ‫‪Title: Wilcoxon Signed-Ranks Test‬‬ ‫‪Sample Median -> 498.2‬‬ ‫‪Test Statistics: T  35. T  70. T  35.‬‬ ‫‪2 - sided PValue -> 0.295776‬‬

‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ ‪:‬‬ ‫اوﻻ ‪ :‬اﻟﻣدﺧﻼت‬

‫اﻟﻘﺎﺋﻣﺔ اﻟﻣﺳﻣﺎه ‪ grades‬وھﻰ‬ ‫‪grades={494.4,510.8,487.5,493.2,502.6,500,495.9,498.2,501.6,‬‬ ‫}‪497.3,492,504.3,499.2,493.5,505.8‬‬ ‫وﺗﺤﺘﻮى ﻋﻠﻰ اﻟﺒﯿﺎﻧﺎت اﻟﺨﺎﺻﺔ ﺑﺎﻟﻤﺜﺎل )‪(٦-٧‬‬ ‫ﺛﺎﻧﯾﺎ اﻟﻣﺧرﺟﺎت ‪:‬‬ ‫ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪npmSignedRanksTest[grades,500‬‬

‫واﻟﻤﺨﺮج ھﻮ‬ ‫‪T  35.‬‬

‫‪Title: Wilcoxon Signed-Ranks Test‬‬ ‫‪Sample Median -> 498.2‬‬ ‫‪Test Statistics: T  35. T  70.‬‬ ‫‪0.295776‬‬

‫>‪- sided PValue -‬‬

‫‪2‬‬

‫ﺣﯿﺚ اﻟوﺳﯾط ﻟﻠﻌﯾﻧﺔ ھو‬ ‫‪498.2‬‬

‫>‪Sample Median -‬‬

‫ﻣﺟﻣوع اﻟرﺗب اﻟﻣوﺟﺑﺔ و ﻣﺟﻣوع اﻟرﺗب اﻟﺳﺎﻟﺑﺔ و ‪ t  min(t  , t  )  35‬ﻧﺣﺻل‬ ‫ﻋﻠﯾﮭﺎ ﻣن‬ ‫‪Test Statistics: T  35. T  70. T  35.‬‬ ‫ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪   0.05‬وﺑﻣﺎ ان ﻗﯾﻣﺔ ‪ p-value‬ھﻰ‬ ‫‪0.295776‬‬

‫>‪- sided PValue -‬‬

‫‪2‬‬

‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻛﺑر ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧﻘﺑل ‪. H 0‬‬

‫)‪ (٥-٧‬اﺧﺗﺑﺎرات ﺗﺗﻌﻠق ﺑﻣﻌﻠﻣﺔ اﻟﻧﺳﺑﺔ‬ ‫‪Tests Concerning Proportion Parameter‬‬ ‫ﺗﻌﺗﺑر اﻟﻧﺳﺑﺔ ﻓﻲ اﻟﻣﺟﺗﻣﻊ ﻣن أھم اﻟﻣﻌﺎﻟم اﻟﺗﻲ ﺗدور ﺣوﻟﮭﺎ ﺗﺳﺎؤﻻت ﻛﺛﯾرة‪،‬أﺣﯾﺎﻧﺎ ً ﻧرﯾد أن‬ ‫ﻧﺧﺗﺑر ھل اﻟﻧﺳﺑﺔ ﻓﻲ اﻟﻣﺟﺗﻣﻊ ﺗﺳﺎوي ﻣﻘدارا ً ﻣﻌﯾﻧﺎ ً أم ﻻ؟ وأﺣﯾﺎﻧﺎ ً ﻧرﯾد ﺗﻘدﯾر ھذه اﻟﻧﺳﺑﺔ‪ .‬ﻓﻲ ھذه‬ ‫اﻟﺣﺎﻟﺔ ﯾﻛون اﻟﻣﺟﺗﻣﻊ ﻣﻘﺳم إﻟﻰ ﻗﺳﻣﯾن وھو ﻣﺎ ﺳﻧﻌﺎﻟﺟﮫ ھﻧﺎ اﻣﺎ إذا ﻛﺎن اﻟﻣﺟﺗﻣﻊ ﻣﻘﺳم ﻷﻛﺛر ﻣن‬ ‫ﻗﺳﻣﯾن ﻓﻘد ﺗم ﻣﻌﺎﻟﺟﺗﮫ ﻣن ﻗﺑل ﺑﺈﺧﺗﺑﺎر ﻣرﺑﻊ ﻛﺎي ﻟﻠﺗﺟﺎﻧس إذا ﻛﺎن اﻟﻣﺟﺗﻣﻊ ﻣﻘﺳم إﻟﻰ ﻗﺳﻣﯾن ﻓﺈﻧﻧﺎ‬ ‫ﺳﻧﺳﺗﺧدم اﺧﺗﺑﺎر ذي اﻟﺣدﯾن ‪ The Binomial Test‬وھو ﯾﻌﺗﻣد ﻋﻠﻰ ﻣﺎ ﯾﺳﻣﻰ ﺑﺗوزﯾﻊ ذي اﻟﺣدﯾن‪.‬‬ ‫‪٥٥٧‬‬

‫ﻧﻔرض أﻧﻧﺎ ﻣﮭﺗﻣﯾن ﺑظﺎھرة ﻣﻘﺳﻣﺔ إﻟﻰ ﻗﺳﻣﯾن وأن اﺣﺗﻣﺎل ﻧﺟﺎح ھذه اﻟظﺎھرة ﺛﺎﺑت وﯾرﻣز‬ ‫ﻟﮫ ﺑﺎﻟرﻣز ‪ p‬واﺣﺗﻣﺎل اﻟﻔﺷل ھو ‪ q‬وأن ‪ . p  q  1‬ﻣن اﻷﻣﺛﻠﺔ ﻋﻠﻰ ذﻟك أﻧﮫ ﻓﻲ دراﺳﺎت اﻟﺳوق‬ ‫ﯾﻛون أﺣﯾﺎﻧﺎ ً ﻣن اﻟﻣﻔﯾد ﻣﻌرﻓﺔ ﻧﺳﺑﺔ اﻟﻌﺎﺋﻼت اﻟذﯾن ﯾﻣﻠﻛون أﺟﮭزة ﺗﻛﯾﯾف وﻣن اﻷﺷﯾﺎء اﻟﮭﺎﻣﺔ أﯾﺿﺎ ً‬ ‫ﻣﻌرﻓﺔ ﻧﺳﺑﺔ اﻷطﻔﺎل اﻟذﯾن طﻌﻣوا ﺿد ﻣرض ﻣﻌﯾن أو ﻧﺳﺑﺔ اﻷﺷﺧﺎص اﻟذﯾن ﯾؤﯾدون ﻗرارا ً ﺳﯾﺎﺳﯾﺎ ً‬ ‫ﻣﻌﯾﻧﺎ ً‪...‬أﻟــﺦ‪ .‬ﻧﻔرض أﻧﻧﺎ أﺧذﻧﺎ ﻋﯾﻧﺔ ﻣن اﻟﺣﺟم ‪ n‬وأن اﻟﻣﺟﺗﻣﻊ اﻟﻣﺳﺣوﺑﺔ ﻣﻧﮫ اﻟﻌﯾﻧﺔ ﻣﻘﺳم إﻟﻰ‬ ‫ﻗﺳﻣﯾن ‪ c1 ,c 2‬ﺑﺣﯾث ﺗﻛون ‪ p‬ھﻲ ﻧﺳﺑﺔ ﻋﻧﺎﺻر اﻟﺟزء ‪ q ، c1‬ھﻲ ﻧﺳﺑﺔ ﻋﻧﺎﺻر اﻟﺟزء ‪ . c2‬إذا‬ ‫ﻓرﺿﻧﺎ أن ﻋدد اﻟﻌﻧﺎﺻر ﻓﻲ اﻟﻌﯾﻧﺔ اﻟﺗﻲ ﺗﻧﺗﻣﻲ إﻟﻰ اﻟﺟزء ‪ c1‬ھو ‪ x‬وأن ﻋدد اﻟﻌﻧﺎﺻر ﻓﻲ اﻟﻌﯾﻧﺔ‬ ‫اﻟﺗﻲ ﺗﻧﺗﻣﻲ إﻟﻰ اﻟﺟزء ‪ c2‬ھو )‪ . (n  x‬ﻟﺣﺳﺎب أﯾﺔ اﺣﺗﻣﺎﻻت ﻓﻲ اﻟﻌﯾﻧﺔ ﻓﺈﻧﻧﺎ ﻧﺳﺗﺧدم ﺗوزﯾﻊ ذي‬ ‫اﻟﺣدﯾن اﻵﺗﻲ‪:‬‬ ‫‪n‬‬ ‫‪x  0,1,2,...,n.‬‬ ‫‪f (x)    p x q n  x ,‬‬ ‫‪x‬‬ ‫ﻋﻠﻰ أﺳﺎس أن ‪ q , p‬ھﻲ ﻧﺳﺑﺔ اﻟﻧﺟﺎح واﻟﻔﺷل ﻋﻠﻰ اﻟﺗواﻟﻲ ﻓﻲ اﻟﻌﯾﻧﺔ‪ .‬اﻵن ﻧرﯾد أن ﻧﺧﺗﺑر ﻓرﺿﺎ ً‬ ‫ﺣول ﻣﻌﻠﻣﺔ اﻟﻣﺟﺗﻣﻊ ‪ p‬ﻓﻌﻠﻰ ﺳﺑﯾل اﻟﻣﺛﺎل ﻧﻔرض أن اﻟﻧﺳﺑﺔ ﻓﻲ اﻟﻣﺟﺗﻣﻊ ﻣﺟﮭوﻟﺔ وﺗﺳﺎوي ‪ p 0‬ﻓﮭل‬ ‫ﯾﻣﻛن اﺧﺗﺑﺎر ذﻟك اﻟﻔرض؟ ﻓﻲ ھذه اﻟﺣﺎﻟﺔ ﻧﻔﺗرض أن ﻟدﯾﻧﺎ ﻣﺗﻐﯾر ﻟﮫ ﺣﺎﻟﺗﯾن ھﻣﺎ ﻧﺟﺎح وﻓﺷل‬ ‫وﯾﺗﻛرر ‪ n‬ﻣن اﻟﻣرات اﻟﻣﺳﺗﻘﻠﺔ وﻋﻠﯾﮫ ﯾﻛون اﺣﺗﻣﺎل اﻟﻧﺟﺎح ﺛﺎﺑت وﻛذﻟك اﺣﺗﻣﺎل اﻟﻔﺷل‪ .‬ﻟﻧﻔرض أن‬ ‫‪ s‬ﺗﻣﺛل ﻋدد ﺣﺎﻻت اﻟﻧﺟﺎح ﻓﺈن ﻧﺳﺑﺔ اﻟﻧﺟﺎح ﻓﻲ اﻟﻌﯾﻧﺔ ﺗﺣﺳب ﻛﺎﻵﺗﻲ‪:‬‬ ‫‪pˆ  s / n.‬‬ ‫ﻓﻰ ھذه اﻟﺣﺎﻟﺔ ﯾﻛون ﻟدﯾﻧﺎ ﺛﻼث أزواج ﻣن اﻟﻔروض ﻛﺎﻵﺗﻲ‪:‬‬ ‫‪. H 0 : p  p 0 , H1 : p  p0‬‬ ‫اﻟﻔرض ‪: A‬‬ ‫‪. H 0 : p  p 0 , H1 : p  p 0‬‬ ‫اﻟﻔرض ‪: B‬‬ ‫‪. H 0 : p  p0 , H1 : p  p 0‬‬ ‫اﻟﻔرض ‪: C‬‬ ‫ﺳﻧرﻣز ﻹﺣﺻﺎﺋﻲ اﻻﺧﺗﺑﺎر ﺑﺎﻟرﻣز )‪ (s‬واﻟذى ﯾﻣﺛل ﻋدد ﺣﺎﻻت اﻟﻧﺟﺎح ﻓﻲ اﻟﻌﯾﻧﺔ وﺳﯾﻛون اﻟﺗوزﯾﻊ‬ ‫اﻟﻌﯾﻧﻲ اﻟذي ﯾﺗﺣﻛم ﻓﻲ اﻟرﻓض أو اﻟﻘﺑول ھو ﺗوزﯾﻊ ذي اﻟﺣدﯾن‪.‬ﺗﺧﺗﻠف ﻗواﻋد اﻟﺣﻛم ﻟﻛل زوج ﻣن‬ ‫اﻟﻔروض اﻟﺳﺎﺑﻘﺔ ﻛﺎﻵﺗﻲ‪:‬‬ ‫اﻟﻔرض ‪ : A‬اﻻﺧﺗﺑﺎر ﻓﻲ ھذه اﻟﺣﺎﻟﺔ ﻣن طرﻓﯾن ﻣﻌﻧﻰ ذﻟك أﻧﮫ ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪ ‬ﯾﻛون‬ ‫اﻹﺣﺗﻣﺎل ﻓﻲ ﻛل طرف ھو ‪  / 2‬وﺑﺎﺳﺗﺧدام ﻣﻠﺣق )‪ (٧‬ﻧﺳﺗﺧرج ‪ s1 , s 2‬واﻟﺗﻲ ﺗﺣﻘق اﻟﻣﻌﺎدﻟﺗﯾن‬ ‫اﻟﺗﺎﻟﯾﺗﯾن‪:‬‬ ‫‪P(X  s1 n, p0 )   / 2,‬‬

‫‪P(X  s 2 n, p0 )   / 2.‬‬ ‫ﯾﻘﺑل ﻓرض اﻟﻌدم إذا ﻛﺎﻧت ‪ s‬اﻟﻣﺣﺳوﺑﺔ ﻣﺣﺻورة ﻣﺎ ﺑﯾن ‪ s1 , s 2‬وﻧرﻓض ﻓﯾﻣﺎ ﻋدا ذﻟك‪.‬‬ ‫اﻟﻔرض ‪ : B‬وﻧﺣدد ﻗﯾﻣﺔ ‪ s1‬اﻟﺗﻲ ﺗﺣﻘق اﻟﻣﻌﺎدﻟﺔ اﻟﺗﺎﻟﯾﺔ ‪:‬‬ ‫‪P(x  s1 n,p0 )   ,‬‬ ‫ﻧرﻓض ﻓرض اﻟﻌدم إذا ﻛﺎﻧت ‪ s‬اﻟﻣﺣﺳوﺑﺔ أﻛﺑر ﻣن ‪ s1‬واﻟﻌﻛس ﺻﺣﯾﺢ‬ ‫اﻟﻔرض ‪ : C‬ﺑﻧﻔس اﻷﺳﻠوب ﺗﺣﺳب ﻗﯾﻣﺔ ‪ s1‬اﻟﺗﻲ ﺗﺣﻘق اﻟﻣﻌﺎدﻟﺔ‪:‬‬ ‫‪P(x  s1 n,p0 )   ,‬‬ ‫‪٥٥٨‬‬

‫وﻧرﻓض ﻓرض اﻟﻌدم إذا ﻛﺎﻧت ‪ s‬أﻛﺑر ﻣن ‪. s1‬‬ ‫ﻋﻧدﻣﺎ ﺗﻛﺑر ‪ n‬ﻛﺑﯾرا ً ﻛﺎﻓﯾﺎ ً ﺑﺷرط أن ﺗﻛون ‪ p‬ﺑﻌﯾدة ﻋن اﻟواﺣد اﻟﺻﺣﯾﺢ أو اﻟﺻﻔر ﻓﺈﻧﮫ ﯾﻣﻛن‬ ‫اﺳﺗﺧدام اﻟﺗﻘرﯾب ﻟﻠﺗوزﯾﻊ اﻟطﺑﯾﻌﻲ ﺣﯾث وﺟد أن ‪:‬‬ ‫‪Z  (s  np0 ) / np0 (1  p0 ),‬‬ ‫ﯾﺗﺑﻊ اﻟﺗوزﯾﻎ اﻟطﺑﯾﻌﻲ اﻟﻣﻌﯾﺎري‪ .‬ﺣﯾث ‪:‬‬ ‫‪s1  np0  z np0 (1  p0 ),‬‬ ‫ﻋﻧدﻣﺎ ﯾﻛون اﻻﺧﺗﺑﺎر ﻣن اﻟطرﻓﯾن ﺑﺎﺳﺗﺧدام ‪ z / 2‬اﻟﻣوﺟﺑﺔ ﻧﺢ‬ ‫ﺳب ‪ s1‬وﺑﺎﺳﺗﺧدام ‪ z / 2‬اﻟﺳﺎﻟﺑﺔ ﻧﺣﺳب ‪ s2‬وﻧﻘﺑل ﻓرض اﻟﻌدم إذا وﻗﻌت ‪ s‬ﺑﯾن ‪ . s1 , s 2‬ﻓﻲ ﺣﺎﻟﺔ‬ ‫اﺧﺗﺑﺎر اﻟﻔرض ‪ B‬ﺗﺳﺗﺧدم ‪ z ‬اﻟﻣوﺟﺑﺔ وﻧرﻓض ﻓرض اﻟﻌدم إذا ﻛﺎﻧت ‪ s‬أﻛﺑر ﻣن ‪ s1‬وﻋﻧد اﺧﺗﺑﺎر‬ ‫اﻟﻔرض ‪ C‬ﺗﺳﺗﺧدم ‪ z ‬اﻟﺳﺎﻟﺑﺔ وﻧرﻓض ﻓرض اﻟﻌدم إذا ﻛﺎﻧت ‪ s‬اﻟﻣﺣﺳوﺑﺔ أﻗل ﻣن ‪. s1‬‬

‫ﻣﺛﺎل)‪(٧-٧‬‬ ‫اﺧﺗﺑر اﻟﻔروض اﻟﺗﺎﻟﯾﺔ‪:‬‬ ‫‪H1 : p  0.6‬‬ ‫‪n  400,‬‬ ‫وذﻟك ﻋﻧد ﻣﺳﺗوى ‪.   0.05‬‬

‫‪H1 : p  0.6,‬‬

‫‪H0 : p  0.6,‬‬

‫اذا ﻛ ﺎن ‪s  228‬‬

‫اﻟﺣــل‪:‬‬ ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ ‪ Mathematica‬وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬ ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫`‪<<Statistics`DiscreteDistributions‬‬ ‫‪upperPSum[n_,p0_,s_]:=Module[{k},‬‬ ‫;]‪bdist=BinomialDistribution[n,p0‬‬ ‫;]‪upbound=PDF[bdist,s‬‬ ‫;]‪onetail=CDF[bdist,s‬‬ ‫;‪twotail=onetail‬‬ ‫;‪k=n‬‬ ‫‪While[And[PDF[bdist,k]<=upbound,k>s],twotail=twotail+PD‬‬ ‫;]‪F[bdist,k];k=k-1‬‬ ‫;]‪twotail=Min[twotail,1‬‬ ‫]}‪{onetail,twotail‬‬ ‫‪lowerPSum[n_,p0_,s_]:=Module[{k},‬‬ ‫;]‪bdist=BinomialDistribution[n,p0‬‬ ‫‪٥٥٩‬‬

upbound=PDF[bdist,s]; onetail=1-CDF[bdist,s]; twotail=onetail; k=0; While[PDF[bdist,k]<=upbound,twotail=twotail+PDF[bdist,k ];k=k+1]; Min[twotail,1]; {onetail,twotail}] npmBinomialPValue[n_,p0_,s_]:=Module[{bdist,pvals,pHat}, bdist=BinomialDistribution[n,p0]; pHat=s/n; If[pHat<=p0,pvals=upperPSum[n,p0,s]]; If[pHat>p0,pvals=lowerPSum[n,p0,s]]; Print["One-Sided PValue -> ",pvals[[1]]]; Print["Two-Sided PValue -> ",pvals[[2]]]] npmBinomialPValue[400,0.6,228] One-Sided PValue -> 0.120488 Two-Sided PValue -> 0.22109

‫ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ npmBinomialPValue[400,0.6,228]

‫ﻧﺣﺻل ﻋﻠﻰ اﻟﻣﺧرج اﻟﺗﺎﻟﻰ‬

One-Sided PValue -> 0.120488 Two-Sided PValue -> 0.22109

‫ ھﻰ‬p-value ‫ وﻗﯾﻣﺔ‬  0.05 ‫ﻻﺧﺘﺒﺎر ﻣﻦ ﺟﺎﻧﺒﯿﻦ وﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ‬ Two-Sided PValue ->

0.22109

H 0 ‫ ﻓﺈﻧﻧﺎ ﻧﻘﺑل‬0.05 ‫وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻛﺑر ﻣن‬ 228 ‫ ھﻰ‬p-value ‫ وﻗﯾﻣﺔ‬pˆ   .57  .6 ‫ﻻﺧﺘﺒﺎر ﻣﻦ ﺟﺎﻧﺐ واﺣﺪ وﺑﻤﺎ ان‬ 400 One-Sided PValue ->

0.120488

. H1 : p  0.6 ‫ وﻋﻠﻰ ذﻟك ﻧﻘﺑل ﻓرض اﻟﻌدم وﻧرﻓض اﻟﻔرض اﻟﺑدﯾل‬  0.05 ‫وھﻲ اﻛﺒﺮ ﻣﻦ‬

٥٦٠

‫)‪ (٦-٧‬اﺧﺗﺑﺎر اﻟدورات ‪Test Runs‬‬ ‫ﯾﻔﯾ د ھ ذا اﻻﺧﺗﺑ ﺎر ﻓ ﻲ ﻣﺟ ﺎﻻت ﻛﺛﯾ رة ﻣﻧﮭ ﺎ اﻟﻣﺷ ﺎﻛل اﻟﺑﯾوﻟوﺟﯾ ﺔ ‪ .‬ﻓﻘ د ﯾرﻏ ب ﺑﺎﺣ ث ﻣ ﺎ ﻓ ﻲ‬ ‫ﻣﻌرﻓﺔ ﻣﺎ إذا ﻛ ﺎن ﻋ دد ﺣ ﺎﻻت اﻹﺻ ﺎﺑﺔ ﺑﻣ رض اﻟﻣﻼرﯾ ﺎ ‪ ،‬ﻣ ﺛﻼ ً ‪ ،‬ﯾﺗﻐﯾ ر ﻋﺷ واﺋﯾﺎ ً ﻣ ن ﺳ ﻧﺔ إﻟ ﻲ ﺳ ﻧﺔ‬ ‫أﺧ رى أم أن ھﻧ ﺎك ﻋواﻣ ل ﻏﯾ ر ﻋﺷ واﺋﯾﺔ ﺗ ؤدى إﻟ ﻰ ﻧﻘ ص أو زﯾ ﺎدة ﻋ دد ﺣ ﺎﻻت اﻹﺻ ﺎﺑﺔ ﺑﮭ ذا‬ ‫اﻟﻣرض‪ .‬ﻹﺟراء اﻻﺧﺗﺑﺎر ﻧﻔﺗرض أن ﻟدﯾﻧﺎ ﻣﺗﺗﺎﺑﻌﺔ ﻣن اﻟﻣﺷﺎھدات اﻟﻣﺳ ﺟﻠﺔ ﺗﺑﻌ ﺎ ً ﻟﺗرﺗﯾ ب ﺣ دوﺛﮭﺎ وأن‬ ‫اﻟﻣﺷﺎھدات ﯾﻣﻛن ﺗﻘﺳﯾﻣﮭﺎ إﻟﻲ ﻧوﻋﯾن ) ﻟﯾﻛن ‪ . ( a , b‬ﯾﻌﺗﻣد ھذا اﻻﺧﺗﺑ ﺎر ﻋﻠ ﻰ ﻣﺗﻐﯾ ر ﯾطﻠ ق ﻋﻠﯾ ﮫ‬ ‫أﺳ م اﻟ دورة ‪ run‬ﺣﯾ ث ﺗﻌ رف ﺑﺄﻧﮭ ﺎ ﻣﺟﻣوﻋ ﺔ اﻷﺣ داث اﻟﻣﺗﺷ ﺎﺑﮭﺔ اﻟﺗ ﻲ ﯾﺳ ﺑﻘﮭﺎ أو ﯾﺗﺑﻌﮭ ﺎ ﻧ وع آﺧ ر‬ ‫ﻣﺧﺎﻟﻔﺎ ﻣن اﻷﺣداث أو ﻻ ﯾﺗﺑﻌﮭﺎ أو ﻻ ﯾﺳﺑﻘﮭﺎ أﯾﺔ أﺣداث‪ .‬ﻋدد اﻹﺣداث ﻓ ﻲ اﻟ دورة ﯾطﻠ ق ﻋﻠﯾﮭ ﺎ ط ول‬ ‫اﻟدورة ) ﯾﻣﻛن أن ﺗﺣﺗوى اﻟدورة ﻋﻠﻰ ﺣدث واﺣد(‪ .‬ﺑﻔرض أن ﻟدﯾﻧﺎ اﻟﺑﯾﺎﻧﺎت اﻟﺗﺎﻟﯾ ﺔ واﻟﺗ ﻲ ﺗ م ﺗوﻟﯾ دھﺎ‬ ‫ﻋﻠﻰ اﻟﺣﺎﺳب اﻵﻟﻲ ‪.‬‬ ‫‪0.1, 0.4, 0.2, 0.8, 0.6, 0.9, 0.3, 0.4, .01, 0.2‬‬ ‫ﺑﻔ رض أن ‪ a‬ﺗﻣﺛ ل اﻟ رﻗم اﻟ ذي أﻗ ل ﻣ ن ‪ 0.5‬و ‪ b‬ﺗﻣﺛ ل اﻟ رﻗم اﻟ ذي أﻛﺑ ر ﻣ ن ‪ 0.5‬واﻟﺗ ﻲ ﺗﻌط ﻰ‬ ‫اﻟﻣﺗﺗﺎﺑﻌﺔ ‪:‬‬ ‫‪aaaa‬‬

‫‪bbb‬‬

‫‪aaa‬‬

‫ﻓﻲ ھذه اﻟﺣﺎﻟﺔ ﻟدﯾﻧﺎ ﺛ ﻼث دورات ‪ .‬ﺳ وف ﻧرﻣ ز ﻟﻌ دد اﻟ دورات ﺑ ﺎﻟرﻣز ‪ ، r‬أي أن ‪ . r  3‬أﯾﺿ ﺎ‬ ‫‪ n1‬ﺳ وف ﺗرﻣ ز ﻟﻌ دد ﻗ ﯾم ‪ a‬و ‪ n2‬ﺳ وف ﺗرﻣ ز ﻟﻌ دد ﻗ ﯾم ‪ . b‬ﻓ رض اﻟﻌ دم واﻟﻔ رض اﻟﺑ دﯾل ﺳ وف‬ ‫ﯾﻛوﻧﺎن ﻋﻠﻰ اﻟﺷﻛل‬ ‫‪ : H 0‬اﻟﻧوﻋﺎن ﯾﻘﻌﺎن ﺑﻌﺷواﺋﯾﺔ ‪.‬‬ ‫‪ : H1‬اﻟﻧوﻋﺎن ﻻ ﯾﻘﻌﺎن ﺑﻌﺷواﺋﯾﺔ ‪.‬‬ ‫ﺑﻔ رض أن ‪ H 0‬ﺻ ﺣﯾﺢ ﻓ ﺈن ‪ r ‬ھ ﻲ ﻗﯾﻣ ﺔ ﻟﻺﺣﺻ ﺎء ‪ . R ‬اﻟﻘ ﯾم اﻟﺣرﺟ ﺔ اﻟﺳ ﻔﻠﻲ ﻟﻺﺣﺻ ﺎء ‪R ‬‬ ‫ﺗﺳﺗﺧرج ﻣ ن اﻟﺟ دول ﻓ ﻲ ﻣﻠﺣ ق )‪ (٩‬واﻟﻘ ﯾم اﻟﺣرﺟ ﺔ اﻟﻌﻠﯾ ﺎ ﻟﻺﺣﺻ ﺎء ‪ R‬ﺗﺳ ﺗﺧرج ﻣ ن اﻟﺟ دول ﻓ ﻲ‬ ‫ﻣﻠﺣ ق )‪ (١٠‬وذﻟ ك ﻋﻧ د ﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ ‪ . n1 ,n 2 ,   0.05‬ﻟ ﺗﻛن ‪ r1‬اﻟﻘﯾﻣ ﺔ اﻟﺳ ﻔﻠﻲ و ‪ r2‬اﻟﻘﯾﻣ ﺔ‬ ‫اﻟﻌﻠﯾ ﺎ ﻋﻧ د ‪ ، n1 ,n 2 ,   0.05‬ﻣﻧطﻘ ﺔ اﻟ رﻓض ﺳ وف ﺗﻛ ون ‪ R   r2‬أو ‪ . R   r1‬إذا وﻗﻌ ت‬ ‫اﻟﻘﯾﻣﺔ ‪ r‬ﻓﻲ ﻣﻧطﻘ ﺔ اﻟ رﻓض ﻧ رﻓض ‪ . H 0‬ﻟﻠﻔ رض اﻟﺑ دﯾل ‪ : H1‬اﻟﻧوﻋ ﺎن ﻻ ﯾﻘﻌ ﺎن ﺑﻌﺷ واﺋﯾﺔ ﻟوﺟ ود‬ ‫‪‬‬ ‫ﻓ ﺈن ﻣﻧطﻘ ﺔ اﻟ رﻓض ‪ . R   r2‬ﻟﻠﻔ رض اﻟﺑ دﯾل ‪: H1‬‬ ‫ﻋ دد ﻛﺑﯾ ر ﻣ ن اﻟ دورات وﻟﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ‬ ‫‪2‬‬ ‫‪‬‬ ‫ﻓ ﺈن ﻣﻧطﻘ ﺔ اﻟ رﻓض‬ ‫اﻟﻧوﻋﺎن ﻻ ﯾﻘﻌﺎن ﺑﻌﺷواﺋﯾﺔ ﻟوﺟود ﻋدد ﻗﻠﯾل ﻣن اﻟدورات وﻟﻣﺳﺗوى ﻣﻌﻧوﯾ ﺔ‬ ‫‪2‬‬ ‫‪ . R   r1‬إذا ﻛﺎﻧ ت ‪ n1 , n 2‬أﻛﺑ ر ﻣ ن ‪ ) 20‬ﻛﻠﯾﮭﻣ ﺎ أو أﯾﮭﻣ ﺎ ( ﻓ ﺈن اﻟﺟ دوﻻن ﻓ ﻲ ﻣﻠﺣ ق )‪ (٩‬و‬ ‫)‪ (١٠‬ﻻ ﯾﺻﻠﺣﺎن ﻟﻼﺳﺗﺧدام و ﻟﻘد وﺟد ﺑﺎﻟﺑرھﺎن أن‪:‬‬ ‫‪ 2n n‬‬ ‫‪‬‬ ‫‪r   1 2  1‬‬ ‫‪ n1  n 2 ‬‬ ‫‪z‬‬ ‫‪.‬‬ ‫) ‪2n1n 2 (2n1n 2  n1  n 2‬‬ ‫)‪(n1  n 2 ) 2 (n1  n 2  1‬‬ ‫ﻗﯾﻣﺔ ﻟﻣﺗﻐﯾر ﻋﺷواﺋﻲ ‪ Z‬ﺗﻘرﯾﺑﺎ ً ﯾﺗﺑﻊ اﻟﺗوزﯾﻊ اﻟطﺑﯾﻌﻲ اﻟﻘﯾﺎﺳﻲ‪.‬‬ ‫‪٥٦١‬‬

‫ﻣﺛﺎل)‪(٨-٧‬‬ ‫إذا ﻛ ﺎن ﻟ دﯾﻧﺎ ‪ 34‬ﺷﺧﺻ ﺎ وﺑﻔ رض أن ‪ M‬ﺗرﻣ ز ﻟﻠ ذﻛر و ‪ F‬ﺗرﻣ ز ﻟﻸﻧﺛ ﻰ وﻛﺎﻧ ت اﻟﻧﺗ ﺎﺋﺞ‬ ‫ﻛﺎﻵﺗﻲ ‪:‬‬ ‫‪FF MMMMMMMMM FF M FF MMMMMMMMM FFFFF MMM F‬‬

‫اﻟﻣطﻠوب ﺗﺣدﯾد ھل اﻟﻌﯾﻧﺔ ﻋﺷواﺋﯾﺔ أم ﻻ ؟ وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪.   0.05‬‬ ‫اﻟﺣــل‪:‬‬ ‫‪ : H0‬اﻟﻌﯾﻧﺔ ﻋﺷواﺋﯾﺔ‪.‬‬ ‫‪ : H1‬اﻟﻌﯾﻧﺔ ﻏﯾر ﻋﺷواﺋﯾﺔ‪.‬‬ ‫ﻣن ﺑﯾﺎﻧﺎت اﻟﻌﯾﻧﺔ ﻧﺟد أن ‪:‬‬ ‫‪ ) n 2  22 , r  9‬ﻋدد ‪ ( M‬و ‪ ) n1  12‬ﻋدد ‪ ( F‬وﺑﻣﺎ أن ‪ n  20‬ﻓﺈﻧﻧﺎ ﻧﺳﺗﺧدم اﻟﺗﻘرﯾب‬ ‫اﻟطﺑﯾﻌﻲ وﻋﻠﻰ ذﻟك ‪:‬‬ ‫‪ 2n n‬‬ ‫‪‬‬ ‫‪r   1 2  1‬‬ ‫‪ n1  n 2 ‬‬ ‫‪z‬‬ ‫) ‪2n1n 2 (2n1n 2  n1  n 2‬‬ ‫)‪(n1  n 2 ) 2 (n1  n 2  1‬‬

‫‪ 2(12)(22) ‬‬ ‫‪9‬‬ ‫‪ 1‬‬ ‫‪12  22‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫]‪2(12)(22)[2(12)(22)  12  22‬‬ ‫)‪(12  22) 2 (12  22  1‬‬ ‫‪9  16.529‬‬ ‫‪‬‬ ‫‪ 2.879.‬‬ ‫‪6.8373702‬‬ ‫ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪   0.05‬و ‪ z.025 = 1.96‬واﻟﻣﺳﺗﺧرﺟﺔ ﻣن ﺟدول اﻟﺗوزﯾﻊ اﻟطﺑﯾﻌﻲ اﻟﻘﯾﺎﺳﻲ ﻓﻲ‬ ‫ﻣﻠﺣق )‪ . (١‬ﻣﻧطﻘﺔ اﻟرﻓض ‪ Z > 1.96‬أو ‪ Z < - 1.96‬وﺑﻣﺎ أن ‪ z  2.579‬ﺗﻘﻊ ﻓﻲ ﻣﻧطﻘﺔ‬ ‫اﻟرﻓض ﻧرﻓض ‪. H 0‬‬ ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ ‪ Mathematica‬وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬ ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫;]‪Off[General::spell1‬‬ ‫`‪<<Statistics`NormalDistribution‬‬ ‫`‪<<Statistics`DataManipulation‬‬ ‫]]‪dropVals[x_]:=If[x!=med,AppendTo[newlist,x‬‬ ‫‪f[y_,m0_]:=Module[{},‬‬ ‫]]‪If[y<m0,0,1‬‬ ‫‪findFirstOne[dlist_]:=Module[{},‬‬ ‫;‪k=1‬‬ ‫;]‪While[dlist[[k]]!=1,k=k+1‬‬ ‫‪٥٦٢‬‬

k] findFirstZero[dlist_]:=Module[{}, k=1; While[dlist[[k]]!=0,k=k+1]; k] findRunOnes[datalist_]:=Module[{runList,runLength}, newVals=datalist; pos=findFirstOne[newVals]; runList=TakeWhile[newVals,equalOneQ]; runLength=Length[runList]; Drop[newVals,{pos,pos+runLength-1}]] findRunZeros[datalist_]:=Module[{runList,runLength}, newVals=datalist; pos=findFirstZero[newVals]; runList=TakeWhile[newVals,equalZeroQ]; runLength=Length[runList]; Drop[newVals,{pos,pos+runLength-1}]] firstQ[dlist_]:=If[dlist[[1]]==1,1,0] pairsOnesFirst[datalist_]:=Module[{}, temp=findRunOnes[datalist]; sumOneRuns=sumOneRuns+1; If[Length[temp]>0,temp=findRunZeros[temp]]; sumZeroRuns=sumZeroRuns+1; ] pairsOnesFirst[datalist_]:=Module[{}, temp=findRunOnes[datalist]; sumOneRuns=sumOneRuns+1; If[Length[temp]>1,temp=findRunZeros[temp]]; sumZeroRuns=sumZeroRuns+1; If[Length[temp]==1,temp={}]] pairsZerosFirst[datalist_]:=Module[{}, temp=findRunZeros[datalist]; sumZeroRuns=sumZeroRuns+1; If[Length[temp]>1,temp=findRunOnes[temp]]; sumOneRuns=sumOneRuns+1; If[Length[temp]==1,temp={}]] oneFirstLoop[dlist_]:=Module[{}, temp=dlist; While[Length[temp]>1,Do[pairsOnesFirst[temp]]]; If[Length[temp]==1,sumOneRuns=sumOneRuns+1]] zeroFirstLoop[dlist_]:=Module[{}, temp=dlist; While[Length[temp]>1,Do[pairsZerosFirst[temp]]]; If[Length[temp]==1,sumZeroRuns=sumZeroRuns+1]] pairOrder[zerosAndOnes_]:=If[firstQ[zerosAndOnes]==1,oneFirs tLoop[zerosAndOnes],zeroFirstLoop[zerosAndOnes]] ٥٦٣

p1[rVal_]:=Module[{fac1,fac2}, fac1=1/Binomial[n1+n2,n1]; fac2=2*Binomial[n1-1,rVal/2-1]*Binomial[n21,rVal/2-1]; fac1*fac2//N] p2[rVal_]:=Module[{fac1,fac2,fac3}, fac1=1/Binomial[n1+n2,n1]; fac2=Binomial[n1-1,(rVal-1)/2]*Binomial[n21,(rVal-3)/2]; fac3=Binomial[n1-1,(rVal-3)/2]*Binomial[n21,(rVal-1)/2]; fac1*(fac2+fac3)//N] pCalc[rVal_]:=Module[{}, psum=0; rtemp=rVal; While[rtemp>=2,psum=psum+prob[rtemp];rtemp--]; psum2=psum; If[tail==2,psum=upperPSum[rVal]]; psum] pCalcRMax[rVal_]:=Module[{}, psum=0; rtemp=rVal; While[rtemp<=rmax,psum=psum+prob[rtemp];rtemp++]; psum2=psum; If[tail==2,psum=lowerPSum[rVal]]; psum] prob[kVal_]:=Module[{}, If[EvenQ[kVal],p1[kVal],p2[kVal]]] upperPSum[rVal_]:=Module[{k}, upbound=prob[rVal]; twotail=psum2; k=rmax; While[And[prob[k]<=upbound,k>rVal],twotail=twotail+prob [k];k=k-1]; twotail=Min[twotail,1]; twotail] lowerPSum[rVal_]:=Module[{k}, upbound=prob[rVal]; twotail=psum2; k=2; While[And[prob[k]<=upbound,k<rVal],twotail=twotail+prob [k];k=k+1]; Min[twotail,1]; twotail] pNormal[rVal_]:=Module[{}, ٥٦٤

muSubR=2*n1*n2/(n1+n2)+1; numer=2*n1*n2*(2*n1*n2-n1-n2); denom=(n1+n2)^2*(n1+n2-1); sigSubRSquare=numer/denom; zStat=(rVal-muSubR)/Sqrt[sigSubRSquare]; dist=NormalDistribution[0,1]; psum=1-CDF[dist,Abs[zStat]]//N; If[tail==2,psum=2*psum]; psum] diff[i_,zeroOneList_]:=If[Abs[zeroOneList[[i+1]]zeroOneList[[i]]]==1,True,False] Convert to Zero's and One's convertToZerosAndOnes[datalist_]:=Module[{}, newlist={0}; med=Median[datalist]; addNonMed=Map[dropVals[#]&,datalist]; total=Length[addNonMed]; nonMedList=Drop[addNonMed[[total]],1]; Map[f[#,med]&,datalist]] Options[npmRunsTest]={sided->2,method->exact}; npmRunsTest[zeroOneList_,opts___]:=Module[{}, tail=sided/. {opts} /. Options[npmRunsTest]; mtype=method/. {opts} /. Options[npmRunsTest]; n1=Count[zeroOneList,0]; n2=Count[zeroOneList,1]; tfTable= Table[diff[i,zeroOneList],{i,1,Length[zeroOneList]-1}]; r=Count[tfTable,True]+1; rCompVal=2*n1*n2/(n1+n2)+1; rmax=Min[n1+n2,2*Min[n1,n2]+1]; If[mtype==exact,If[r<=rCompVal,pval=pCalc[r],pval=pCalc RMax[r]]]; If[mtype==approx,pval=pNormal[r]]; Print["Number of Runs -> ",r]; If[tail==2,Print["Two-Sided PValue -> ",pval]]; If[tail==1,Print["One-Sided PValue -> ",pval]]] dropVals[x_]:=If[x!=med,AppendTo[newlist,x]] f[y_,m0_]:=Module[{}, If[y<m0,0,1]] convertToZerosAndOnes[datalist_]:=Module[{}, newlist={0}; med=Median[datalist]; addNonMed=Map[dropVals[#]&,datalist]; total=Length[addNonMed]; nonMedList=Drop[addNonMed[[total]],1]; Map[f[#,med]&,datalist]] ٥٦٥

‫‪ZeroOnes={0,0,1,1,1,1,1,1,1,1,1,0,0,1,0,0,1,1,1,1,1,1,1,1,1,‬‬ ‫}‪0,0,0,0,0,1,1,1,0‬‬ ‫‪{0,0,1,1,1,1,1,1,1,1,1,0,0,1,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0‬‬ ‫}‪,1,1,1,0‬‬ ‫]‪npmRunsTest[ZeroOnes‬‬ ‫‪Number of Runs -> 9‬‬ ‫‪Two-Sided PValue -> 0.00547009‬‬ ‫]‪npmRunsTest[ZeroOnes,method->approx‬‬ ‫‪Number of Runs -> 9‬‬ ‫‪Two-Sided PValue -> 0.00398311‬‬ ‫]‪npmRunsTest[ZeroOnes,sided->1‬‬ ‫‪Number of Runs -> 9‬‬ ‫‪One-Sided PValue -> 0.00364762‬‬

‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ ‪:‬‬ ‫اوﻻ ‪ :‬اﻟﻣدﺧﻼت‬

‫وﺑﻣﺎ ان اﻟﺑﯾﺎﻧﺎت وﺻﻔﯾﺔ ﻓﺳوف ﻧﺣول ھذه اﻟﺑﯾﺎﻧﺎت إﻟﻰ ‪ 0‬او ‪ 1‬ﺣﯾث ﯾﻌطﻰ ‪ 0‬إﻟﻰ ‪F‬‬ ‫)ﻋﻠﻰ ﺳﺑﯾل اﻟﻣﺛﺎل( وﺗﻌطﻰ ‪1‬إﻟﻰ ‪ M‬ﻓﻧﺣﺻل ﻋﻠﻰ اﻟﻘﺎﺋﻣﺔ اﻟﻣﺳﻣﺎه ‪ ZeroOnes‬وھﻰ ‪:‬‬ ‫‪ZeroOnes={0,0,1,1,1,1,1,1,1,1,1,0,0,1,0,0,1,1,1,1,1,1,1,1,1,‬‬ ‫}‪0,0,0,0,0,1,1,1,0‬‬

‫ﺛﺎﻧﯾﺎ ‪ :‬اﻟﻣﺧرﺟﺎت‬ ‫ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪npmRunsTest[ZeroOnes‬‬

‫واﻟﻣﺧرج ھو‬ ‫‪0.00547009‬‬

‫‪Number of Runs -> 9‬‬ ‫>‪Two-Sided PValue -‬‬

‫‪ r ‬ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻛﺎﻟﺗﺎﻟﻰ‬ ‫‪Number of Runs -> 9‬‬

‫ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪   0.05‬وﻗﯾﻣﺔ ‪ p-value‬ھﻰ‬ ‫‪0.00547009‬‬

‫>‪Two-Sided PValue -‬‬

‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻗل ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧرﻓض ‪. H 0‬‬ ‫وﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧب واﺣد ﻧﺳﺗﺧدم اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪npmRunsTest[ZeroOnes,sided->1‬‬

‫واﻟﻣﺧرج ھو‬ ‫‪0.00364762‬‬ ‫‪ r ‬ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻛﺎﻟﺗﺎﻟﻰ‬ ‫‪Number of Runs -> 9‬‬

‫ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪   0.05‬وﻗﯾﻣﺔ ‪ p-value‬ھﻰ‬ ‫‪٥٦٦‬‬

‫‪Number of Runs -> 9‬‬ ‫>‪One-Sided PValue -‬‬

‫>‪One-Sided PValue -‬‬

‫‪0.00364762‬‬

‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧب واﺣد وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻗل ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧرﻓض ‪. H0‬‬

‫وﺑﻣﺎ أن ‪ n  20‬ﻓﺈﻧﻧﺎ ﻧﺳﺗﺧدم اﻟﺗﻘرﯾب اﻟطﺑﯾﻌﻲ و ﻧﺣﺻل ﻋﻠﻰ اﻟﻣﺧرﺟﺎت ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪npmRunsTest[ZeroOnes,method->approx‬‬

‫واﻟﻣﺧرج ھو‬ ‫‪0.00398311‬‬

‫‪Number of Runs -> 9‬‬ ‫>‪Two-Sided PValue -‬‬

‫‪ r ‬ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻛﺎﻟﺗﺎﻟﻰ‬ ‫‪Number of Runs -> 9‬‬

‫ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪   0.05‬وﻗﯾﻣﺔ ‪ p-value‬ھﻰ‬ ‫‪0.00398311‬‬

‫>‪Two-Sided PValue -‬‬

‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻗل ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧرﻓض ‪. H0‬‬

‫ﻣﺛﺎل)‪(٩-٧‬‬ ‫ﻧﻔ رض أن ﻟ دﯾﻧﺎ ﻋﯾﻧ ﮫ ﻣ ن ‪ ٢٢‬ﺷ ﺧص وﻛﻧ ﺎ ﻧرﻣ ز ﻟﻠﺷ ﺧص اﻟ ذﻛر ﻓ ﻲ اﻟﻌﯾﻧ ﺔ ﺑ ﺎﻟرﻣز )واﺣ د (‬ ‫وﻟﻸﻧﺛﻰ ﺑﺎﻟرﻣز )ﺻﻔر( وﻛﺎﻧت ﻟدﯾﻧﺎ اﻟﻧﺗﺎﺋﺞ اﻵﺗﯾﺔ‪:‬‬ ‫‪000‬‬

‫‪1111‬‬

‫‪000‬‬

‫‪111‬‬

‫‪0‬‬

‫‪11111‬‬

‫‪000‬‬

‫اﻟﻣطﻠوب ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪   0.05‬ﺗﺣدﯾد ھل اﻟﻌﯾﻧﺔ ﻋﺷواﺋﯾﺔ أم ﻻ ؟‬

‫اﻟﺣــل‪:‬‬ ‫ﻓرض اﻟﻌدم ‪ H0‬اﻟﻌﯾﻧﺔ ﻋﺷواﺋﯾﺔ ‪،‬اﻟﻔرض اﻟﺑدﯾل ‪ H1‬اﻟﻌﯾﻧﺔ ﻏﯾر ﻋﺷواﺋﯾﺔ‪.‬‬ ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﻧﻔس اﻟﺑرﻧﺎﻣﺞ اﻟﺧﺎص ﺑﺎﻟﻣﺛﺎل اﻟﺳﺎﺑق وذﻟك ﺑﺷرح اﻟﻣدﺧﻼت‬ ‫واﻟﻤﺨﺮﺟﺎت ﻟﮭﺬا اﻟﺒﺮﻧﺎﻣﺞ‬ ‫}‪ZeroOnes={0,0,0,1,1,1,1,1,0,1,1,1,0,0,0,1,1,1,1,0,0,0‬‬ ‫}‪{0,0,0,1,1,1,1,1,0,1,1,1,0,0,0,1,1,1,1,0,0,0‬‬ ‫]‪npmRunsTest[ZeroOnes‬‬ ‫‪Number of Runs -> 7‬‬ ‫‪Two-Sided PValue -> 0.0437519‬‬ ‫]‪npmRunsTest[ZeroOnes,method->approx‬‬ ‫‪Number of Runs -> 7‬‬ ‫‪Two-Sided PValue -> 0.0304864‬‬ ‫]‪npmRunsTest[ZeroOnes,sided->1‬‬ ‫‪Number of Runs -> 7‬‬ ‫‪One-Sided PValue -> 0.0241724‬‬

‫اوﻻ ‪ :‬اﻟﻣدﺧﻼت‬

‫‪٥٦٧‬‬

‫اﻟﻘﺎﺋﻣﺔ‬

‫اﻟﻣﺳﻣﺎه ‪ ZeroOnes‬وھﻲ‪:‬‬ ‫}‪ZeroOnes={0,0,0,1,1,1,1,1,0,1,1,1,0,0,0,1,1,1,1,0,0,0‬‬ ‫}‪{0,0,0,1,1,1,1,1,0,1,1,1,0,0,0,1,1,1,1,0,0,0‬‬

‫ﺛﺎﻧﯾﺎ ‪ :‬اﻟﻣﺧرﺟﺎت‬ ‫ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪npmRunsTest[ZeroOnes‬‬

‫و اﻟﻣﺧرج ھو‬ ‫‪0.0437519‬‬

‫‪Number of Runs -> 7‬‬ ‫>‪Two-Sided PValue -‬‬

‫‪ r ‬ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻟﻣﺧرج‬ ‫‪Number of Runs -> 7‬‬

‫ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪   0.05‬وﻗﯾﻣﺔ ‪ p-value‬ھﻰ‬ ‫‪0.0437519‬‬

‫>‪Two-Sided PValue -‬‬

‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻗل ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧرﻓض ‪. H0‬‬ ‫وﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧب واﺣد ﻧﺳﺗﺧدم اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪npmRunsTest[ZeroOnes,sided->1‬‬

‫و اﻟﻣﺧرج ھو‬ ‫‪0.0241724‬‬

‫‪Number of Runs -> 7‬‬ ‫>‪One-Sided PValue -‬‬

‫واﻟﻣﺧرج ھو‬ ‫‪0.0304864‬‬

‫‪Number of Runs -> 7‬‬ ‫>‪Two-Sided PValue -‬‬

‫ﺣﯾث ‪ r  7‬ﺣﯾث ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻛﺎﻟﺗﺎﻟﻰ‬ ‫‪7‬‬

‫>‪Number of Runs -‬‬

‫ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪   0.05‬وﻗﯾﻣﺔ ‪ p-value‬ھﻰ‬ ‫‪0.0304864‬‬

‫>‪Two-Sided PValue -‬‬

‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻗل ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧرﻓض ‪. H0‬‬

‫ﻣﺛﺎل)‪(١٠-٧‬‬ ‫ﺑﻔرض أن ﻟدﯾﻧﺎ اﻟﺑﯾﺎﻧﺎت اﻟﺗﺎﻟﯾﺔ ‪:‬‬ ‫‪47.93,49.34,39.43,26.89,31.45,41.94,40.72,46.24,38.19,21.94‬‬ ‫‪,30.08,26.89,18.48,34.65,20.23,21.96,15.41,23.46,52.95,19.4‬‬ ‫‪,29.72,21.26,27.25,34.79,38.35,38.92,35.87,28.49‬‬ ‫اﻟﻣطﻠوب ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪   0.05‬ﺗﺣدﯾد ھل ‪٥٦٨‬‬ ‫اﻟﻌﯾﻧﺔ ﻋﺷواﺋﯾﺔ أم ﻻ ؟‬

:‫اﻟﺣــل‬ .‫ اﻟﻌﯾﻧﺔ ﻏﯾر ﻋﺷواﺋﯾﺔ‬H1 ‫ اﻟﻌﯾﻧﺔ ﻋﺷواﺋﯾﺔ واﻟﻔرض اﻟﺑدﯾل‬H 0 ‫ﻓرض اﻟﻌدم‬ ‫ وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬Mathematica ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ‬ . ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت‬ Off[General::spell1]; <<Statistics`NormalDistribution` <<Statistics`DataManipulation` dropVals[x_]:=If[x!=med,AppendTo[newlist,x]] f[y_,m0_]:=Module[{}, If[y<m0,0,1]] findFirstOne[dlist_]:=Module[{}, k=1; While[dlist[[k]]!=1,k=k+1]; k] findFirstZero[dlist_]:=Module[{}, k=1; While[dlist[[k]]!=0,k=k+1]; k] findRunOnes[datalist_]:=Module[{runList,runLength}, newVals=datalist; pos=findFirstOne[newVals]; runList=TakeWhile[newVals,equalOneQ]; runLength=Length[runList]; Drop[newVals,{pos,pos+runLength-1}]] findRunZeros[datalist_]:=Module[{runList,runLength}, newVals=datalist; pos=findFirstZero[newVals]; runList=TakeWhile[newVals,equalZeroQ]; runLength=Length[runList]; Drop[newVals,{pos,pos+runLength-1}]] firstQ[dlist_]:=If[dlist[[1]]==1,1,0] pairsOnesFirst[datalist_]:=Module[{}, temp=findRunOnes[datalist]; sumOneRuns=sumOneRuns+1; If[Length[temp]>0,temp=findRunZeros[temp]]; sumZeroRuns=sumZeroRuns+1; ] pairsOnesFirst[datalist_]:=Module[{}, temp=findRunOnes[datalist]; sumOneRuns=sumOneRuns+1; If[Length[temp]>1,temp=findRunZeros[temp]]; sumZeroRuns=sumZeroRuns+1; ٥٦٩

If[Length[temp]==1,temp={}]] pairsZerosFirst[datalist_]:=Module[{}, temp=findRunZeros[datalist]; sumZeroRuns=sumZeroRuns+1; If[Length[temp]>1,temp=findRunOnes[temp]]; sumOneRuns=sumOneRuns+1; If[Length[temp]==1,temp={}]] oneFirstLoop[dlist_]:=Module[{}, temp=dlist; While[Length[temp]>1,Do[pairsOnesFirst[temp]]]; If[Length[temp]==1,sumOneRuns=sumOneRuns+1]] zeroFirstLoop[dlist_]:=Module[{}, temp=dlist; While[Length[temp]>1,Do[pairsZerosFirst[temp]]]; If[Length[temp]==1,sumZeroRuns=sumZeroRuns+1]] pairOrder[zerosAndOnes_]:=If[firstQ[zerosAndOnes]==1,oneFirs tLoop[zerosAndOnes],zeroFirstLoop[zerosAndOnes]] p1[rVal_]:=Module[{fac1,fac2}, fac1=1/Binomial[n1+n2,n1]; fac2=2*Binomial[n1-1,rVal/2-1]*Binomial[n21,rVal/2-1]; fac1*fac2//N] p2[rVal_]:=Module[{fac1,fac2,fac3}, fac1=1/Binomial[n1+n2,n1]; fac2=Binomial[n1-1,(rVal-1)/2]*Binomial[n21,(rVal-3)/2]; fac3=Binomial[n1-1,(rVal-3)/2]*Binomial[n21,(rVal-1)/2]; fac1*(fac2+fac3)//N] pCalc[rVal_]:=Module[{}, psum=0; rtemp=rVal; While[rtemp>=2,psum=psum+prob[rtemp];rtemp--]; psum2=psum; If[tail==2,psum=upperPSum[rVal]]; psum] pCalcRMax[rVal_]:=Module[{}, psum=0; rtemp=rVal; While[rtemp<=rmax,psum=psum+prob[rtemp];rtemp++]; psum2=psum; If[tail==2,psum=lowerPSum[rVal]]; psum] prob[kVal_]:=Module[{}, If[EvenQ[kVal],p1[kVal],p2[kVal]]] upperPSum[rVal_]:=Module[{k}, ٥٧٠

upbound=prob[rVal]; twotail=psum2; k=rmax; While[And[prob[k]<=upbound,k>rVal],twotail=twotail+prob [k];k=k-1]; twotail=Min[twotail,1]; twotail] lowerPSum[rVal_]:=Module[{k}, upbound=prob[rVal]; twotail=psum2; k=2; While[And[prob[k]<=upbound,k<rVal],twotail=twotail+prob [k];k=k+1]; Min[twotail,1]; twotail] pNormal[rVal_]:=Module[{}, muSubR=2*n1*n2/(n1+n2)+1; numer=2*n1*n2*(2*n1*n2-n1-n2); denom=(n1+n2)^2*(n1+n2-1); sigSubRSquare=numer/denom; zStat=(rVal-muSubR)/Sqrt[sigSubRSquare]; dist=NormalDistribution[0,1]; psum=1-CDF[dist,Abs[zStat]]//N; If[tail==2,psum=2*psum]; psum] diff[i_,zeroOneList_]:=If[Abs[zeroOneList[[i+1]]zeroOneList[[i]]]==1,True,False] Convert to Zero's and One's convertToZerosAndOnes[datalist_]:=Module[{}, newlist={0}; med=Median[datalist]; addNonMed=Map[dropVals[#]&,datalist]; total=Length[addNonMed]; nonMedList=Drop[addNonMed[[total]],1]; Map[f[#,med]&,datalist]] Options[npmRunsTest]={sided->2,method->exact}; npmRunsTest[zeroOneList_,opts___]:=Module[{}, tail=sided/. {opts} /. Options[npmRunsTest]; mtype=method/. {opts} /. Options[npmRunsTest]; n1=Count[zeroOneList,0]; n2=Count[zeroOneList,1]; tfTable= Table[diff[i,zeroOneList],{i,1,Length[zeroOneList]-1}]; r=Count[tfTable,True]+1; rCompVal=2*n1*n2/(n1+n2)+1; ٥٧١

rmax=Min[n1+n2,2*Min[n1,n2]+1]; If[mtype==exact,If[r<=rCompVal,pval=pCalc[r],pval=pCalc RMax[r]]]; If[mtype==approx,pval=pNormal[r]]; Print["Number of Runs -> ",r]; If[tail==2,Print["Two-Sided PValue -> ",pval]]; If[tail==1,Print["One-Sided PValue -> ",pval]]] dropVals[x_]:=If[x!=med,AppendTo[newlist,x]] f[y_,m0_]:=Module[{}, If[y<m0,0,1]] convertToZerosAndOnes[datalist_]:=Module[{}, newlist={0}; med=Median[datalist]; addNonMed=Map[dropVals[#]&,datalist]; total=Length[addNonMed]; nonMedList=Drop[addNonMed[[total]],1]; Map[f[#,med]&,datalist]] totalpayrolls={47.93,49.34,39.43,26.89,31.45,41.94,40.72,46. 24,38.19,21.94,30.08,26.89,18.48,34.65,20.23,21.96,15.41,23. 46,52.95,19.4,29.72,21.26,27.25,34.79,38.35,38.92,35.87,28.4 9}; ZeroOnes=convertToZerosAndOnes[totalpayrolls] {1,1,1,0,1,1,1,1,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,1,1,1,0} npmRunsTest[ZeroOnes] Number of Runs -> 10 Two-Sided PValue -> 0.0824696 npmRunsTest[ZeroOnes,method->approx] Number of Runs -> 10 Two-Sided PValue -> 0.0541266 npmRunsTest[ZeroOnes,sided->1] Number of Runs -> 10 One-Sided PValue -> 0.0412348

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬: ‫اوﻻ‬

‫ وھﻰ‬totalpayrolls ‫ﻧدﺧل اﻟﻘﺎﺋﻣﺔ اﻟﻣﺳﻣﺎه‬ totalpayrolls={47.93,49.34,39.43,26.89,31.45,41.94,40.72,46. 24,38.19,21.94,30.08,26.89,18.48,34.65,20.23,21.96,15.41,23. 46,52.95,19.4,29.72,21.26,27.25,34.79,38.35,38.92,35.87,28.4 9}; : ‫ﺛﺎﻧﯾﺎ اﻟﻣﺧرﺟﺎت‬ .‫ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻛﻤﺎ ﻓﻰ اﻟﻤﺜﺎل اﻟﺴﺎﺑﻖ‬ ( ‫ ( اﺧﺗﺑﺎر اﻹﺷﺎرة ﻟﻌﯾﻧﺗﯾن ﻣرﺗﺑطﺗﯾن ) ﻋﯾﻧﺔ ﻣزدوﺟﺔ‬٨-٧ ) ٥٧٢

‫‪The Sign –Test for Two Related Sample‬‬ ‫ﯾﻣﻛن ﺗﻌدﯾل اﺧﺗﺑﺎر اﻹﺷﺎرة ﻟﻌﯾﻧﺔ واﺣدة واﺳﺗﺧداﻣﮫ ﻓﻲ ﺣﺎﻟﺔ اﻟﻌﯾﻧﺗﯾن اﻟﻣرﺗﺑطﺗﯾن ﻋﻧدﻣﺎ‬ ‫ﻻﯾﺗﺣﻘق اﻻﻋﺗدال ‪ .‬ﺑﻔرض أن ﻟدﯾﻧﺎ ‪ n‬ﻣن أزواج اﻟﻣﺷﺎھدات اﻟﻣﺳﺗﻘﻠﺔ‬ ‫‪ (xi , yi ) ; i  1,2,...,n‬ﻓﺈﻧﻧﺎ ﻧﺳﺗﺑدل ﻛل زوج ﻣن اﻟﻣﺷﺎھدات ﺑﺈﺷﺎرة ﻣوﺟﺑﺔ إذا ﻛﺎﻧت‬ ‫اﻟﻣﺷﺎھدة اﻷوﻟﻲ أﻛﺑر ﻣن اﻟﻣﺷﺎھدة اﻟﺛﺎﻧﯾﺔ وﺑﺈﺷﺎرة ﺳﺎﻟﺑﺔ إذا ﻛﺎﻧت اﻟﻣﺷﺎھدة اﻷوﻟﻰ أﺻﻐر ﻣن‬ ‫اﻟﻣﺷﺎھدة اﻟﺛﺎﻧﯾﺔ وﻧﮭﻣل اﻟزوج اﻟذي ﻓﯾﮫ اﻟﻣﺷﺎھدات ﻣﺗﺳﺎوﯾﺗﺎن‪ .‬ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪H 0 :  D  0‬‬ ‫ﺿد اﻟﻔرض اﻟﺑدﯾل ‪ H1 :  D  0‬أو اﻟﻔرض اﻟﺑدﯾل ‪ H1 :  D  0‬أو اﻟﻔرض اﻟﺑدﯾل‬ ‫‪ H1 :  D  0‬ﻧﺗﺑﻊ ﻧﻔس اﻟﺧطوات اﻟﻣﺑﯾﻧﺔ ﻓﻲ اﺧﺗﺑﺎر اﻻﺷﺎرة ﻟﻌﯾﻧﺔ واﺣدة ‪.‬‬

‫ﻣﺛﺎل)‪(١١-٧‬‬ ‫ﯾﻌطﻰ اﻟﺟدول اﻟﺗﺎﻟﻰ ﻛﻣﯾﺔ ﻣرﻛب ﻓﻲ دم ﻋﯾﻧﺔ ﻋﺷواﺋﯾﺔ ﻣن ‪ 21‬ﺣﯾوان ﻗﺑل وﺑﻌد إﻋطﺎﺋﮭم دواء‬ ‫ﻟﺗدﻋﯾم اﻟﻧﻘص ﻓﻲ دورة ﻣﺎ‪ .‬اﻟﻣطﻠوب اﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪ H 0 :  D  0‬ﺿد اﻟﻔرض اﻟﺑدﯾل‬ ‫‪ H1 :  D  0‬وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪.   0.05‬‬ ‫اﻟﺣﯾوان‬ ‫ﻗﺑل ‪xi‬‬ ‫ﺑﻌد ‪yi‬‬ ‫إﺷﺎرة ) ‪(x i  yi‬‬ ‫‬‫‬‫‪0‬‬ ‫‪+‬‬ ‫‪+‬‬ ‫‬‫‬‫‪+‬‬ ‫‪+‬‬ ‫‬‫‬‫‬‫‪+‬‬ ‫‪+‬‬ ‫‬‫‬‫‪+‬‬ ‫‪+‬‬ ‫‪+‬‬ ‫‪+‬‬ ‫‪-‬‬

‫‪2.5‬‬ ‫‪1.2‬‬ ‫‪2.9‬‬ ‫‪3.1‬‬ ‫‪3.1‬‬ ‫‪1.1‬‬ ‫‪1.5‬‬ ‫‪4.1‬‬ ‫‪2.1‬‬ ‫‪2.4‬‬ ‫‪1.3‬‬ ‫‪2.8‬‬ ‫‪3.5‬‬ ‫‪3.6‬‬ ‫‪1.1‬‬ ‫‪1.6‬‬ ‫‪4.2‬‬ ‫‪2.2‬‬ ‫‪2.5‬‬ ‫‪1.3‬‬ ‫‪1.3‬‬

‫‪2.6‬‬ ‫‪1.5‬‬ ‫‪2.9‬‬ ‫‪2.0‬‬ ‫‪2.3‬‬ ‫‪1.5‬‬ ‫‪1.6‬‬ ‫‪3.1‬‬ ‫‪1.4‬‬ ‫‪2.5‬‬ ‫‪1.4‬‬ ‫‪2.9‬‬ ‫‪2.4‬‬ ‫‪2.1‬‬ ‫‪1.3‬‬ ‫‪1.7‬‬ ‫‪3.2‬‬ ‫‪1.5‬‬ ‫‪2.1‬‬ ‫‪1.1‬‬ ‫‪1.5‬‬ ‫‪٥٧٣‬‬

‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪13‬‬ ‫‪14‬‬ ‫‪15‬‬ ‫‪16‬‬ ‫‪17‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪21‬‬

:‫اﻟﺣــل‬ ‫( ﻣﻌطﺎة ﻓﻲ اﻟﺟدول اﻟﺳﺎﺑق‬xi , yi ) ; i  1,2,...,n ‫إ ِﺷﺎرات اﻟﻔروق ﺑﯾن أزواج اﻟﻣﺷﺎھدات‬ ‫ ﻓﺈن ﻋدد اﻹﺷﺎرات اﻟﻣوﺟﺑﺔ‬n  20 ‫وﺑﻌد إھﻣﺎل اﻟﻔرق اﻟﻣﺳﺎوي ﻟﻠﺻﻔر وﺗﻌدﯾل ﺣﺟم اﻟﻌﯾﻧﺔ إﻟﻰ‬ : ‫ إذا ﻛﺎﻧت‬  0.05 ‫ ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ‬H 0 ‫ ﻧرﻓض‬. k  10 P(K  k n,0.5)   . : ‫ﻧﺣﺳب‬ 10

P(K  10 20,0.5)   b(x;20,0.5)  0.588. x 0

‫ وﺑﻣﺎ‬. p  0.5 , n  20 ‫( ﻋﻧد‬١) ‫ ﺗﺳﺗﺧرج ﻣن ﺟدول ذي اﻟﺣدﯾن ﻓﻲ ﻣﻠﺣق‬0.588 ‫ﺣﯾث أن اﻟﻘﯾﻣﺔ‬ . H 0 ‫ ﻓﺈﻧﻧﺎ ﻧﻘﺑل‬0.05 ‫ أﻛﺑر ﻣن‬0.588 ‫أن‬ ‫ وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬Mathematica ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ‬ . ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت‬ Off[General::spell1]; <<Statistics`DiscreteDistributions` f[y_,m0_]:=Module[{}, If[y<m0,-1,If[y>m0,1,0]]] Options[npmSignTest]={sided->2}; npmSignTest[data_,m0_,opts___]:=Module[{m,n,signs,s,equalNum ber,tail,med,pval}, m=Length[data]; signs=Map[f[#,m0]&,data]; s=Count[signs,1]; equalNumber=Count[signs,0]; n=m-equalNumber; tail=sided/. {opts} /. Options[npmSignTest]; If[s<=n/2,pval=N[CDF[BinomialDistribution[n,1/2],s]],pv al=1-N[CDF[BinomialDistribution[n,1/2],s]]]; If[tail==2,pval=2*pval]; med=Median[data]//N; Print["Title: Sign Test"]; Print["Estimate: Sample Median -> ",med]; Print["Test Statistic: Number of Pluses is ",s]; Print["Distribution: BinomialDistribution[",n,",1/2]"]; Print[tail," - sided p-value -> ",pval]]; ٥٧٤

women={2.5,1.2,2.9,3.1,3.1,1.1,1.5,4.1,2.1,2.4,1.3,2.8,3.5,3 .6,1.1,1.6,4.2,2.2,2.5,1.3,1.3}; men={2.6,1.5,2.9,2,2.3,1.5,1.6,3.1,1.4,2.5,1.4,2.9,2.4,2.1,1 .3,1.7,3.2,1.5,2.1,1.1,1.5}; df[i_,list1_,list2_]:=list1[[i]]-list2[[i]] diffTable=Table[df[i,women,men],{i,1,Length[women]}] {-0.1,-0.3,0.,1.1,0.8,-0.4,-0.1,1.,0.7,-0.1,-0.1,0.1,1.1,1.5,-0.2,-0.1,1.,0.7,0.4,0.2,-0.2} npmSignTest[diffTable,0] Title: Sign Test Estimate: Sample Median -> 0. Test Statistic: Number of Pluses is 10 Distribution: BinomialDistribution[ 20 ,1/2] 2 - sided p-value -> 1.1762 npmSignTest[diffTable,0,sided1] Title: Sign Test Estimate: Sample Median -> 0. Test Statistic: Number of Pluses is 10 Distribution: BinomialDistribution[ 20 ,1/2] 1 - sided p-value -> 0.588099

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬

‫اﻟﺗﺎﻟﯾﺗﺎن‬

women ‫ واﻟﻘﺎﺋﻣﺔ اﻟﻣﺳﻣﺎه‬women ‫اﻟﻘﺎﺋﻣﺔ اﻟﻣﺳﻣﺎه‬ women={2.5,1.2,2.9,3.1,3.1,1.1,1.5,4.1,2.1,2.4,1.3,2.8,3.5,3 .6,1.1,1.6,4.2,2.2,2.5,1.3,1.3}; men={2.6,1.5,2.9,2,2.3,1.5,1.6,3.1,1.4,2.5,1.4,2.9,2.4,2.1,1 .3,1.7,3.2,1.5,2.1,1.1,1.5};

‫ اﻟﻣﺧرﺟﺎت‬: ‫ﺛﺎﻧﯾﺎ‬ ‫ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ npmSignTest[diffTable,0]

‫واﻟﻤﺨﺮج ھﻮ‬ Title: Sign Test Estimate: Sample Median -> 0. Test Statistic: Number of Pluses is 10 Distribution: BinomialDistribution[ 20 ,1/2] 2 - sided p-value -> 1.1762

‫ﺣﯿﺚ اﻟوﺳﯾط ﻟﻠﻌﯾﻧﺔ ھو‬ Estimate: Sample Median ->

‫وﻋدد اﻻﺷﺎرات اﻟﻣوﺟﺑﺔ ھو‬ Test Statistic: Number of Pluses is

‫ ھﻰ‬p-value ‫وﻗﯾﻣﺔ‬ 2

- sided p-value ->

1.1762 ٥٧٥

‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻛﺑر ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧﻘﺑل ‪. H 0‬اﻣﺎ إذا ﻛﺎن اﻻﺧﺗﺑﺎر‬ ‫ﻣن ﺟﺎﻧب واﺣد اى اﻟﻣطﻠوب اﺧﺗﺑﺎر ‪ H 0 :  D  0‬ﺿد اﻟﻔرض اﻟﺑدﯾل ‪ H1 :  D  0‬او‬ ‫‪) H1 :  D  0‬وﻟﻠﻌﻠم اﻟﺑرﻧﺎﻣﺞ ﻻ ﯾﺣدد ھل اﻟﺧﺗﺑﺎر ﻣن اﻟﯾﻣﯾن او ﻣن اﻟﯾﺳﺎر( وذﻟك ﻋﻧد ﻣﺳﺗوى‬ ‫ﻣﻌﻧوﯾﺔ ‪.   0.05‬‬ ‫ﻓﺈﻧﻧﺎ ﻧﺳﺗﺧدم اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪npmSignTest[diffTable,0,sided1‬‬

‫واﻟﻣﺧرج ھو‬ ‫‪Title: Sign Test‬‬ ‫‪Estimate: Sample Median -> 0.‬‬ ‫‪Test Statistic: Number of Pluses is 10‬‬ ‫]‪Distribution: BinomialDistribution[ 20 ,1/2‬‬ ‫‪1 - sided p-value -> 0.588099‬‬

‫وﻗﯾﻣﺔ ‪ p-value‬ھﻰ‬ ‫‪0.588099‬‬

‫>‪- sided p-value -‬‬

‫‪1‬‬

‫وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻛﺑر ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧﻘﺑل ‪. H 0‬‬

‫)‪ (٩ -٧‬اﺧﺗﺑﺎر ‪Mann – Whitney – Wilcoxon‬‬ ‫ﯾﺷﺗرط ﻓﻲ اﺧﺗﺑﺎر ‪ t‬اﻟ ذي ﯾﺧ ص اﻟﻔ رق ﺑ ﯾن ﻣﺗوﺳ طﻲ ﻣﺟﺗﻣﻌ ﯾن ‪ ،‬اﻟ ذي ﺗﻧﺎوﻟﻧ ﺎه ﻓ ﻲ اﻟﻔﺻ ل‬ ‫اﻟﺳﺎدس‪ ،‬أن اﻟﻣﺟﺗﻣﻌﯾن اﻟﻠذﯾن اﺧﺗرﻧﺎ ﻣﻧﮭﻣ ﺎ اﻟﻌﯾﻧﺗ ﯾن ﯾﺗﺑﻌ ﺎن ﺗوزﯾﻌ ﺎ ً طﺑﯾﻌﯾ ﺎ ً ‪ .‬ﻋﻧ دﻣﺎ ﻻ ﯾﺗ وﻓر ھ ذا‬ ‫اﻟﺷرط ﻓﺈن اﺧﺗﺑﺎر ‪ Mann-Whitney‬ﯾﻛون اﻟﺑدﯾل ‪.‬‬ ‫ﺗﺗﻛ ون اﻟﺑﯾﺎﻧ ﺎت اﻟﻼزﻣ ﺔ ﻟﻠﺗﺣﻠﯾ ل ﻣ ن ﻋﯾﻧ ﺔ ﻋﺷ واﺋﯾﺔ ﻣ ن اﻟﺣﺟ م ‪ n1‬ﻣ ن اﻟﻣﺷ ﺎھدات‬ ‫‪ x1, x 2 ,..., x n‬ﻣن اﻟﻣﺟﺗﻣ ﻊ اﻷول اﻟﻣﺗﺻ ل ‪ .‬أﯾﺿ ﺎ ﻧﺧﺗ ﺎر ﻋﯾﻧ ﺔ ﻋﺷ واﺋﯾﺔ أﺧ رى ﻣ ن اﻟﺣﺟ م ‪n 2‬‬ ‫ﻣن اﻟﻣﺷﺎھدات ‪ y1, y2 ,..., yn 2‬ﻣن اﻟﻣﺟﺗﻣﻊ اﻟﺛﺎﻧﻲ اﻟﻣﺗﺻل وﯾﺷﺗرط أن ﺗﻛون اﻟﻌﯾﻧﺗ ﯾن ﻣﺳ ﺗﻘﻠﺗﯾن‪.‬‬ ‫ﻓرض اﻟﻌدم واﻟﻔرض اﻟﺑدﯾل ﺳوف ﯾﻛوﻧﺎن ﻋﻠﻰ اﻟﺷﻛل ‪:‬‬ ‫‪ : H 0‬اﻟﻣﺟﺗﻣﻌﯾن ﻟﮭﻣﺎ ﻧﻔس اﻟﺗوزﯾﻊ ‪.‬‬ ‫‪ : H1‬ﻗﯾم ‪ xs‬ﺗﺗﺟﮫ ﻷن ﺗﻛون أﺻﻐر ﻣن ﻗﯾم ‪. ys‬‬ ‫ﻹﺟ راء اﻻﺧﺗﺑ ﺎر ﻧﻘ وم ﺑ دﻣﺞ ﻣﺷ ﺎھدات اﻟﻌﯾﻧﺗ ﯾن ﻣﻌ ﺎ ﻓ ﻲ ﻋﯾﻧ ﺔ واﺣ دة ﺛ م ﻧﻘ وم ﺑﺗرﺗﯾ ب‬ ‫اﻟﻣﺷﺎھدات ﺗﺻ ﺎﻋدﯾﺎ ً ﻓﻧﻌط ﻲ اﻟرﺗﺑ ﺔ ‪ 1‬ﻷﺻ ﻐر ﻣﺷ ﺎھدة واﻟرﺗﺑ ﺔ ‪ 2‬ﻟﻠﻣﺷ ﺎھدة اﻟﺗ ﻲ ﺗﻠﯾﮭ ﺎ ﻓ ﻲ اﻟﻌﯾﻧ ﺔ‬ ‫وھﻛ ذا ﺣﺗ ﻰ اﻟﻣﺷ ﺎھدة اﻷﺧﯾ رة واﻟﺗ ﻲ ﺗﻣﺛ ل أﻛﺑ ر ﻗﯾﻣ ﺔ ﺣﯾ ث ﺗﻌط ﻰ اﻟرﺗﺑ ﺔ ‪ . n1  n 2‬إذا ظﮭ رت‬ ‫ﻣﺷﺎھدات ﻣﺗﺳﺎوﯾﺔ ﻓﻲ اﻟﻌﯾﻧﺔ ) ﺗداﺧﻼت ( ﻓﺈﻧﻧﺎ ﻧرﺗب اﻟﻣﺷﺎھدات ﻛﻣﺎ ﻟو أﻧﮭﺎ ﻟﯾﺳ ت ﻓﯾﮭ ﺎ ﻣﺷ ﺎھدات‬ ‫ﻣﺗﺳﺎوﯾﺔ ﻓﻲ اﻟﻌﯾﻧﺔ ﺛم ﻧﺣﺳب اﻟوﺳط اﻟﺣﺳﺎﺑﻲ ﻟرﺗ ب اﻟﻣﺷ ﺎھدات ﻓ ﻲ ﻓﺋ ﺔ اﻟﻣﺷ ﺎھدات اﻟﻣﺗﺳ ﺎوﯾﺔ ﻓ ﻲ‬ ‫اﻟﻘﯾﻣﺔ وﻧﻌﺗﺑر اﻟوﺳط اﻟﺣﺳﺎﺑﻲ رﺗﺑﺔ ﻟﻛل ﻣﺷﺎھدة ﻓﻲ اﻟﻔﺋﺔ ‪ .‬ﻧﺣﺳب اﻟﻘﯾﻣﺔ ‪:‬‬ ‫)‪n (n  1‬‬ ‫‪w s 1 1‬‬ ‫‪,‬‬ ‫‪2‬‬ ‫ﺣﯾث ‪ s‬ﺗﻣﺛل ﻣﺟﻣوع اﻟرﺗ ب ﻟﻠﻌﯾﻧ ﺔ اﻟﻣﺧﺗ ﺎرة ﻣ ن اﻟﻣﺟﺗﻣ ﻊ اﻷول و ‪ w‬ﻗﯾﻣ ﺔ ﻟﻺﺣﺻ ﺎء ‪ W‬وذﻟ ك‬ ‫ﺗﺣت ﻓرض أن ‪ H 0‬ﺻﺣﯾﺢ ‪ .‬ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪ ‬ﻓ ﺈن ﻣﻧطﻘ ﺔ اﻟ رﻓض ‪ W  w ‬ﺣﯾ ث ‪ w ‬ھ ﻲ‬ ‫‪٥٧٦‬‬

‫اﻟﻘﯾﻣ ﺔ اﻟﺣرﺟ ﺔ ﻟﻺﺣﺻ ﺎء ‪ W‬وﺗﺳ ﺗﺧرج ﻣ ن اﻟﺟ دول ﻓ ﻲ ﻣﻠﺣ ق )‪ (١١‬ﻋﻧ د ‪ n1 , n2‬وﻣﺳ ﺗوﯾﺎت‬ ‫ﻣﻌﻧوﯾ ﺔ ﻣﺧﺗﻠﻔ ﺔ‪ .‬ﻟﻠﻔ رض اﻟﺑ دﯾل ‪ : H0‬ﻗ ﯾم ‪ xs‬ﺗﺗﺟ ﮫ ﻷن ﺗﻛ ون أﻛﺑ ر ﻣ ن ﻗ ﯾم ‪ ys‬ﻓ ﺈن ﻣﻧطﻘ ﺔ‬ ‫اﻟرﻓض ‪ W  w1‬ﺣﯾث ‪ w1‬ﺗﺣﺳب ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬ ‫‪w1  n1n 2  w  ,‬‬ ‫ﻟﻠﻔ رض اﻟﺑ دﯾل ‪ : H1‬اﻟﻣﺟﺗﻣﻌ ﺎن ﯾﺧﺗﻠﻔ ﺎن ﺑﺎﻟﻧﺳ ﺑﺔ ﻟﻠﻣوﻗ ﻊ ‪ ،‬ﻓ ﺈن ﻣﻧطﻘ ﺔ اﻟ رﻓض ‪ W  w ‬أو‬ ‫‪2‬‬

‫‪1‬‬

‫‪ W  w ‬ﺣﯾث ‪:‬‬ ‫‪2‬‬

‫‪ n1n 2  w  .‬‬ ‫‪2‬‬

‫‪‬‬ ‫‪2‬‬

‫‪w‬‬

‫‪1‬‬

‫ﻣﺛﺎل)‪(١٢-٧‬‬ ‫ﯾﻌط ﻰ اﻟﺟ دول اﻟﺗ ﺎﻟﻰ أزﻣﻧ ﺔ اﻟﻔﺷ ل ﻟﻧ وﻋﯾن ﻣ ن اﻷﺟﮭ زة اﻹﻟﻛﺗروﻧﯾ ﺔ واﻟﻣطﻠ وب اﺧﺗﺑ ﺎر‬ ‫ﻓ رض اﻟﻌ دم ‪ : H0‬اﻟﻣﺟﺗﻣﻌ ﯾن ﻟﮭﻣ ﺎ ﻧﻔ س اﻟﺗوزﯾ ﻊ ﺿ د اﻟﻔ رض اﻟﺑ دﯾل ‪ : H1‬اﻟﻣﺟﺗﻣﻌ ﯾن‬ ‫ﻟﯾس ﻟﮭﻣﺎ ﻧﻔس اﻟﺗوزﯾﻊ ) ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪.(   0.05‬‬ ‫‪7‬‬ ‫‪120 14‬‬ ‫‪62‬‬ ‫‪47 225 71 246 21‬‬ ‫‪x 23 261 87‬‬ ‫‪y 55 320 56 104 220 239 47 246 176 182 33‬‬

‫اﻟﺣــل‪:‬‬ ‫ﻣن اﻟﺟدول اﻟﺗﺎﻟﻰ ﯾﺗم ﺣﺳﺎب ﻣﺟﻣوع اﻟرﺗب ﻟﻠﻌﯾﻧﺔ اﻷوﻟﻲ وھﻲ ‪. s  124‬‬ ‫‪x‬‬ ‫‪87‬‬ ‫‪12‬‬

‫‪x‬‬ ‫‪71‬‬ ‫‪11‬‬ ‫‪y‬‬ ‫‪320‬‬ ‫‪23‬‬

‫‪x‬‬ ‫‪62‬‬ ‫‪10‬‬ ‫‪x‬‬ ‫‪261‬‬ ‫‪22‬‬

‫‪y‬‬ ‫‪Y‬‬ ‫‪55‬‬ ‫‪56‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪x‬‬ ‫‪Y‬‬ ‫‪246 246‬‬ ‫‪20.5 20.5‬‬

‫‪x‬‬ ‫‪47‬‬ ‫‪6.5‬‬ ‫‪y‬‬ ‫‪239‬‬ ‫‪19‬‬

‫‪y‬‬ ‫‪47‬‬ ‫‪6.5‬‬ ‫‪x‬‬ ‫‪225‬‬ ‫‪18‬‬

‫‪x‬‬ ‫‪23‬‬ ‫‪4‬‬ ‫‪y‬‬ ‫‪182‬‬ ‫‪16‬‬

‫‪y‬‬ ‫‪33‬‬ ‫‪5‬‬ ‫‪y‬‬ ‫‪220‬‬ ‫‪17‬‬

‫‪x‬‬ ‫‪21‬‬ ‫‪3‬‬ ‫‪y‬‬ ‫‪176‬‬ ‫‪15‬‬

‫‪x‬‬ ‫‪14‬‬ ‫‪2‬‬ ‫‪x‬‬ ‫‪120‬‬ ‫‪14‬‬

‫وﻋﻠﻰ ذﻟك ﻓﺈن ‪:‬‬ ‫)‪n1 (n1  1‬‬ ‫‪2‬‬ ‫)‪12(12  1‬‬ ‫‪ 124 ‬‬ ‫‪2‬‬ ‫‪ 124  78  46.‬‬ ‫‪w s‬‬

‫ﻣن اﻟﺟدول ﻓﻲ ﻣﻠﺣق )‪ (١١‬ﻓﺈن ‪ w 0.025  34‬ﻋﻧد ‪ n 2  11, n1  12‬وﺑﻣﺎ أن ‪:‬‬ ‫‪w   n1n 2  w  ,‬‬ ‫‪2‬‬

‫‪2‬‬

‫‪٥٧٧‬‬

‫‪1‬‬

‫‪x‬‬ ‫‪7‬‬ ‫‪1‬‬ ‫‪y‬‬ ‫‪104‬‬ ‫‪13‬‬

‫اﻟرﺗب‬ ‫اﻟرﺗب‬

‫ﻓﺈن ‪:‬‬ ‫‪w 0.975  (12)(11)  34  98 .‬‬ ‫ﻣﻧطﻘ ﺔ اﻟ رﻓض ‪ W  98‬أو ‪ W  34‬وﺑﻣ ﺎ أن ‪ w  46‬ﺗﻘ ﻊ ﻓ ﻲ ﻣﻧطﻘ ﺔ اﻟﻘﺑ ول ﻧﻘﺑ ل ‪. H 0‬‬ ‫ﻋﻧدﻣﺎ ‪ n 2 ,n1‬أﻛﺑ ر ﻣ ن ‪ 20‬ﻓﺈﻧﻧ ﺎ ﻻ ﻧﺳ ﺗطﯾﻊ اﺳ ﺗﺧدام اﻟﺟ دول ﻓ ﻲ ﻣﻠﺣ ق )‪ (١١‬وﻓ ﻲ ھ ذه اﻟﺣﺎﻟ ﺔ‬ ‫ﯾﻣﻛن ﺣﺳﺎب اﻟﻘﯾﻣﺔ ‪:‬‬ ‫‪nn‬‬ ‫‪w 1 2‬‬ ‫‪2‬‬ ‫‪z‬‬ ‫‪.‬‬ ‫‪(n1n 2 )(n1  n 2  1) /12‬‬

‫واﻟﺗﻲ ﺗﻣﺛل ﻗﯾﻣﺔ ﻟﻠﻣﺗﻐﯾر اﻟﻌﺷواﺋﻲ ‪Z‬وھو ﺗﻘرﯾﺑﺎ ً ﯾﺗﺑﻊ اﻟﺗوزﯾﻊ اﻟطﺑﯾﻌ ﻲ اﻟﻘﯾﺎﺳ ﻲ وذﻟ ك ﺗﺣ ت ﻓ رض‬ ‫أن ‪ H 0‬ﺻﺣﯾﺢ ‪.‬‬ ‫ﺑﺎﻟﻧﺳﺑﺔ ﻟﻣﺷﻛﻠﺔ اﻟﺗداﺧﻼت واﻟﺗﻲ ﻗد ﺗﺣ دث داﺧل ﻛ ل ﻣﺟﻣوﻋ ﺔ أو ﺑ ﯾن ﻗ ﯾم اﻟﻣﺟﻣ وﻋﺗﯾن ﻓﻘ د ﺗ م‬ ‫إﺛﺑﺎت أن اﻟﺗداﺧﻼت داﺧل اﻟﻣﺟﻣوﻋﺔ ﻟﯾس ﻟﮭﺎ ﺗﺄﺛﯾر ﻋﻠﻰ ﻗﯾﻣﺔ اﻹﺣﺻﺎء وﻟﻛ ن وﺟ ود ﺗ داﺧﻼت ﺑ ﯾن‬ ‫اﻟﻣﺟﻣ وﻋﺗﯾن ﯾ ؤﺛر ﻋﻠ ﻰ اﻟﻧﺗ ﺎﺋﺞ ‪ .‬ﻓ ﻲ ھ ذه اﻟﺣﺎﻟ ﺔ ﻻ ﺑ د ﻣ ن ﻋﻣ ل ﺗﺻ ﺣﯾﺢ ﻟﻘﯾﻣ ﺔ ‪ z‬ﺑﻔ رض أن ‪u‬‬ ‫ﺗرﻣز ﻟﻌدد اﻟﺗداﺧﻼت ﻟرﺗب ﻣﻌطﺎة ﻓﺈن ﻣﻌﺎﻣل اﻟﺗﺻﺣﯾﺢ ﻟﻠﺗداﺧﻼت ﯾﺣﺳب ﻣن اﻟﺻﯾﻐﺔ اﻟﺗﺎﻟﯾﺔ ‪:‬‬ ‫)‪n1n 2 (u 3  u‬‬ ‫‪,‬‬ ‫)‪12(n1  n 2 )(n1  n 2  1‬‬ ‫واﻟﺗﻲ ﺗطرح ﻣن اﻟﻣﻘﺎم ﻓﻲ ﺻﯾﻐﺔ ‪ z‬ﺗﺣت اﻟﺟذر ‪ .‬وﻋﻠﻰ ذﻟك اﻟﻣﻘﺎم ﻓﻲ ﺻﯾﻐﺔ ‪ z‬ﯾﺻﺑﺢ ‪:‬‬

‫)‪(n1n 2 )(n1  n 2  1‬‬ ‫)‪n1n 2 (u 3  u‬‬ ‫‪‬‬ ‫‪.‬‬ ‫‪12‬‬ ‫)‪12(n1  n 2 )(n1  n 2  1‬‬

‫ﻣﺛﺎل)‪(١٣-٧‬‬ ‫ﻟﻣﻘﺎرﻧﺔ اﻟﻘﻠوﯾﺔ ﻓﻰ ﺑرﻛﺗﯾن )ﻣﻘﺎﺳﺔ ﺑﺎﻟﻣﻠﯾﺟرام ﻟﻛل ﻟﺗر( أﺧذت ﻋﯾﻧﺗﯾن ﻣن اﻟﺣﺟم ‪ 5‬ﻣ ن ﻛ ل‬ ‫ﺑرﻛﺔ‪ .‬أﺳﺗﺧدم اﺧﺗﺑﺎر ‪ Wilcoxon‬ﻟﺗﻘدﯾر ﻣﺎ إذا ﻛﺎن ھﻧ ﺎك ﻓ رق ﻣﻌﻧ وى ﻓ ﻰ اﻟﻘﻠوﯾ ﺔ ﺑ ﯾن‬ ‫اﻟﺑرﻛﺗﯾن‪:‬‬ ‫‪102‬‬ ‫‪116‬‬ ‫‪122‬‬ ‫‪112‬‬ ‫‪104‬‬ ‫اﻟﺑرﻛﺔ ‪1‬‬ ‫‪108‬‬ ‫‪117‬‬ ‫‪115‬‬ ‫‪120‬‬ ‫‪105‬‬ ‫اﻟﺑرﻛﺔ ‪2‬‬

‫اﻟﺣــل‪:‬‬ ‫‪ w .250  3‬اﻟﻣﺳﺗﺧرﺟﮫ ﻣن اﻟﺟدول ﻓﻰ ﻣﻠﺣق )‪ (١١‬ﻋﻧد ‪n1  5 ,‬‬

‫‪٥٧٨‬‬

‫‪n2  5 ,‬‬

‫ وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬Mathematica ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ‬ ‫وﻟﻠﻌﻠﻢ ﻓﺈن اﻟﺒﺮﻧﺎﻣﺞ ﯾﺼﻠﺢ ﻓﻘﻂ ﻋﻨﺪ ﻋﺪم وﺟﻮد ﺗﺪاﺧﻼت‬. ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت‬

list1={102,116,122,112,104}; list2={108,117,115,120,105}; Off[General::spell1]; <<Statistics`DataManipulation` <<Statistics`NormalDistribution` rank[j_,xlist_]:=Module[{}, k=1; flag=0; xsort=Sort[xlist]; poslist=Position[xsort,xlist[[j]]]; k=poslist[[1,1]]; m=Length[poslist]; Sum[val,{val,k,k+m-1}]/m//N] reorder[list1_,list2_]:=Module[{len1,len2,newlist}, len1=Length[list1]; len2=Length[list2]; If[len1<=len2,newlist={list1,list2},newlist={list2,list 1}]; newlist] w-probability Clear[n,p] n[k_,n1_,n2_]:=1 /; k===n1*(n1+1)/2 && n2===0 n[k_,n1_,n2_]:=0 /;k!=n1*(n1+1)/2 && n2===0 n[k_,n1_,n2_]:=1 /; k===0 && n1===0 n[k_,n1_,n2_]:=0 /; k!=0 && n1===0 n[k_,n1_,n2_]:=0 /; k<n1*(n1+1)/2 n[k_,n1_,n2_]:=0 /; k>n1*(n1+2*n2+1)/2 n[k_,n1_,n2_]:=n[k,n1,n2]=n[k,n1,n2-1]+n[k-n1-n2,n1-1,n2] p[w_,n1_,n2_]:=Sum[n[k,n1,n2]/Binomial[n1+n2,n1],{k,n1*(n1+1 )/2,w}]//N Other Functions exactCalc[w_]:=Module[{}, If[w<=mn,onetail=p[w,n1,n2],onetail=1-p[w,n1,n2]]; If[tail==1,val=onetail,val=Min[2*onetail,1]]; val] normalApprox[w_,both_]:=Module[{fcount,tsum,numer,one,two,zS tat,dist,onetail,val}, u=w-(n1*(n1+1)/2); ٥٧٩

fcount=Frequencies[both]; tsum=Sum[fcount[[j,1]]^3fcount[[j,1]],{j,1,Length[fcount]}]; numer=u-n1*n2/2; one=n1*n2*(n1+n2+1)/12; two=n1*n2*tsum/(12*(n1+n2)*(n1+n2-1)); zStat=numer/Sqrt[one-two]; dist=NormalDistribution[0,1]; onetail=1-CDF[dist,Abs[zStat]]; If[tail==1,val=onetail,val=2*onetail]; N[val]] Options[npmMannWhitneyTest]={sided->2,mthd->approx}; Clear[npmMannWhitneyTest] npmMannWhitneyTest[list1_,list2_,opts___]:=Module[{}, tail=sided/. {opts} /. Options[npmMannWhitneyTest]; mtype=mthd/. {opts} /. Options[npmMannWhitneyTest]; sortList=reorder[list1,list2]; n1=Length[sortList[[1]]]; n2=Length[sortList[[2]]]; m1=Median[sortList[[1]]]//N; m2=Median[sortList[[2]]]//N; min=n1*(n1+1)/2; max=n1*(n1+2*n2+1)/2; mn=n1*(n1+n2+1)/2; var=n1*n2*(n1+n2+1)/12; both=Join[sortList[[1]],sortList[[2]]]; rank1=Table[{both[[k]],rank[k,both]},{k,1,Length[both]} ]; wTestStat=Sum[rank1[[j,2]],{j,1,Length[sortList[[1]]]}] ; u=wTestStat-(n1*(n1+1)/2); If[mtype==approx,pval=normalApprox[wTestStat,both]]; If[mtype==exact,pval=exactCalc[wTestStat]]; Print["Title: Mann-Whitney Test"]; Print["Sample Medians: ", m1,", ",m2]; Print["Test Statistic: ",u]; If[mtype==approx,Print["Distribution: Normal Approximation"]];If[mtype==exact,Print["Distribution: Exact"]]; Print[tail," - Sided PValue -> ",pval]] npmMannWhitneyTest[list1,list2] Title: Mann-Whitney Test ٥٨٠

Sample Medians: 112. , 115. Test Statistic: 10. Distribution: Normal Approximation 2 - Sided PValue -> 0.601508 npmMannWhitneyTest[list1,list2,mthd->exact] Title: Mann-Whitney Test Sample Medians: 112. , 115. Test Statistic: 10. Distribution: Exact 2

- Sided PValue ->

0.690476

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬ : ‫ﻗﺎﺋﻣﺔ ﺧﺎﺻﺔ ﺑﺎﻟﻣﺟﻣوﻋﺔ اﻻوﻟﻰ‬ list1={102,116,122,112,104};

: ‫ﻗﺎﺋﻣﺔ ﺧﺎﺻﺔ ﺑﺎﻟﻣﺟﻣوﻋﺔ اﻟﺛﺎﻧﯾﺔ‬ list2={108,117,115,120,105};

: ‫ﺛﺎﻧﯾﺎ اﻟﻣﺧرﺟﺎت‬ ‫ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ npmMannWhitneyTest[list1,list2]

‫واﻟﻤﺨﺮج ھﻮ‬ Title: Mann-Whitney Test Sample Medians: 112. , 115. Test Statistic: 10. Distribution: Normal Approximation 2 - Sided PValue -> 0.601508

‫ﺣﯿﺚ اﻟوﺳﯾط ﻟﻠﻌﯾﻧﺔ اﻻوﻟﻰ واﻟﺛﺎﻧﯾﺔ ﻋﻠﻰ اﻟﺗواﻟﻰ ھﻰ‬ Sample Medians:

112. ,

115.

‫ ﺑﺎﺳﺗﺧدام اﻟﺗوزﯾﻊ اﻟطﺑﯾﻌﻰ ھو‬w ‫واﻻﺣﺻﺎء‬ Test Statistic: 10. Distribution: Normal Approximation

‫ ھﻰ‬p-value ‫وﻗﯾﻣﺔ‬ 2

- Sided PValue ->

0.601508

. H 0 ‫ ﻓﺈﻧﻧﺎ ﻧﻘﺑل‬0.05 ‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻛﺑر ﻣن‬

‫ اﻟﻣﺿﺑوط ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ ﻣﻊ اﻟﻣﺧرج‬w ‫واﻻﺣﺻﺎء‬ npmMannWhitneyTest[list1,list2,mthd->exact] Title: Mann-Whitney Test Sample Medians: 112. , 115. Test Statistic: 10. Distribution: Exact 2

- Sided PValue ->

0.690476

Kruskal – Wallis ‫ ( اﺧﺗﺑﺎر‬١٠-٧) ٥٨١

‫ﯾﻌﺗﺑ ر اﺧﺗﺑ ﺎر ‪ Kruskal – Wallis‬ﻣ ن أﻛﺛ ر اﻻﺧﺗﺑ ﺎرات اﻟﻼﻣﻌﻠﻣﯾ ﺔ ﺷ ﯾوﻋﺎ ً ﻹﺟ راء ﺗﺣﻠﯾ ل‬ ‫اﻟﺗﺑﺎﯾن ﻓﻲ ﺣﺎﻟﺔ اﻟﺗﺻﻧﯾف اﻷﺣﺎدي‪ .‬ﺗﺗﻛون اﻟﺑﯾﺎﻧﺎت اﻟﻼزﻣ ﺔ ﻟﻠﺗﺣﻠﯾ ل ﻣ ن ‪ k‬ﻣ ن اﻟﻌﯾﻧ ﺎت اﻟﻌﺷ واﺋﯾﺔ‬ ‫ﻣن اﻟﺣﺟم ‪ n1, n 2 ,...,n k‬ﻋﻠﻰ أن ﺗﻛون اﻟﻣﺷﺎھدات ﻣﺳﺗﻘﻠﺔ ﺳواء ﺑﯾن أو داﺧل اﻟﻣﻌﺎﻟﺟﺎت ﻛﻣ ﺎ أن‬ ‫اﻟﻣﺟﺗﻣﻌ ﺎت اﻟﺗ ﻲ اﺧﺗﯾ رت ﻣﻧﮭ ﺎ اﻟﻌﯾﻧ ﺎت ﺗﻛ ون ﻣ ن اﻟﻧ وع اﻟﻣﺗﺻ ل ‪ .‬ﻓ رض اﻟﻌ دم واﻟﻔ رض اﻟﺑ دﯾل‬ ‫ﺳوف ﯾﻛوﻧﺎن ﻋﻠﻰ اﻟﺷﻛل ‪:‬‬ ‫‪ : H 0‬اﻟﺗوزﯾﻌﺎت ﻟﻠﻣﺟﺗﻣﻌﺎت اﻟﺗﻲ ﻋددھﺎ ‪ k‬ﻣﺗطﺎﺑﻘﺔ ‪.‬‬ ‫‪ : H1‬اﻟﺗوزﯾﻌﺎت ﻟﻠﻣﺟﺗﻣﻌﺎت اﻟﺗﻲ ﻋددھﺎ ‪ k‬ﻟﯾس ﻟﮭﺎ ﻧﻔس اﻟوﺳﯾط ‪.‬‬ ‫ﻓﻲ ھذا اﻻﺧﺗﺑﺎر ﻧﻘوم ﺑدﻣﺞ ﻣﺷﺎھدات اﻟﻌﯾﻧﺎت ﻓﻲ ﻋﯾﻧﺔ واﺣدة وإﻋطﺎء رﺗ ب ﻟﮭ ذه اﻟﻣﺷ ﺎھدات‬ ‫ﺣﯾ ث ‪ R i‬ﺗرﻣ ز إﻟ ﻰ ﻣﺟﻣ وع اﻟرﺗ ب ﻟﻠﻣﺷ ﺎھدات اﻟﺗ ﻲ ﺗﻧﺗﻣ ﻲ إﻟ ﻰ اﻟﻣﺟﻣوﻋ ﺔ رﻗ م ‪ i‬واﻟﺗ ﻲ ﻋ دد‬ ‫‪k‬‬

‫ﻣﺷﺎھداﺗﮭﺎ ‪ ni‬ﺣﯾث أن ‪ N   n i‬ﺗﻣﺛل اﻟﻌدد اﻟﻛﻠ ﻲ ﻟﻠﻣﺷ ﺎھدات اﻟﻧﺎﺗﺟ ﺔ ﻣ ن ‪ k‬ﻣ ن اﻟﻌﯾﻧ ﺎت ‪.‬‬ ‫‪i 1‬‬

‫ﻗﯾﻣﺔ اﻹﺣﺻﺎء اﻟذي ﯾﻌﺗﻣد ﻋﻠﯾﮫ ﻗرارﻧﺎ ھو ‪:‬‬ ‫‪2‬‬

‫‪k‬‬ ‫‪12‬‬ ‫‪1‬‬ ‫‪n i (N  1) ‬‬ ‫‪h‬‬ ‫‪R‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪i‬‬ ‫‪ ,‬‬ ‫‪N(N  1) i 1 n i ‬‬ ‫‪2‬‬ ‫واﻟﺗ ﻲ ﯾﻣﻛ ن ﺗﺑﺳ ﯾطﮫ‬ ‫ﺑﺎﻟﺻﯾﻐﺔ اﻟﺗﺎﻟﯾﺔ ‪:‬‬ ‫‪k R2‬‬ ‫‪12‬‬ ‫‪i‬‬ ‫‪h‬‬ ‫‪ 3(N  1).‬‬ ‫‪‬‬ ‫‪N(N  1) i 1 n i‬‬ ‫ﺣﯾ ث أن ‪ h‬ھ و ﻗﯾﻣ ﺔ ﻟﻺﺣﺻ ﺎء ‪ H‬ﺑ ﺎﻓﺗراض أن ‪ H 0‬ﺻ ﺣﯾﺢ ‪.‬ﺗﺳ ﺗﺧرج اﻟﻘ ﯾم اﻟﺣرﺟ ﺔ ﻟﻺﺣﺻ ﺎء‬ ‫‪ H‬ﻣن اﻟﺟدول ﻓﻲ ﻣﻠﺣق )‪ (١٢‬ﻋﻧد ﻣﺳﺗوﯾﺎت ﻣﺧﺗﻠﻔﺔ ﻣن اﻟﻣﻌﻧوﯾ ﺔ ﺑﺷ رط أن ‪n i  5,i  1, 2,3‬‬ ‫‪ .‬إذا ﻛﺎﻧت اﻟﻘﯾﻣﺔ اﻟﻣﺣﺳوﺑﺔ اﻛﺑر ﻣن اﻟﻘﯾﻣﺔ اﻟﺣرﺟﺔ ﻧرﻓض ﻓ رض اﻟﻌ دم‪ .‬إذا ﻛ ﺎن ﻋ دد اﻟﻌﯾﻧ ﺎت أو‬ ‫ﻋ دد اﻟﻣﺷ ﺎھدات داﺧ ل ﻛ ل ﻋﯾﻧ ﺔ ﻏﯾ ر ﻣﺗ وﻓر ﻓ ﻲ اﻟﺟ دول ﻓﻘ د وﺟ د ﺑﺎﻟﺑرھ ﺎن أن ‪ H‬ﺗﻘرﯾﺑ ﺎ ً ﺗﺗﺑ ﻊ‬ ‫‪ 2‬ﺗﺣ ت ﺷ رط أن ﻋ دد اﻟﻣﺷ ﺎھدات ﻓ ﻲ ﻛ ل ﻋﯾﻧ ﺔ ﻻ ﺗﻘ ل ﻋ ن ‪ 5‬أي أن‬ ‫ﺗوزﯾ ﻊ‬

‫‪ . n i  5,i  1, 2,..., k‬ﻟﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ ‪ ‬ﻓ ﺈن ﻣﻧطﻘ ﺔ اﻟ رﻓض ‪ X2  2‬ﺣﯾ ث أن ‪2‬‬ ‫ﺗﺳﺗﺧرج ﻣن ﺟدول ﺗوزﯾﻊ ‪ 2‬ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪ ‬ودرﺟ ﺎت ﺣرﯾ ﺔ ‪ . k  1‬ﻧ رﻓض ‪ H 0‬إذا‬ ‫وﻗﻌت ‪ h‬اﻟﻣﺣﺳوﺑﺔ ﻓﻲ ﻣﻧطﻘﺔ اﻟرﻓض‪.‬‬ ‫ﻓﻲ ﺣﺎﻟﺔ وﺟود ﺗداﺧﻼت ﻻ ﺑد ﻣن ﺗﺻﺣﯾﺢ اﻹﺣﺻﺎء ‪ H‬إﻟﻰ اﻹﺣﺻﺎء ‪ H ‬ﺣﯾث ‪:‬‬ ‫‪H‬‬ ‫) ‪(u 3i  u i‬‬ ‫‪H  ,‬‬ ‫‪C 1‬‬ ‫‪C‬‬ ‫‪N3  N‬‬ ‫ﺣﯾث ‪ ui‬ھو ﻋدد اﻟﻘﯾم ﻓﻲ ﻛل ﻓﺋﺔ ﺑﮭﺎ ﻗﯾم ﻣﺗﺳﺎوﯾﺔ و ‪ N‬ھﻲ اﻟﻘﯾﻣﺔ اﻟﻧﺎﺗﺟ ﺔ ﻣ ن دﻣ ﺞ اﻟﻌﯾﻧ ﺎت اﻟﺗ ﻲ‬ ‫‪k‬‬

‫ﻋددھﺎ ‪ k‬أي أن ‪ . N   n i‬ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪ ‬ﻓﺈن ﻣﻧطﻘﺔ اﻟرﻓض ‪ H  2‬ﺣﯾث‪:‬‬ ‫‪i 1‬‬

‫‪٥٨٢‬‬

‫‪2‬‬

‫‪ ‬ﺗﺳ ﺗﺧرج ﻣ ن ﺟ دول ﺗوزﯾ ﻊ ‪ ‬ﻓ ﻲ ﻣﻠﺣ ق )‪ (٣‬ﺑ درﺟﺎت ﺣرﯾ ﺔ ‪. k  1‬إذا وﻗﻌ ت ‪h‬‬ ‫اﻟﻣﺣﺳوﺑﺔ )‪ (h   h‬ﻓﻲ ﻣﻧطﻘ ﺔ اﻟ رﻓض ﻧ رﻓض ‪ . H0‬إذا ﻛﺎﻧ ت اﻟﻘﯾم اﻟﻣﺗداﺧﻠ ﺔ ﻗﻠﯾﻠ ﺔ ﻓ ﻲ اﻟﻌﯾﻧ ﺎت‬ ‫ﻓﺈن ‪ H ‬ﺗﻘﺗرب ﻣن ‪. H‬‬

‫ﻣﺛﺎل)‪(١٤-٧‬‬ ‫ﻓﻲ دراﺳﺔ ﻋﻠﻰ ﻧوع ﻣﻌﯾن ﻣن اﻟطﺣﺎﻟب ﻓﻲ ﺑﺣﯾرة ﺻﻐﯾرة ﻗﺎم ﺑﺎﺣث ﺑﻣﻘﺎرﻧ ﺔ اﻟﻧﺳ ﺑﺔ اﻟﻣﺋوﯾ ﺔ ﻟﻌ دد‬ ‫اﻟطﺣﺎﻟ ب ﻣ ن ﻧ وع ﻣ ﺎ ﻓ ﻲ ﻣﻧطﻘ ﺔ ﻣﺣ ددة ﺧ ﻼل ﻓﺻ ل اﻟرﺑﯾ ﻊ وﻣﻧﺗﺻ ف اﻟﺻ ﯾف وآﺧ ر اﻟﺧرﯾ ف‬ ‫واﻟﺑﯾﺎﻧﺎت ﻣﻌطﺎة ﻓﻲ اﻟﺟدول‬ ‫اﻟﺗﺎﻟﻰ ‪:‬‬ ‫‪ 9.6 11.2 11.6 11.7 12.8 12.9 15.8 22.7 24.6 32.5‬اﻟرﺑﯾﻊ ‪%‬‬ ‫‪32.8‬‬

‫‪26.4‬‬

‫‪25.6‬‬

‫‪24.6‬‬

‫‪21.1‬‬

‫‪9.6‬‬

‫‪9.2‬‬

‫‪7.6‬‬

‫‪4.8‬‬

‫‪11.7‬‬

‫‪11.3‬‬

‫‪10.7‬‬

‫‪10.2‬‬

‫‪9.5‬‬

‫‪8.8‬‬

‫‪8.0‬‬

‫‪7.1‬‬

‫‪6.5‬‬

‫‪ 5.4‬آﺧر اﻟﺧرﯾف‬ ‫‪%‬‬

‫ﻣﻧﺗﺻف‬ ‫اﻟﺻﯾف ‪%‬‬

‫اﻟﺣــل‪:‬‬ ‫اﻟﻧﺗﺎﺋﺞ اﻟﻼزﻣﺔ ﻟﺣﺳﺎب ‪ h‬ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ‬ ‫‪Ri‬‬

‫‪11.5 15 17 18.5 20 21 22 24 25.5 29 203.5‬‬ ‫اﻟرﺑﯾﻊ‬ ‫‪1 5.5 5.5‬‬ ‫‪9 11.5 23 25.5 27 28 30‬‬ ‫‪166‬‬ ‫ﻣﻧﺗﺻف‬ ‫اﻟﺻﯾف‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪7‬‬ ‫‪8 10 13 14 16 18.5 95.5‬‬ ‫آﺧر اﻟﺧرﯾف‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪ R1 R 2 R 32 ‬‬ ‫‪12‬‬ ‫‪h‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫)‪  3(N  1‬‬ ‫‪N(N  1)  n1 n 2 n 3 ‬‬

‫‪12  203.52 1662 95.52 ‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪  3(31)  7.759.‬‬ ‫‪30(31)  10‬‬ ‫‪10‬‬ ‫‪10 ‬‬ ‫وﻟوﺟود ﺗداﺧﻼت ﻻﺑد ﻣن ﺗﺣوﯾل ‪ h‬إﻟﻰ ‪ . h‬ﻧﺣﺳب ‪ u 3  u‬ﻟﻛل ﺗداﺧل ﺛم ﻧﺣﺳب‬ ‫)‪  (u3  u‬ﻛﺎﻟﺗﺎﻟﻲ ‪:‬‬ ‫‪(8  2)  (8  2)  (8  2)  (8  2)  24 .‬‬ ‫ﻣﻌﺎﻣل اﻟﺗﺻﺣﯾﺢ ﯾﺣﺳب ﻣن اﻟﺻﯾﻐﺔ اﻟﺗﺎﻟﯾﺔ ‪:‬‬

‫‪٥٨٣‬‬

‫)‪(u 3  u‬‬ ‫‪1 3‬‬ ‫‪N N‬‬ ‫‪24‬‬ ‫‪1‬‬ ‫‪ 0.9991,‬‬ ‫‪27000  30‬‬ ‫وﻋﻠﻰ ذﻟك ‪:‬‬

‫‪7.759‬‬ ‫‪ 7.76599.‬‬ ‫‪0.9991‬‬ ‫‪2‬‬ ‫‪ .05  5.992‬واﻟﻣﺳ ﺗﺧرﺟﺔ ﻣ ن ﺟ دول ﺗوزﯾ ﻊ ‪ 2‬ﻓ ﻲ ﻣﻠﺣ ق )‪ (٣‬ﻋﻧ د درﺟ ﺎت ﺣرﯾ ﺔ‬ ‫‪ . k  1  3  1  2‬ﻣﻧطﻘ ﺔ اﻟ رﻓض ‪ . X2 > 5.992‬وﺑﻣ ﺎ أن ‪ h   7.76599‬ﺗﻘ ﻊ ﻓ ﻲ ﻣﻧطﻘ ﺔ‬ ‫اﻟرﻓض ﻧرﻓض ‪. H 0‬‬ ‫‪h ‬‬

‫ﻣﺛﺎل)‪(١٥-٧‬‬ ‫اﺳﺗﺧدﻣت ﺛﻼﺛﺔ ط رق ﺗﻌﻠﯾﻣﯾ ﺔ ﻣﺧﺗﻠﻔ ﺔ ﻟﺗﻌﻠ ﯾم ﺛﻼﺛ ﺔ ﻣﺟﻣوﻋ ﺎت ﻣﺗﺷ ﺎﺑﮭﺔ ﻣ ن اﻟطﻠﺑ ﺔ وﻛﺎﻧ ت‬ ‫درﺟﺎت اﻻﻣﺗﺣﺎن اﻟﻧﮭﺎﺋﻲ ﻣﻌطﺎة ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ‬ ‫اﻟﻣﺟﻣوﻋﺔ ‪A‬‬ ‫‪20‬‬ ‫‪37‬‬ ‫‪39‬‬ ‫‪41‬‬ ‫‪45‬‬ ‫‪53‬‬

‫‪48‬‬

‫‪46‬‬

‫‪43‬‬

‫اﻟﻣﺟﻣوﻋﺔ ‪B‬‬

‫‪44‬‬

‫‪38‬‬

‫‪31‬‬

‫اﻟﻣﺟﻣوﻋﺔ ‪C‬‬

‫ھل ﺗوﺟد ﻓروق ﻣﻌﻧوﯾﺔ ﺑﯾن اﻟطرق اﻟﺛﻼﺛﺔ ؟وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪.   0.05‬‬

‫اﻟﺣــل‪:‬‬ ‫‪ : H0‬ﺗوزﯾﻌﺎت اﻟﻣﺟﺗﻣﻌﺎت اﻟﺛﻼﺛﺔ اﻟﺗﻲ اﺧﺗﯾرت ﻓﯾﮭﺎ اﻟﻌﯾﻧﺎت اﻟﺛﻼﺛﺔ ﻣﺗطﺎﺑﻘﺔ ‪.‬‬ ‫‪ : H1‬ﺗوزﯾﻌﺎت اﻟﻣﺟﺗﻣﻌﺎت اﻟﺛﻼﺛﺔ ﻟﯾس ﻟﮭﺎ ﻧﻔس اﻟوﺳﯾط‪.‬‬ ‫ﻟﻠﺣﺻول ﻋﻠﻰ اﻟﻘﯾﻣﺔ ‪ h‬ﺗم دﻣﺞ اﻟﺛﻼث ﻋﯾﻧﺎت ﻣﻌﺎ ووﺿﻊ رﺗب ﻟﻠﻌﯾﻧﺔ اﻟﻣﺷﺗرﻛﺔ وﺗﺣدﯾد رﺗب ﻛل‬ ‫ﻋﯾﻧﺔ واﻟﻧﺗﺎﺋﺞ ﻣﻌطﺎة ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ‪.‬‬

‫‪Ri‬‬ ‫‪24‬‬ ‫‪40‬‬ ‫‪14‬‬

‫‪9‬‬

‫‪6‬‬ ‫‪12‬‬

‫اﻟرﺗــب‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪10 11‬‬ ‫‪4‬‬ ‫‪8‬‬

‫وﻋﻠﻰ ذﻟك ﻓﺈن ﻗﯾﻣﺔ ‪ h‬ھﻲ ‪:‬‬

‫‪٥٨٤‬‬

‫‪1‬‬ ‫‪7‬‬ ‫‪2‬‬

‫‪A‬‬ ‫‪B‬‬ ‫‪C‬‬

h

k R 12 i  3(N  1)  N(N  1) i 1 n i

 R12 R 22 R 32  12       3(N  1) N(N  1)  n1 n 2 n 3  1  242 402 142        3(13) 13  5 4 3  1    .580.53   39  13   44.66  39  5.66. ) ‫ واﻟﻣﺳ ﺗﺧرﺟﺔ ﻣ ن اﻟﺟ دول ﻓ ﻲ ﻣﻠﺣ ق‬H ‫ اﻟﻘﯾﻣ ﺔ اﻟﺣرﺟ ﺔ ﻟﻺﺣﺻ ﺎء‬  0.05 ‫ﻟﻣﺳﺗوى ﻣﻌﻧوﯾ ﺔ‬ ‫ وﺑﻣ ﺎ‬H  5.6308 ‫ ﻣﻧطﻘ ﺔ اﻟ رﻓض‬. n1  5 , n 2  4 , n 3  3 ‫ ﻋﻧ د‬5.6308 ‫ ( ھ ﻲ‬١٢ . H 0 ‫ ﺗﻘﻊ ﻓﻲ ﻣﻧطﻘﺔ اﻟرﻓض ﻧرﻓض‬h  5.66 ‫أن‬ ‫ وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬Mathematica ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ‬ . ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت‬ Off[General::spell1]; <<Statistics`DataManipulation` <<Statistics`NormalDistribution` rank[j_,xlist_]:=Module[{}, k=1; flag=0; xsort=Sort[xlist]; poslist=Position[xsort,xlist[[j]]]; k=poslist[[1,1]]; m=Length[poslist]; Sum[val,{val,k,k+m-1}]/m//N] Options[npmKruskalWallisTest]={mthd->approx};

٥٨٥

npmKruskalWallisTestsamples_, opts___ : Moduleflag, mtype  mthd . opts . OptionsnpmKruskalWallisTest; flag  1; medianVals  Map Median, samples; all  Flattensamples; rank1  Tableallk, rankk, all, k, 1, Lengthall; n  Lengthall; niVals  MapLength, samples; k  LengthniVals; totali_ : SumniValsj, j, 1, i; tj_, vals_ : Tablevalsi, i, totalj  1  1, totalj  1  niValsj; withRanks  Tabletj, rank1, j, 1, LengthniVals; rankIm_ : SumwithRanksm, j, 2, j, 1, LengthwithRanks m; rankVals  TablerankIj, j, 1, k; h N 12 nn  1SumrankValsj ^2 niValsj, j, 1, k  3n  1; approxPValh_ : Modulefcount, tsum, fcount  Frequenciesall; tsum  Sumfcountj, 1^3  fcountj, 1, j, 1, Lengthfcount; c  N1  tsum  n^3  n; hc  h c; 1 ٥٨٦ k  1, hc; CDFChiSquareDistribution If And mtype  exact, MaxniVals  5, flag  500; If mtype  approx, pval  approxPValh; If And mtype  exact, flag  1,

altposition[h_,statTab_]:=Module[{}, diff[x_]:=Abs[h-x]; lessThanQ[x_]:=If[x<0.0005,True,False]; statVals=Transpose[statTab][[1]]; chopdfList=Map[diff,statVals]//Chop; tfTable=Map[lessThanQ,chopdfList]; posSet=Position[tfTable,True]] heat1={20,37,39,41,45}; heat2={43,46,48,53}; heat3 ={31,38,44}; threeheats={heat1,heat2,heat3}; npmKruskalWallisTest[threeheats,mthd->exact] Title: Kruskal Wallis Test Sample Medians: {39,47,38} Test Statistic: 5.65641 Distribution: Exact PValue -> exactCalc[5.65641] npmKruskalWallisTest[threeheats] Title: Kruskal Wallis Test Sample Medians: {39,47,38} Test Statistic: 5.65641 Distribution: Chi Square PValue ->

0.0591189

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬ : ‫ﻗﺎﺋﻣﺔ ﺧﺎﺻﺔ ﺑﺎﻟﻣﺟﻣوﻋﺔ اﻻوﻟﻰ‬ heat1={20,37,39,41,45};

: ‫ﻗﺎﺋﻣﺔ ﺧﺎﺻﺔ ﺑﺎﻟﻣﺟﻣوﻋﺔ اﻟﺛﺎﻧﯾﺔ‬ heat2={43,46,48,53};

: ‫ﻗﺎﺋﻣﺔ ﺧﺎﺻﺔ ﺑﺎﻟﻣﺟﻣوﻋﺔ اﻟﺛﺎﻟﺛﺔ‬ heat3 ={31,38,44};

: ‫ﺛﺎﻧﯾﺎ اﻟﻣﺧرﺟﺎت‬ ‫ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ npmKruskalWallisTest[threeheats]

‫واﻟﻣﺧرج ھو‬ Title: Kruskal Wallis Test Sample Medians: {39,47,38} Test Statistic: 5.65641 Distribution: Chi Square PValue ->

0.0591189

٥٨٧

‫ﺣﯿﺚ اﻟوﺳﯾط ﻟﻠﻌﯾﻧﺔ اﻻوﻟﻰ واﻟﺛﺎﻧﯾﺔ واﻟﺛﺎﻟﺛﺔ ھﻣﺎ ﻣن‬ ‫}‪{39,47,38‬‬

‫‪Sample Medians:‬‬

‫واﻻﺣﺻﺎء اﻟﻣﻘدر ﺑﺎﺳﺗﺧدام ﺗوزﯾﻊ ﻣرﺑﻊ ﻛﺎى ھو‬ ‫‪5.65641‬‬

‫‪Test Statistic:‬‬

‫وﻗﯾﻣﺔ ‪ p-value‬ھﻰ‬ ‫‪0.0591189‬‬

‫>‪PValue -‬‬

‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻛﺑر ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧﻘﺑل ‪ . H 0‬ﯾﻼﺣظ ان اﻟﻘرار ھﻧﺎ‬ ‫ﯾﺧﺗﻠف ﻋن اﻟذى ﺗم اﻟﺣﺻول ﻋﻠﯾﮫ ﯾدوﯾﺎ وﻗد ﯾرﺟﻊ ذﻟك اﻟﻰ ﻋﻣﻠﯾﺔ اﻟﺗﻘرﯾب‪.‬‬

‫)‪ (١١-٧‬اﺧﺗﺑﺎر ﻓرﯾدﻣﺎن ﻟﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﻟﻠرﺗب ﻓﻲ اﺗﺟﺎھﯾن‪:‬‬ ‫ھو اﺧﺗﺑﺎر ﻻ ﻣﻌﻠﻣﻲ ﯾﻧﺎظر اﻻﺧﺗﺑﺎر اﻟﻣﻌﻠﻣﻲ ﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﻓﻲ اﺗﺟﺎھﯾن‪ ،‬ﻧﻘوم ﻓﯾﮫ ﺑﺎﻟﺣﺳﺎﺑﺎت‬ ‫ﺑﺎﺳﺗﺧدام اﻟرﺗب اﻟﺗﻲ ﺗﻌطﻰ ﻟﻠﻣﺷﺎھدات ﻓﻲ ﻛل ﻗطﺎع‪.‬‬ ‫اﻟطرﯾﻘﺔ اﻟﻣﻘدﻣﺔ ﺑواﺳطﺔ ﻓرﯾدﻣﺎن ﯾﻣﻛﻧﻧﺎ اﺳﺗﺧداﻣﮭﺎ ﻋﻧدﻣﺎ ﯾﻛون ﻣن ﻏﯾر اﻟﻣرﻏوب ﻓﯾﮫ‬ ‫اﺳﺗﺧدام اﻻﺧﺗﺑﺎر اﻟﻣﻌﻠﻣﻲ ﻟﺗﺣﻠﯾل اﻟﺗﺑﺎﯾن ﻓﻲ اﺗﺟﺎھﯾن‪ ،‬ﻣﺛﻼ ﻗد ﻻ ﯾرﻏب اﻟﺑﺎﺣث ﻓﻲ ﻓرض أن‬ ‫ﻣﺟﺗﻣﻌﺎت اﻟﻌﯾﻧﺎت ﺗﺗوزع طﺑﯾﻌﯾﺎ‪ ،‬وھو ﺷرط اﻻﺳﺗﺧدام اﻟﺻﺣﯾﺢ ﻟﻼﺧﺗﺑﺎر اﻟﻣﻌﻠﻣﻲ‪ ،‬أو رﺑﻣﺎ ﺗﻛون‬ ‫اﻟرﺗب ھﻲ اﻟﺗﺣﻠﯾل اﻟوﺣﯾد اﻟﻣﺗوﻓر‪.‬‬

‫اﻟﺷروط‪:‬‬ ‫‪ /١‬ﻧﻔرض أن ﻟدﯾﻧﺎ ﺗﺻﻣﯾم ﻗطﺎﻋﺎت ﻛﺎﻣﻠﺔ اﻟﻌﺷواﺋﯾﺔ ﺗﺣﺗوي ﻋﻠﻰ ‪ b‬ﻋﯾﻧﺔ ﻣﺳﺗﻘﻠﺔ ﺑﺎﻟﺗﺑﺎدل )ﻗطﺎﻋﺎت(‬ ‫ﺣﺟﻣﮭﺎ ‪ ، k‬ﺗﻛون اﻟﺑﯾﺎﻧﺎت ﻣﻌروﺿﺔ ﻛﻣﺎ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪ ،‬ﺣﯾث اﻟرﻣز ‪ Xij‬ﯾﻣﺛل اﻟﻣﺷﺎھد ة‬ ‫رﻗم ‪ j‬ﻓﻲ اﻟﺻف رﻗم ‪ ، i‬وﺗﻣﺛل اﻷﻋﻣدة اﻟﻣﻌﺎﻟﺟﺎت أﻣﺎ اﻟﺻﻔوف ﻓﺗﻣﺛل اﻟﻘطﺎﻋﺎت ‪ ،‬وﻏﺎﻟﺑﺎ ﻣﺎ‬ ‫ﺗوﺿﻊ ﻓﻲ اﻟﺻﻔوف ﺑﻌض اﻟﻣﻌﺎﻟم ﻏﯾر اﻟﻣﮭﻣﺔ ﻣﺛل اﻟزﻣن‪ ،‬وﯾﻛون اھﺗﻣﺎﻣﻧﺎ ھو اﻟﻔرق ﺑﯾن‬ ‫اﻷﻋﻣدة وﺳﻧﮭﻣل اﻟﻔرق ﺑﯾن اﻟﺻﻔوف‪ ،‬وﻧﻔرض أن اﻟﻣﻌﺎﻟﺟﺎت ﺗﺗوزع ﻋﺷواﺋﯾﺎ داﺧل ﻛل ﻗطﺎع‬ ‫وﯾﻛون ھدﻓﻧﺎ ﻣﻌرﻓﺔ ھل ھﻧﺎك ﻓرق ﺑﯾن اﻟﻣﻌﺎﻟﺟﺎت أم ﻻ‪ ،‬وإن وﺟد ﻧﺳﺗﺧدم أﺳﻠوب اﻟﻣﻘﺎرﻧﺎت‬ ‫اﻟﻣﺗﻌددة ﻟﻣﻌرﻓﺔ أي اﻟﻣﻌﺎﻟﺟﺎت اﻟﺳﺑب ﻓﻲ وﺟود ھذا اﻟﻔرق‪،‬وﺗﻌﺑﯾر اﻟﻣﻌﺎﻟﺟﺎت ﻟﮫ ﻣﻌﻧﻰ ﻋﺎم ﺟدا‪:‬‬ ‫رﺑﻣﺎ ﻧﻌﻧﻲ ﺑﮫ ﻣﻌﺎﻟﺟﮫ ﺑﺎﻟﻣﻌﻧﻰ اﻟﻌﺎدي ﻟﻠﻛﻠﻣﺔ‪ ،‬أو رﺑﻣﺎ ﻧﻌﻧﻲ ﺑﮫ ﺑﻌض اﻟﺣﺎﻻت اﻷﺧرى ﻣﺛل‬ ‫اﻟﺣﺎﻟﺔ اﻻﺟﺗﻣﺎﻋﯾﺔ‪ ،‬اﻻﻗﺗﺻﺎدﯾﺔ أو اﻟﻣﺳﺗوى اﻟﺗﻌﻠﯾﻣﻲ‪.‬‬ ‫‪ /٢‬ﯾﺟب أن ﯾﻛون اﻟﻣﺗﻐﯾر اﻟﻣﮭم )ﺗﺣت اﻟدراﺳﺔ( ﻣﺗﺻل‪.‬‬ ‫‪ /٣‬ﯾﺟب أن ﻻ ﯾوﺟد ﺗﻔﺎﻋل ﺑﯾن اﻟﻘطﺎﻋﺎت واﻟﻣﻌﺎﻟﺟﺎت)ﺑﯾن اﻟﺻﻔوف واﻷﻋﻣدة(‪.‬‬ ‫‪ /٤‬اﻟﻣﺷﺎھدات داﺧل ﻛل ﻗطﺎع ﻣن اﻟﻣﻣﻛن أن ﺗﺣول إﻟﻰ رﺗب ﺗﺻﺎﻋدﯾﺔ‪.‬‬ ‫‪k‬‬

‫…‬

‫‪j‬‬

‫‪….‬‬

‫‪3‬‬

‫‪2‬‬

‫‪1‬‬

‫اﻟﻣﻌﺎﻟﺟﺔ‬ ‫اﻟﻘطﺎع‬

‫‪٥٨٨‬‬

‫‪x1k‬‬ ‫‪x 2k‬‬ ‫‪x 3k‬‬

‫‪x ik‬‬

‫‪x1j‬‬ ‫‪x2 j‬‬ ‫‪x3j‬‬

‫‪x13‬‬ ‫‪x 23‬‬ ‫‪x 33‬‬

‫‪x12‬‬ ‫‪x 22‬‬ ‫‪x 32‬‬

‫‪x11‬‬ ‫‪x 21‬‬ ‫‪x 31‬‬

‫‪x i3‬‬

‫‪x i2‬‬

‫‪x i1‬‬

‫‪x b3‬‬

‫‪x b2‬‬

‫‪x b1‬‬

‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪.‬‬ ‫‪.‬‬

‫‪x ij‬‬

‫‪x‬‬

‫‪bk‬‬ ‫‪x bj‬‬ ‫ﻹﺟراء اﻻﺧﺗﺑﺎر ﻧﺿﻊ اﻟﻔروض اﻟﺗﺎﻟﯾﺔ ‪:‬‬

‫‪i‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪b‬‬

‫‪ : H 0‬ﺗﺄﺛﯾر اﻟﻣﻌﺎﻟﺟﺎت داﺧل اﻟﻘطﺎﻋﺎت ﻣﺗﻣﺎﺛل‪.‬‬ ‫‪ : H1‬ﻋﻠﻰ اﻷﻗل واﺣد ﻣن اﻟﻣﻌﺎﻟﺟﺎت ﯾﻣﯾل إﻟﻰ إﻋطﺎء ﻗﯾم أﻛﺑر ﻣن ﻣﻌﺎﻟﺟﮫ أﺧرى ﻋﻠﻰ اﻷﻗل‪).‬ﺗﺄﺛﯾر‬ ‫اﻟﻣﻌﺎﻟﺟﺎت ﻏﯾر ﻣﺗطﺎﺑق(‪.‬‬ ‫اﻟﺧطوة اﻷوﻟﻰ ﻓ ﻲ ﺣﺳ ﺎب إﺣﺻ ﺎﺋﻲ اﻻﺧﺗﺑ ﺎر ھ ﻲ ﺗﺣوﯾ ل اﻟﻣﺷ ﺎھدات اﻷﺻ ﻠﯾﺔ إﻟ ﻰ رﺗ ب )ھ ذه‬ ‫اﻟﺧطوة ﺑﺎﻟطﺑﻊ ﻟﯾﺳت ﺿرورﯾﺔ إذا ﻛﺎﻧت اﻟﻣﺷﺎھدات اﻷﺻﻠﯾﺔ رﺗﺑﺎ (‪.‬‬ ‫إذا ﻛ ﺎن ‪ H 0‬ﺻ ﺣﯾﺢ أي أن ﻛ ل اﻟﻣﻌﺎﻟﺟ ﺎت ﻟﮭ ﺎ ﺗ ﺄﺛﯾر ﻣﺗط ﺎﺑق ﻓ ﺄن اﻟرﺗﺑ ﺔ اﻟﺗ ﻲ ﺗظﮭ ر ﻓ ﻲ ﻋﻣ ود‬ ‫ﻣﻌﯾن ﺗﻛون ﻣﺟرد ﺻدﻓﺔ‪ ،‬ﺑﻧﺎء ﻋﻠﻰ ھذا ﻓﺈﻧ ﮫ ﻋﻧ دﻣﺎ ‪ H 0‬ﺻ ﺣﯾﺢ ﻻ رﺗ ب ﺻ ﻐﯾرة وﻻ ﻛﺑﯾ رة ﯾﺟ ب أن‬ ‫ﺗﻣﯾ ل إﻟ ﻰ اﻟظﮭ ور ﻓ ﻲ ﻋﻣ ود ﻣﻌ ﯾن‪ ،‬وھ ذا ﯾﻌﻧ ﻲ أن اﻟرﺗ ب ﻓ ﻲ أي ﻗط ﺎع ﯾﺟ ب أن ﺗﻛ ون ﻣوزﻋ ﺔ‬ ‫ﻋﺷواﺋﯾﺎ ﻋﻠﻰ اﻷﻋﻣدة )اﻟﻣﻌﺎﻟﺟ ﺎت (‪ ،‬أﻣ ﺎ إذا ﻛ ﺎن ‪ H 0‬ﺧ ﺎطﻰء ﻓﻧﺗوﻗ ﻊ اﻧﻌ دام ﻟﻠﻌﺷ واﺋﯾﺔ ﻓ ﻲ اﻟﺗوزﯾ ﻊ‬ ‫‪،‬ﻓﺈذا ﻛﺎﻧت إﺣدى اﻟﻣﻌﺎﻟﺟﺎت أﻓﺿل ﻣن ﻏﯾرھﺎ ﻓﺈﻧﻧﺎ ﻧﺗوﻗﻊ رﺗب ﻛﺑﯾرة ﻟﮭذه اﻟﻣﻌﺎ ﻟﺟﺔ )اﻟﻌﻣود( ‪.‬‬ ‫اﻟﺧط وة اﻟﺛﺎﻧﯾ ﺔ ﻓ ﻲ ﺣﺳ ﺎب إﺣﺻ ﺎﺋﻲ اﻻﺧﺗﺑ ﺎر ھ ﻲ إﯾﺟ ﺎد ﻣﺟ ﺎﻣﯾﻊ اﻟرﺗ ب ﻓ ﻲ ﻛ ل ﻋﻣ ود وﯾرﻣ ز‬ ‫ﻟﻣﺟﻣوع اﻟرﺗب ﻓﻲ اﻟﻌﻣود رﻗم ‪ i‬ﺑﺎﻟرﻣز ‪ ، R i‬إذا ﻛﺎﻧ ت ‪ H 0‬ﺻ ﺣﯾﺣﺔ ﻧﺗوﻗ ﻊ أن ﺗﻛ ون اﻟﻣﺟ ﺎﻣﯾﻊ إﻟ ﻰ‬ ‫ﺣد ﻣﺎ ﻣﺗﻘﺎرﺑﺔ ﻓﻲ اﻟﺣﺟم‪ ،‬ﺣﯾث أن اﻟﺗﻘﺎرب ﯾﺟﻌﻠﻧﺎ ﻧﺳﺗطﯾﻊ إرﺟﺎع اﻟﻔ رق ﻟﻠﺻ دﻓﺔ‪ ،‬ﺑﯾﻧﻣ ﺎ ﻋﻧ دﻣﺎ ﯾﻛ ون‬ ‫اﺣد اﻟﻣﺟﺎﻣﯾﻊ ﻣﺧﺗﻠف ﻣﺎ ﻓﯾﮫ اﻟﻛﻔﺎﯾ ﺔ ﻓ ﻲ اﻟﺣﺟ م ﻋﻠ ﻰ اﻷﻗ ل ﻋ ن أﺣ د اﻟﻣﺟ ﺎﻣﯾﻊ اﻷﺧ رى‪ ،‬ﻓﺈﻧﻧ ﺎ ﻧ رﻓض‬ ‫إرﺟ ﺎع اﻻﺧ ﺗﻼف ﻟﻠﺻ دﻓﺔ وﺣ دھﺎ‪ ،‬ﻣﻣ ﺎ ﯾﻌﻧ ﻲ أن ھﻧ ﺎك ﺳ ﺑب آﺧ ر ﻟﻼﺧ ﺗﻼف أي أن ﻓ رض اﻟﻌ دم‬ ‫ﺧﺎطﺊ‪.‬‬ ‫اﻻﺧﺗﻼف ﻓﻲ اﻟﺣﺟم ﺑﯾن ﻣﺟﺎﻣﯾﻊ اﻟرﺗب ﺑدرﺟﺔ ﻛﺎﻓﯾﺔ ﯾﻌطﻲ ارﺗﻔﺎﻋﺎ ﻓ ﻲ ﻗﯾﻣ ﺔ إﺣﺻ ﺎﺋﻲ اﻻﺧﺗﺑ ﺎر‬ ‫ﺑدرﺟﺔ ﻛﺑﯾرة ﻛﻔﺎﯾﺔ أﯾﺿﺎ ﻣﻣﺎ ﯾﺳﺑب ﻓﻲ رﻓض ‪. H 0‬‬ ‫ﺻﯾﻐﺔ إﺣﺻﺎﺋﻲ اﻻﺧﺗﺑﺎر ھﻲ ‪:‬‬ ‫‪k‬‬ ‫‪12‬‬ ‫‪2‬‬ ‫‪ R j  3b(k  1).‬‬ ‫‪bk(k  1) j1‬‬

‫‪٥٨٩‬‬

‫‪ 2r ‬‬

‫ﻋﻧدﻣﺎ ﺗﻛون ﻛﻼ ﻣن ‪ b‬و‪ k‬ﺻﻐﯾرة ﻧﻘﺎ رن اﻟﻘﯾﻣ ﺔ اﻟﻣﺣﺳ وﺑﺔ ‪ 2r‬ﻣ ﻊ اﻟﻘﯾﻣ ﺔ اﻟﺣرﺟ ﺔ اﻟﻣﺳ ﺗﺧرﺟﺔ ﻣ ن‬ ‫اﻟﺟدول اﻟﺧﺎص ﺑﺎﺧﺗﺑﺎر ﻓرﯾ دﻣﺎن ﺑﻣﻌﻠوﻣﯾ ﺔ ﻛ ل ﻣ ن ‪ b‬و ‪ k‬و ‪ ، ‬ﻓ ﺈذا ﻛﺎﻧ ت اﻟﻘﯾﻣ ﺔ اﻟﻣﺣﺳ وﺑﺔ أﻛﺑ ر‬ ‫‪2‬‬ ‫ﻣن أو ﺗﺳﺎوي اﻟﻘﯾﻣﺔ اﻟﺟدوﻟﯾﺔ ‪ ، r‬ﻓﺈﻧﻧﺎ ﻧرﻓض ‪ H 0‬ﻋﻧد ﻣﺳﺗو ى ﻣﻌﻧوﯾﺔ ‪. ‬‬ ‫ﻋﻧدﻣﺎ ﻻ ﺗﺷﻣل ﺟداول ﻓرﯾدﻣﺎن ﻋﻠ ﻰ ﻗﯾﻣ ﺔ ‪ k‬و‪ b‬أﺣ دھﻣﺎ أو ﻛﻼھﻣ ﺎ ﻓﺈﻧﻧ ﺎ ﻧﺳ ﺗﺧدم ﺟ داول ﻣرﺑ ﻊ ﻛ ﺎي‬

‫ﻹﯾﺟ ﺎد اﻟﻘﯾﻣ ﺔ اﻟﺣرﺟ ﺔ ﻋﻧ د درﺟ ﺔ ﺣرﯾ ﺔ)‪ (k-1‬وﻧ رﻓض ‪ H0‬ﻋﻧ د ﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ ‪ ‬إذا ﻛﺎﻧ ت ‪2r‬‬ ‫‪2‬‬ ‫‪  (1‬ﺑدرﺟﺎت ﺣرﯾﺔ )‪.(k-1‬‬ ‫اﻟﻣﺣﺳوﺑﺔ ﻣن اﻟﺑﯾﺎﻧﺎت أﻛﺑر ﻣن أو ﺗﺳﺎوي اﻟﻘﯾﻣﺔ اﻟﺟدوﻟﯾﺔ ﻟـ ) ‪‬‬

‫ﻋﻧدﻣﺎ ﺗظﮭر اﻟﺗداﺧﻼت ﯾﻣﻛﻧﻧﺎ ﺗﻌدﯾل إﺣﺻﺎﺋﻲ اﻻﺧﺗﺑﺎر ﻟﯾﺄﺧذ اﻟﺗداﺧﻼت ﻓ ﻲ اﻟﺣﺳ ﺎب ﻣﻣ ﺎ ﯾ ؤدي‬ ‫‪b‬‬ ‫‪Ti‬‬ ‫‪ 1  ‬ﺣﯾ ث أن‬ ‫إﻟ ﻰ ﺗﺣﺳ ﯾن اﻟﻧﺗ ﺎﺋﺞ ودﻗ ﺔ اﻻﺧﺗﺑ ﺎر‪ ،‬وﯾﻛ ون ذﻟ ك ﺑﻘﺳ ﻣﺔ ‪ 2‬ﻋﻠ ﻰ‬ ‫‪2‬‬ ‫)‪i 1 bk(k  1‬‬ ‫‪3‬‬

‫‪ ، Ti   ti   t i‬و ‪ = t i‬ﻋدد اﻟﻣﺷﺎھدات اﻟﻣﺗداﺧﻠﺔ ﻟرﺗﺑﺔ ﻣﻌﯾﻧﺔ ﻓﻲ اﻟﻘط ﺎع رﻗ م ‪) i‬ﻣ ﻊ ﻣﻼﺣظ ﺔ‬ ‫أﻧﻧﺎ ﻧﮭﺗم ﺑﺎﻟﺑﯾﺎﻧﺎت اﻟﻣﺗداﺧﻠﺔ ﻓﻲ اﻟﻘطﺎع ﻓﻘط( ‪.‬‬ ‫وﻷن اﻟﺗﺻﺣﯾﺢ ﻓﻲ ﺣﺎﻟﺔ اﻟﺗداﺧﻼت ﯾﺟﻌل اﻟﻘﯾﻣﺔ اﻟﻣﺣﺳوﺑﺔ أﻛﺑر ﻓﻼ ﯾﻛون ﻣن اﻟﺿروري ﺗﺻﺣﯾﺢ‬ ‫ﻗﯾﻣﺔ إﺣﺻﺎﺋﻲ اﻻﺧﺗﺑﺎر ﻋﻧدﻣﺎ ﺗﻛون ﻛﺑﯾرة ﺑدرﺟﺔ ﻛﺎﻓﯾﺔ ﺗﺳﺑب رﻓض ﻓرض اﻟﻌدم‪.‬‬ ‫ﻋﻧد رﻓض ﻓرض اﻟﻌ دم وﻗﺑ ول اﻟﻔ رض اﻟﺑ دﯾل ﻓ ﺈن ھ دﻓﻧﺎ ھ و ﺗﺣدﯾ د أي اﻟﻣﻌﺎﻟﺟ ﺎت ھ ﻲ اﻟﺳ ﺑب‬ ‫ﻓ ﻲ وﺟ ود ھ ذا اﻟﻔ رق وﻟﮭ ذا ﻧﺟ ري اﺧﺗﺑ ﺎر اﻟﻣﻘﺎرﻧ ﺎت اﻟﻣﺗﻌ ددة ‪،‬ﺣﯾ ث أن اﻟﺣ د اﻷﻋﻠ ﻰ ﻟﻠﻔ رق ﺑ ﯾن‬ ‫ﻣﺟﺎﻣﯾﻊ اﻟرﺗب ھو‪:‬‬

‫)‪bk(k  1‬‬ ‫‪,‬‬ ‫‪6‬‬

‫‪z‬‬

‫ﺣﯾث أن ‪ z‬ھﻲ اﻟﻘﯾﻣﺔ اﻟﻣﺳﺗﺧرﺟﺔ ﻣن ﺟدول اﻟﺗوزﯾﻊ اﻟطﺑﯾﻌﻲ اﻟﻘﯾﺎﺳﻲ ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ‪:‬‬ ‫‪1  ‬‬ ‫‪.‬‬ ‫)‪k(k  1‬‬

‫ﻣﺛﺎل)‪(١٦-٧‬‬ ‫طﺑﻘت ﺛﻼث طرق ﻟﻔﺣص ﻣﻘدار ﺧﻣﯾ رة اﻟﻧﺷ ﺎ ﻋﻠ ﻰ ﻋ دة ﻣرﺿ ﻰ ﻣﺻ ﺎﺑون ﺑﺈﻟﺗﮭ ﺎب اﻟﺑﻧﻛرﯾ ﺎس‬ ‫وﻛﺎﻧت اﻟﻧﺗﺎﺋﺞ ﻛﻣﺎ ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪ ،‬وﻧرﻏب ﻓﻲ ﻣﻌرﻓﺔ ﻣﺎ إذا ﻛﺎﻧت ھذه اﻟﺑﯾﺎﻧ ﺎت ﺗﺷ ﯾر إﻟ ﻰ‬ ‫اﺧﺗﻼف ﺑﯾن اﻟطرق اﻟﺛﻼﺛﺔ أم ﻻ‪.‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫طرﯾﻘﺔ اﻟﻔﺣص‬ ‫اﻟﺷﺧص‬ ‫‪6120‬‬ ‫‪2410‬‬ ‫‪2210‬‬ ‫‪2060‬‬ ‫‪1400‬‬ ‫‪249‬‬ ‫‪224‬‬ ‫‪208‬‬ ‫‪227‬‬

‫‪3210‬‬ ‫‪1040‬‬ ‫‪647‬‬ ‫‪570‬‬ ‫‪445‬‬ ‫‪156‬‬ ‫‪155‬‬ ‫‪99‬‬ ‫‪70‬‬

‫‪٥٩٠‬‬

‫‪4000‬‬ ‫‪1600‬‬ ‫‪1600‬‬ ‫‪1200‬‬ ‫‪840‬‬ ‫‪352‬‬ ‫‪224‬‬ ‫‪200‬‬ ‫‪184‬‬

‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪9‬‬

‫اﻟﺣــل‪:‬‬ ‫‪ : H 0‬اﻟطرق اﻟﺛﻼﺛﺔ ﺗﻌطﻲ ﻧﺗﺎﺋﺞ ﻣﺗطﺎﺑﻘﺔ‪.‬‬ ‫‪ : H1‬ﯾوﺟد ﻋﻠﻰ اﻷﻗل ﻣﻌﺎﻟﺟﺔ ﻣﺧﺗﻠﻔﺔ‪.‬‬ ‫اﻷﺷﺧﺎص ﻓﻲ ھذا اﻟﻣﺛﺎل ھم اﻟﻘطﺎﻋﺎت إذن ‪ ، b  9‬ﺣﯾث إﻧﻧﺎ ﻧﺟري اﻟﻔﺣص ﻟﻛل ﺷﺧص ﺑﺛﻼﺛﺔ‬ ‫طرق إذن ﻟدﯾﻧﺎ ‪ ، k  3‬ﻋﻧدﻣﺎ ﻧﺳﺗﺑدل اﻟﻘﯾﺎﺳﺎت اﻷﺻﻠﯾﺔ اﻟﻣﻌطﺎة ﻓﻲ اﻟﺟدول اﻟﺳﺎﺑق ﺑرﺗﺑﺗﮭﺎ‬ ‫ﻧﺣﺻل ﻋﻠﻰ اﻟﺟدول اﻟﺗﺎﻟﻰ‪.‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫طرﯾﻘﺔ اﻟﻔﺣص‬ ‫اﻟﺷﺧص‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪6‬‬ ‫‪3‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪7‬‬ ‫‪2.5‬‬ ‫‪1‬‬ ‫‪2.5‬‬ ‫‪8‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪9‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪R A  19.5‬‬ ‫‪R B  9 R C  25.5‬‬ ‫إذن ﯾﺻﺑﺢ ﻟدﯾﻧﺎ إﺣﺻﺎﺋﻲ اﻻﺧﺗﺑﺎر) ﺑدون ﺗﺻﺣﯾﺢ اﻟﺗداﺧﻼت(‪:‬‬

‫‪12‬‬ ‫‪((19.5)2  92  (25.5)2 )  3(9)(3  1)  15.5.‬‬ ‫)‪9(3)(3  1‬‬ ‫وﺑﻣﻘﺎرﻧﺔ ھذه اﻟﻘﯾﻣﺔ ﻣﻊ اﻟﻘﯾﻣﺔ اﻟﻣﺳﺗﺧرﺟﺔ ﻣن ﺟداول ﻣرﺑﻊ ﻛﺎي ﻋﻧد درﺟﺔ ﺣرﯾﺔ )‪ (2‬وھﻲ ‪5.992‬‬ ‫ﻋﻧد ‪   0.05‬ﻧﺟد أن اﻟﻘﯾﻣﺔ اﻟﻣﺣﺳوﺑﺔ أﻛﺑر ﻣن اﻟﻘﯾﻣﺔ اﻟﺟدوﻟﯾﺔ ‪ ،‬ﻣﻣﺎ ﯾﻌﻧﻲ إﻧﻧﺎ ﻧرﻓض ﻓرض‬ ‫اﻟﻌدم وﻧﺳﺗﻧﺗﺞ أن طرق اﻟﺗﺣﻠﯾل اﻟﺛﻼﺛﺔ ﻻ ﺗﻌطﻲ ﻧﻔس اﻟﻧﺗﺎﺋﺞ‪.‬‬ ‫وﻟﻣﻌرﻓﺔ أي اﻟﻣﻌﺎﻟﺟﺎت ھﻲ اﻟﺳﺑب ﻓﻲ وﺟود ھذا اﻟﻔرق ﻧﻘوم ﺑﺎﻻﺗﻲ‪:‬‬ ‫ﺣﯾث أن ﻣﺳﺗوى اﻟﻣﻌﻧوﯾﺔ ‪   0.01‬ﻓﺈن‪:‬‬ ‫‪0.01‬‬ ‫‪1  ‬‬ ‫‪ 0.02,‬‬ ‫‪k(k  1) ‬‬ ‫)‪3(2‬‬ ‫وﻣن ﺟدول اﻟﺗوزﯾﻊ اﻟطﺑﯾﻌﻲ اﻟﻘﯾﺎﺳﻲ ﻧﺟد أن ‪ z  2.05‬وﺑﺎﻟﺗﺎﻟﻲ ﯾﻛون اﻟﺣد اﻷﻋﻠﻰ ھو‪:‬‬ ‫‪2r ‬‬

‫)‪2 bk(k  1) / 6 = 2.05 9(3)(4‬‬

‫‪ 8.697.‬‬ ‫‪6‬‬ ‫ﻧﺣﺳب اﻟﻔروق اﻟﻣطﻠﻘﺔ ﺑﯾن ﻣﺟﺎﻣﯾﻊ اﻟرﺗب وھﻲ‪:‬‬

‫‪R A  R B  10.5.‬‬ ‫‪R A  RC  6‬‬ ‫‪R B  R C  16.5‬‬ ‫ﻧﺟد أن اﻟﻔرق ﺑﯾن اﻟﻣﻌﺎﻟﺟﺔ اﻷوﻟﻰ واﻟﺛﺎﻧﯾﺔ أﻛﺑر ﻣن اﻟﺣد اﻷﻋﻠﻰ‪ ،‬وﻛ ذﻟك اﻟﻔ رق ﺑ ﯾن اﻟﻣﻌﺎﻟﺟ ﺔ اﻟﺛﺎﻧﯾ ﺔ‬ ‫واﻟﺛﺎﻟﺛﺔ أﯾﺿﺎ أﻛﺑر ﻣن اﻟﺣد اﻷﻋﻠﻰ ‪ ،‬وﻟﻛن ﻻﯾوﺟد ﻓرق ﺑﯾن اﻟطرﯾﻘﺔ ‪. C , A‬‬ ‫‪٥٩١‬‬

‫ وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬Mathematica ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ‬ . ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت‬ Off[General::spell1]; <<Statistics`DataManipulation` <<Statistics`NormalDistribution` <<DiscreteMath`Combinatorica` rank[j_,xlist_]:=Module[{}, k=1; flag=0; xsort=Sort[xlist]; poslist=Position[xsort,xlist[[j]]]; k=poslist[[1,1]]; m=Length[poslist]; Sum[val,{val,k,k+m-1}]/m//N] Options[npmFriedmanTest]={mthd->chiSquare}; npmFriedmanTest[blocks_,opts___]:=Module[{}, mtype=mthd/. {opts} /. Options[npmFriedmanTest]; flag=1; b=Length[blocks]; k=Length[blocks[[1]]]; pairOfVals={k,b}; possiblePairs={{3,2},{3,3},{3,4},{3,5},{3,6},{3,7},{3,8 },{3,9},{3,10},{3,11},{3,12},{3,13},{3,14},{3,15},{4,2},{4,3 },{4,4},{4,5},{4,6},{4,7},{4,8},{5,3}}; tSumsForLists[vec_]:=Sum[vec[[j,1]]^3vec[[j,1]],{j,1,Length[vec]}]; If[And[MemberQ[possiblePairs,pairOfVals]==False,mtype== exact],flag=200]; If[flag==200,Print["The values of k = ",k," and b = ",b," exceed the limitations of this procedure to compute an exact PValue. An appromimate value may be determined using the Chi Square Distribution or F Distribution."]]; If[k==2,Print["For two treatments (k=2) use the Sign Test or Wilcoxon Signed-Ranks Test."]]; If[k==2,flag=100]; If[b==2,Print["For two blocks (b=2) use an analysis based on the Spearman rank correlation coefficient."]]; If[b==2,flag==100]; If[flag<100,trans=Transpose[blocks]]; If[flag<100,medianVals=Map[Median,trans]//N]; If[flag<100,allVals=Flatten[blocks]]; If[flag<100,fcount=Map[Frequencies,blocks]];

٥٩٢

If[flag<100,tb=Table[tSumsForLists[fcount[[i]]],{i,1,Le ngth[fcount]}]]; If[flag<100,tsum=Sum[tb[[j]],{j,1,Length[fcount]}]]; rankBlock[blk_]:=Table[{blk[[k]],rank[k,blk]},{k,1,Leng th[blk]}]; If[flag<100,rankBlockTable=Table[rankBlock[blocks[[j]]] ,{j,1,b}]]; rowTotal[i_]:=Sum[rankBlockTable[[j,i,2]],{j,1,Length[r ankBlockTable]}]; If[flag<100,rTotals=Table[rowTotal[i],{i,1,Length[rankB lockTable[[1]]]}]]; If[And[flag<100,mtype==exact],testStat=12/(b k(k+1))*Sum[rTotals[[j]]^2,{j,1,Length[rTotals]}]-3 b(k+1)]; If[And[flag<100,mtype==exact],pVal=exactTable[testStat, k,b]]; If[And[flag<100,mtype==chiSquare],testStat=(12/(b k(k+1)) Sum[rTotals[[j]]^2,{j,1,Length[rTotals]}]-3 b(k+1))/(1-tsum/(b(k^3-k)))]; fcStat:=Module[{chisq}, chisq=(12/(b k(k+1)) Sum[rTotals[[j]]^2,{j,1,Length[rTotals]}]-3 b(k+1))/(1tsum/(b(k^3-k))); (b-1) chisq/(b(k-1)-chisq)]; If[And[flag<100,mtype==Fdist],testStat=fcStat]; If[And[flag<100,mtype==chiSquare],pVal=1CDF[ChiSquareDistribution[k-1],testStat]]; If[And[flag<100,mtype==Fdist],pVal=1CDF[FRatioDistribution[k-1,(k-1)(b-1)],testStat]]; If[flag<100,Print["Title: Friedman Test"]]; If[flag<100,Print["Sample Medians: ",medianVals]]; If[flag<100,Print["Test Statistic: ",testStat]]; If[And[flag<100,mtype==exact],Print["Distribution: Exact"]]; If[And[flag<100,mtype==chiSquare],Print["Distribution: ChiSquare[",k-1,"]"]]; If[And[flag<100,mtype==Fdist],Print["Distribution: FRatioDistribution[",(k-1)*(b-1),"]"]]; If[flag<100,Print["PValue: ",pVal]]] exactTable[t_,k_,b_]:=Module[{wStat}, ٥٩٣

wStat=t/(b (k-1)); If[And[k==3,b==2],tb=tabk3b2]; If[And[k==3,b==3],tb=tabk3b3]; If[And[k==3,b==4],tb=tabk3b4]; If[And[k==3,b==5],tb=tabk3b5]; If[And[k==3,b==6],tb=tabk3b6]; If[And[k==3,b==7],tb=tabk3b7]; If[And[k==3,b==8],tb=tabk3b8]; If[And[k==3,b==9],tb=tabk3b9]; If[And[k==3,b==10],tb=tabk3b10]; If[And[k==3,b==11],tb=tabk3b11]; If[And[k==3,b==12],tb=tabk3b12]; If[And[k==3,b==13],tb=tabk3b13]; If[And[k==3,b==14],tb=tabk3b14]; If[And[k==3,b==15],tb=tabk3b15]; If[And[k==4,b==2],tb=tabk4b2]; If[And[k==4,b==3],tb=tabk4b3]; If[And[k==4,b==4],tb=tabk4b4]; If[And[k==4,b==5],tb=tabk4b5]; If[And[k==4,b==6],tb=tabk4b6]; If[And[k==4,b==7],tb=tabk4b7]; If[And[k==4,b==8],tb=tabk4b8]; If[And[k==5,b==3],tb=tabk5b3]; eps=10^-3;check[x_,tab_,m_]:=If[Abs[Chop[xtab[[m,1]]]]<eps,True,False]; tfTab=Table[check[wStat,tb,m],{m,1,Length[tb]}]; If[MemberQ[tfTab,True],pos=Position[tfTab,True][[1,1]], pos=Length[tfTab]]; tb[[pos,2]]] Exact Tables emotions={{4000,3210,6120},{1600,1040,2410},{1600,647,2210}, {1200,570,2060},{840,445,1400},{352,156,249},{224,155,224},{ 200,99,208},{184,70,227}}; npmFriedmanTest[emotions] Title: Friedman Test Sample Medians: {840.,445.,1400.} Test Statistic: 15.9429 Distribution: ChiSquare[ 2 ] PValue: 0.000345186 npmFriedmanTest[emotions,mthd->exact] Title: Friedman Test Sample Medians: {840.,445.,1400.} Test Statistic: 15.5 Distribution: Exact PValue: 0. npmFriedmanTest[emotions,mthd->Fdist] ٥٩٤

‫‪Title: Friedman Test‬‬ ‫}‪Sample Medians: {840.,445.,1400.‬‬ ‫‪Test Statistic: 62.‬‬ ‫] ‪Distribution: FRatioDistribution[ 16‬‬ ‫‪PValue: 2.91029  108‬‬

‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ ‪:‬‬ ‫اوﻻ ‪ :‬اﻟﻣدﺧﻼت‬ ‫ﻗﺎﺋﻣﺔ ﺧﺎﺻﺔ ﺑﺎﻟﻣﺟﻣوﻋﺔ اﻻوﻟﻰ ‪:‬‬ ‫;}‪heat1={20,37,39,41,45‬‬

‫ﻗﺎﺋﻣﺔ ﺧﺎﺻﺔ ﺑﺎﻟﻣﺟﻣوﻋﺔ اﻟﺛﺎﻧﯾﺔ ‪:‬‬ ‫;}‪heat2={43,46,48,53‬‬

‫ﻗﺎﺋﻣﺔ ﺧﺎﺻﺔ ﺑﺎﻟﻣﺟﻣوﻋﺔ اﻟﺛﺎﻟﺛﺔ ‪:‬‬ ‫;}‪heat3 ={31,38,44‬‬

‫ﺛﺎﻧﯾﺎ ‪ :‬اﻟﻣﺧرﺟﺎت‬ ‫ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫]‪npmKruskalWallisTest[threeheats‬‬

‫واﻟﻤﺨﺮج ھﻮ‬ ‫‪Title: Kruskal Wallis Test‬‬ ‫}‪Sample Medians: {39,47,38‬‬ ‫‪Test Statistic: 5.65641‬‬ ‫‪Distribution: Chi Square‬‬ ‫‪0.0591189‬‬

‫>‪PValue -‬‬

‫ﺣﯿﺚ اﻟوﺳﯾط ﻟﻠﻌﯾﻧﺔ اﻻوﻟﻰ واﻟﺛﺎﻧﯾﺔ واﻟﺛﺎﻟﺛﺔ ھﻣﺎ‬ ‫}‪{39,47,38‬‬

‫‪Sample Medians:‬‬

‫‪5.65641‬‬

‫‪Test Statistic:‬‬

‫واﻻﺣﺻﺎء اﻟﻣﻘدر ﺑﺎﺳﺗﺧدام ﺗوزﯾﻊ ﻣرﺑﻊ ﻛﺎى ھو‬ ‫وﻗﯾﻣﺔ ‪ p-value‬ھﻰ‬ ‫‪0.0591189‬‬

‫>‪PValue -‬‬

‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻗل ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧرﻓض ‪. H 0‬‬ ‫وﺑﻧﻔس اﻟطرﯾﻘﺔ ﯾﻣﻛن ﺗﻔﺳﯾر اﻻواﻣر اﻻﺧرى واﻟﺗﻰ ﺗﺷرح طرق اﺧرى ‪.‬‬

‫)‪(١٢-٧‬اﺧﺗﺑﺎر ﻛوﻛران ﻟﻠﻌﯾﻧﺎت اﻟﻣرﺗﺑطﺔ‬ ‫اﺧﺗﺑ ﺎر ﻛ وﻛران ﯾﺗﻌﻠ ق ﺑ ﺄﻛﺛر ﻣ ن ﻋﯾﻧﺗ ﯾن ﻏﯾ ر ﻣﺳ ﺗﻘﻠﺗﯾن ‪ ،‬وﻣ ن اﻟﻧﺎﺣﯾ ﺔ اﻟﻌﻣﻠﯾ ﺔ ﯾﺻ ﻌب ﻋﻠ ﻰ‬ ‫اﻟﺑﺎﺣ ث أن ﯾﺣﻘ ق ﺷ رط اﻻﺳ ﺗﻘﻼل ﺑ ﯾن وداﺧ ل اﻟﻣﻌﺎﻟﺟ ﺎت وﯾﻛ ون أﺳ ﮭل ﻟ ﮫ أن ﺗﻛ ون ﻟدﯾ ﮫ وﺣ دات‬ ‫ﻣﺗﺷﺎﺑﮭﺔ ﺗﻣﺎﻣﺎ ً ﻓﯾﻣﺎ ﺑﯾﻧﮭﺎ وﯾﺗم ﺗﻘﺳﯾﻣﮭﺎإﻟﻰ ﻣﺟﻣوﻋﺎت وﯾﺳﺗﺧدم ﻓﻲ ﻛل ﻣﺟﻣوﻋﺔ ﻣﻌﺎﻟﺟ ﺔ ﻣﺧﺗﻠﻔ ﺔ وﺑﻌ د‬ ‫اﻧﺗﮭ ﺎء اﻟﺗﺟرﺑ ﺔ ﻧﺑﺣ ث ﻋ ن ﺗ ﺄﺛﯾر اﻟﻣﻌﺎﻟﺟ ﺎت ھ ل ھ و ﻣﺧﺗﻠ ف أو ﻣﺗﺳ ﺎوي‪ ،‬ﻓ ﻲ ھ ذه اﻟﺣﺎﻟ ﺔ ﺷ رط‬ ‫ﻻﺳﺗﻘﻼل ﯾﻛون ﻏﯾ ر ﻣﺗﺣﻘ ق واﻟﻧﺗ ﺎﺋﺞ ﻓ ﻲ ﺣﺎﻟﺗﻧ ﺎ ھ ذه ﺗﺗﺷ ﺎﺑﮫ ﻣ ﻊ ﺗﺻ ﻣﯾم اﻟﻘطﺎﻋ ﺎت اﻟﻛﺎﻣﻠ ﺔ اﻟﻌﺷ واﺋﯾﺔ‬

‫‪٥٩٥‬‬

‫ﺣﯾث ﺗﻣﺛل ﻓﯾﮭﺎ اﻷﻋﻣدة اﻟﻣﻌﺎﻟﺟ ﺎت وﯾﻣﺛ ل اﻟﺻ ﻔوف ﻣﺟﻣوﻋ ﺎت ﻣ ن اﻟﻌﯾﻧ ﺎت اﻟﻐﯾ ر ﻣﺳ ﺗﻘﻠﺔ وﻏﺎﻟﺑ ﺎ ً ﻣ ﺎ‬ ‫ﺗﺿﻊ ﻓﻲ اﻟﺻﻔوف ﺑﻌض اﻟﻣﻌﺎﻟم اﻟﻐﯾر ﻣﮭﻣﺔ ﻣﺛل ﻋﺎﻣل اﻟزﻣن‪) ،‬اﻟﯾوم‪ ،‬اﻟﺷﮭر‪ ،‬اﻟﺳﻧﺔ(‪.‬‬ ‫ﻋﻧدﻣﺎ ﯾﻛون ﻟدﯾﻧﺎ ﺗﺻﻣﯾم اﻟﻘطﺎﻋﺎت اﻟﻛﺎﻣﻠﺔ اﻟﻌﺷواﺋﯾﺔ وﻋﻧدﻣﺎ ﺗﻛون وﺣدة اﻟﻘﯾ ﺎس اﺳ ﻣﯾﺔ ﻓﺈﻧ ﮫ ﯾﻣﻛ ن‬ ‫اﺧﺗﺑﺎر ھل ھﻧﺎك ﻓرﻗﺎ ً ﺑﯾن اﻟﻣﻌﺎﻟﺟﺎت أم ﻻ وذﻟك ﺑﺎﺳﺗﺧدام اﺧﺗﺑﺎر ﻛوﻛران‪ .‬وﯾﺟرى اﻻﺧﺗﺑﺎر ﻛﺎﻟﺗﺎﻟﻲ‪:‬‬ ‫ﯾﺟب ﺗﻘﺳم اﻟوﺣدات إﻟﻰ أﻋﻣدة وﺻﻔوف وﻋﻠﯾﻧﺎ ﻓرض أن اﻟﻣﺗﻐﯾر ﻣﺳ ﺗﻣر وأن اﻟﻣﺷ ﺎھدات ﻓ ﻲ ﻛ ل‬ ‫ﻗطﺎع ﻗﺎﺑﻠﺔ ﻟﻠﺗﺣوﯾل إﻟﻰ رﺗب ﺗﺻ ﺎﻋدﯾﺔ وأن ﻻ ﯾﻛ ون ھﻧ ﺎك ﺗﻔﺎﻋ ل ﺑ ﯾن اﻟﺻ ﻔوف واﻷﻋﻣ دة ﺑﺎﻹﺿ ﺎﻓﺔ‬ ‫إﻟ ﻰ أن اﻟﺑﯾﺎﻧ ﺎت ﻓ ﻲ ﻛ ل ﻣﻌﺎﻟﺟ ﺔ ﺗﻘ ﺎس ﺑﻣﻘﯾ ﺎس اﺳ ﻣﻲ )ﺻ ﻔر أو واﺣ د( وﺳ ﻧرﻣز ﻟﻣﺟﻣ وع اﻟﺻ ف‬ ‫واﻟﻌﻣود رﻗم ‪ i‬ﺑﺎﻟرﻣز ‪ Ci ، R i‬ﻋﻠﻰ اﻟﺗواﻟﻲ‪.‬‬ ‫ﻣﺟﻣوع اﻟﺻﻔوف‬

‫‪R1‬‬

‫‪C‬‬ ‫‪x1C‬‬

‫‪R2‬‬

‫‪x 2C‬‬

‫‪R3‬‬

‫‪x 3C‬‬

‫‪.‬‬ ‫‪.‬‬

‫‪....‬‬ ‫‪.....‬‬ ‫‪.....‬‬ ‫‪.....‬‬ ‫‪.....‬‬ ‫‪.....‬‬

‫اﻟﻣﻌﺎﻟﺟﺔ‬ ‫‪3‬‬ ‫‪x13‬‬

‫‪2‬‬ ‫‪x12‬‬

‫‪1‬‬ ‫‪x11‬‬

‫‪1‬‬

‫‪x 23‬‬

‫‪x 22‬‬

‫‪x 21‬‬

‫‪2‬‬

‫‪x 33‬‬

‫‪x 32‬‬

‫‪x 31‬‬

‫‪3‬‬

‫‪.‬‬

‫‪.‬‬ ‫‪.‬‬

‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬

‫‪xr2‬‬

‫‪x r1‬‬

‫‪r‬‬

‫‪C2‬‬

‫‪C1‬‬

‫ﻣﺟﻣوع اﻷﻋﻣدة‬

‫‪.‬‬

‫‪.‬‬ ‫‪x r3‬‬ ‫‪x rc‬‬ ‫اﻟﻣﺟﻣوع ‪R r‬‬ ‫‪C3‬‬ ‫‪CC‬‬ ‫اﻹﺟﻣﺎﻟﻲ ‪N ‬‬ ‫ﻓرض اﻟﻌدم ‪ : H 0‬اﻟﻣﻌﺎﻟﺟﺎت ﻣﺗﺳﺎوﯾﺔ اﻟﺗﺄﺛﯾر‪.‬‬ ‫اﻟﻔرض اﻟﺑدﯾل ‪ : H1‬اﻟﻣﻌﺎﻟﺟﺎت ﻟﯾس ﻟﮭﺎ ﻧﻔس اﻟﺗﺄﺛﯾر‪.‬‬ ‫ﺗﺣﺳب اﻟﻘﯾﻣﺔ اﻟﺗﺎﻟﯾﺔ ‪:‬‬

‫اﻟﻌﯾﻧﺔ‬

‫‪C‬‬

‫‪C(C  1)  C2J  (C  1)N 2‬‬ ‫‪J 1‬‬

‫‪.‬‬ ‫‪R 2I‬‬

‫‪r‬‬

‫‪Q‬‬

‫‪CN  ‬‬ ‫‪I 1‬‬

‫ﺣﯾث ‪N  R i  Ci‬‬ ‫ﺑﺎﺳ ﺗﺧدام ﺟ دول ﻣرﺑ ﻊ ﻛ ﺎي وﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ ‪ ‬ﻧﺳ ﺗﺧرج ﻗﯾﻣ ﺔ ﺟدوﻟﯾ ﮫ وﻧ رﻓض اﻟﻌ دﻣﻲ إذا‬ ‫ﻛﺎﻧت اﻟﻣﺣﺳوﺑﺔ أﻛﺑر ﻣن اﻟﺟدوﻟﯾﮫ‪.‬‬ ‫ﻧﻼﺣظ أن ﻗﯾم داﻟﺔ اﻻﺧﺗﺑﺎر ﺗﻌﺗﻣد ﻋﻠﻰ ﻣﺟﻣ وع اﻷﻋﻣ دة )أي ﻣﺟﻣ وع اﻟﻌﻧﺎﺻ ر اﻟﺗ ﻲ ﻛ ل ﻣﻧﮭ ﺎ ‪(1‬‬ ‫وھ ذا ﯾﻌﻧ ﻲ أن اﻟﻘ ﯾم اﻟﺻ ﻔرﯾﺔ ﻻ ﺗ دﺧلﻓ ﻲ اﻟﺣﺳ ﺎﺑﺎت‪ ،‬ﻟ ذا إذا ﻛ ﺎن ھﻧ ﺎك ﻗطﺎﻋ ﺎ ً ﻛ ﺎﻣﻼ ً ﻛﻠ ﮫ أﺻ ﻔﺎر‬ ‫ﻓﺳ وف ﯾﻠﻐ ﻰ ﺗ ﺄﺛﯾر ھ ذا اﻟﻘط ﺎع )اﻟﺻ ف( ﻛﻠﯾ ﺔ‪ ،‬وﻓ ﻲ ھ ذه اﻟﺣﺎﻟ ﺔ ﺗﺣﺳ ب ﻗﯾﻣ ﺔ ‪ p-value‬ﺑﺎﺳ ﺗﺧدام‬ ‫اﻟﺗوزﯾﻊ اﻟﻣﺿﺑوط وذﻟك ﻓﻲ ﺣﺎﻟﺔ اﻟﺟداول اﻟﺗﻲ ﺗﺣﺗوي ﻋﻠﻰ ﺻﻔوف وأﻋﻣدة ﻗﻠﯾﻠﺔ‪.‬‬ ‫أﻣﺎ إذا ﻛﺎن ‪ r  c  24‬و ‪ r  4‬ﻓﺈﻧﮫ ﻗﯾﻣﺔ ‪ p-value‬ﻣﻣﻛن أن ﺗﻘرب ﻋن طرﯾق ﺗوزﯾﻊ ﻣرﺑ ﻊ ﻛ ﺎي‬ ‫ﺑدرﺟﺔ ﺣرﯾﺔ ‪.c-1‬‬ ‫‪٥٩٦‬‬

‫ﻣﺛﺎل)‪(١٧-٧‬‬ ‫ﻧﻔرض أن ﻟدﯾﻧﺎ أرﺑﻌﺔ ﻣﻌﺎﻟﺟﺎت ‪ X,Y,J,K‬ﻣوزﻋﮫ ﻋﻠﻰ ﺳﺗﺔ ﻗطﺎﻋﺎت‪:‬‬ ‫‪X‬‬ ‫‪Y‬‬ ‫‪J‬‬ ‫‪K‬‬ ‫اﻟﺟﻣوع‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪12‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪4‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪2‬‬

‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪5‬‬

‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬

‫اﻟﻣﺟﻣوع‬

‫ﺑﺎﺳﺗﺧدام اﺧﺗﺑﺎر ﻛوﻛران اﺧﺗﺑر‪:‬‬ ‫ﻓرض اﻟﻌدم ‪ : H 0‬اﻟﻣﻌﺎﻟﺟﺎت ﻟﮭﺎ ﻧﻔس اﻟﺗﺄﺛﯾر‬ ‫واﻟﻔرض اﻟﺑدﯾل ‪ : H1‬ﺗوﺟد ﻣﻌﺎﻟﺟﺔ واﺣدة ﻋﻠﻰ اﻷﻗل ﻣﺧﺗﻠﻔﺔ‪.‬‬

‫اﻟﺣــل‪:‬‬ ‫‪4(4  1)(52  12  22  42 )  (4  1)(12) 2 120‬‬ ‫‪Q‬‬ ‫‪‬‬ ‫‪ 6.‬‬ ‫‪20‬‬ ‫) ‪4(12)  (12  32  22  32  22  12‬‬ ‫ﺣﯾث ) ‪ (12‬ھﻲ ﻋدد ﻣرات ﺗﻛرار اﻟﻌﻧﺻر )‪.(1‬‬ ‫وﺣﯾث أن ﺣﺎﺻل ﺿرب اﻟﺻﻔوف ﺑﺎﻷﻋﻣدة ﯾﺳﺎوى ‪ 24‬و ﺑﻣﻘﺎرﻧﮫ اﻟﻘﯾﻣﺔ اﻟﻣﺣﺳوﺑﺔ ﻟـ ‪ Q‬ﻣﻊ اﻟﻘﯾﻣﺔ‬ ‫اﻟﺟدوﻟﯾﺔ ﻋﻧد درﺟﺎت ﺣرﯾﺔ ‪ ، 3  1  4‬وﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪ 0.05‬وھﻰ ‪ 7.815‬وﺑﻣﺎ أن اﻟﻘﯾﻣﺔ‬ ‫اﻟﻣﺣﺳوﺑﺔ اﺻﻐر ﻣن اﻟﺟدوﻟﯾﮫ ﻓﺈﻧﻧﺎ ﻧﻘﺑل ﻓرض اﻟﻌدم أى أن اﻟﻣﻌﺎﻟﺟﺎت ﻛﻠﮭﺎ ﻟﮭﺎ ﻧﻔس اﻟﺗﺄﺛﯾر‪.‬‬

‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ ‪ Mathematica‬وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬ ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫;]‪Off[General::spell1‬‬ ‫`‪<<Statistics`NormalDistribution‬‬ ‫`‪<<Statistics`DataManipulation‬‬ ‫`‪<<DiscreteMath`Combinatorica‬‬ ‫;}‪Options[npmCochransQTest]={mthd->approx‬‬ ‫‪npmCochransQTest[freqList_,opts___]:=Module[{r,rVals,temp},‬‬ ‫;]‪mtype=mthd/. {opts} /. Options[npmCochransQTest‬‬ ‫;‪flag=1‬‬ ‫;]]]‪c=Length[freqList[[1‬‬ ‫‪If[c<3,Print["This procedure does not work for‬‬ ‫‪less than three treatments (columns). For two treatments,‬‬ ‫;]]"‪use the the McNemar Test.‬‬ ‫;]‪If[c<3,flag=100‬‬ ‫‪٥٩٧‬‬

If[flag<100,temp={}]; addterm[i_]:=If[And[Sum[freqList[[i,j]],{j,1,c}]>0,Sum[ freqList[[i,j]],{j,1,c}]<c],AppendTo[temp,freqList[[i]]]]; If[flag<100,Table[addterm[i],{i,1,Length[freqList]}]]; If[flag<100,r=Length[temp]]; rowTotal[i_,numList_]:=Sum[numList[[i,j]],{j,1,c}]; If[flag<100,rVals=Table[rowTotal[i,temp],{i,1,r}]]; colTotal[j_,numList_]:=Sum[numList[[i,j]],{i,1,r}]; If[flag<100,cVals=Table[colTotal[j,temp],{j,1,c}]]; If[flag<100,n=Apply[Plus,rVals]]; If[flag<100,qStat=(c(c1)Sum[cVals[[j]]^2,{j,1,c}]-(c-1)n^2)/(c nSum[rVals[[i]]^2,{i,1,r}])//N]; If[c==3,pVal=c3test[temp,c,qStat,mtype]]; If[c==4,pVal=c4test[temp,c,qStat,mtype]]; If[c>4,pVal=1-CDF[ChiSquareDistribution[c1],qStat]]; If[flag<100,Print["Title: Cochran Q Test"]]; If[flag<100,Print["Test Statistic: ",qStat]]; If[flag<100,Print["Column Totals: ",cVals]]; If[And[flag<100,mtype==exact],Print["Distribution: Exact"]]; If[And[flag<100,mtype==approx],Print["Distribution: Chi Square[",c-1,"]"]]; If[flag<100,Print["PValue: ",pVal]] ]; c3test[tempList_,c_,qStat_,mtype_]:=Module[{p}, If[mtype==approx,p=1-CDF[ChiSquareDistribution[c1],qStat]]; If[mtype==exact,p=cochranExact3[tempList,qStat]]; p ]; c4test[tempList_,c_,qStat_,mtype_]:=Module[{p}, If[mtype==approx,p=1CDF[ChiSquareDistribution[c-1],qStat]]; If[mtype==exact,p=cochranExact4[tempList,qStat]]; p]; cochranExact3[modList_,q_]:=Module[{}, tmptotals={0,0}; ٥٩٨

cntQ[i_,k_,numList_]:=If[Sum[numList[[k,j]],{j,1,c}]==i ,tmptotals[[i]]=tmptotals[[i]]+1]; Table[cntQ[i,k,modList],{k,1,Length[modList]},{i,1,c1}]; {n1,n2}={tmptotals[[1]],tmptotals[[2]]}; compList=Table[Compositions[tmptotals[[i]],3],{i,1,c1}]; allCombs=Table[{compList[[1,i]],compList[[2,j]]},{i,1,L ength[compList[[1]]]},{j,1,Length[compList[[2]]]}]; flatList=Flatten[allCombs,1]; pValForComb3[v_,n1_,n2_]:=Module[{n1Vals,n2Vals,p1,p2}, n1Vals=v[[1]]; n2Vals=v[[2]]; p1=n1!/(3^n1Product[n1Vals[[i]]!,{i,1,Length[n1Vals]}]); p2=n2!/(3^n2 Product[n2Vals[[i]]!,{i,1,Length[n2Vals]}]); N[p1 p2]]; qValForComb3[v_,c_,n1_,n2_]:=Module[{}, one=Sum[(v[[1,j]]-v[[2,j]])^2,{j,1,c}]1/3*(n1-n2)^2; 6*one/(2(n1+n2))//N]; qValTab=Map[qValForComb3[#,c,n1,n2]&,flatList]; qValsList=Intersection[qValTab]; pValTab=Map[pValForComb3[#,n1,n2]&,flatList]; pAndqTable=Table[{qValTab[[i]],pValTab[[i]]},{i,1,Lengt h[qValTab]}]; places=Map[Position[qValTab,#]&,qValsList]; flatPlaces=Table[places[[i]]//Flatten,{i,1,Length[place s]}]; tabWithQandPVal=Table[{qValsList[[i]],Sum[pValTab[[flat Places[[i,k]]]],{k,1,Length[flatPlaces[[i]]]}]},{i,1,Length[ flatPlaces]}]; ij=Position[tabWithQandPVal,q]; Sum[tabWithQandPVal[[k,2]],{k,ij[[1,1]],Length[tabWithQ andPVal]}]] cochranExact4[modList_,q_]:=Module[{}, tmptotals={0,0,0}; ٥٩٩

cntQ[i_,k_,numList_]:=If[Sum[numList[[k,j]],{j,1,4}]==i ,tmptotals[[i]]=tmptotals[[i]]+1]; Table[cntQ[i,k,modList],{k,1,Length[modList]},{i,1,3}]; {n1,n2,n3}={tmptotals[[1]],tmptotals[[2]],tmptotals[[3] ]}; compList1=Compositions[n1,4]; compList2=Compositions[n2,6]; compList3=Compositions[n3,4]; comboTab=Table[{compList1[[i]],compList2[[j]],compList3 [[k]]},{i,1,Length[compList1]},{j,1,Length[compList2]},{k,1, Length[compList3]}]; flatComboTab=Flatten[comboTab,1]; flat2Tab=Flatten[flatComboTab,1]; qAndpValForComb4[v_,c_,n1_,n2_,n3_]:=Module[{n1Vals,n2V als,n3Vals,p1,p2,p3}, n1Vals=v[[1]]; n2Vals=v[[2]]; n3Vals=v[[3]]; cVal[1]=n1Vals[[1]]+n2Vals[[1]]+n2Vals[[2]]+n2Vals[[3]]+n3n3Vals[[1]]; cVal[2]=n1Vals[[2]]+n2Vals[[1]]+n2Vals[[4]]+n2Vals[[5]] +n3-n3Vals[[2]]; cVal[3]=n1Vals[[3]]+n2Vals[[2]]+n2Vals[[4]]+n2Vals[[6]] +n3-n3Vals[[3]]; cVal[4]=n1Vals[[4]]+n2Vals[[3]]+n2Vals[[5]]+n2Vals[[6]] +n3-n3Vals[[4]]; n=n1+2 n2+3 n3; p1=n1!/(4^n1Product[n1Vals[[i]]!,{i,1,Length[n1Vals]}]); p2=n2!/(6^n2 Product[n2Vals[[i]]!,{i,1,Length[n2Vals]}]); p3=n3!/(4^n3Product[n3Vals[[i]]!,{i,1,Length[n3Vals]}]) ; p=N[p1 p2 p3]; q4=(12 Sum[cVal[i]^2,{i,1,4}]-3 n^2)/(3 n1+4 n2+3 n3)//N; {q4,p}]; ٦٠٠

qAndpTab4=Map[qAndpValForComb4[#,c,n1,n2,n3]&,flat2Tab] ; qVals=Transpose[qAndpTab4][[1]]; qValsList=Intersection[qVals]; pValTab=Transpose[qAndpTab4][[2]]; places=Map[Position[qVals,#]&,qValsList]; flatPlaces=Table[places[[i]]//Flatten,{i,1,Length[place s]}]; tabWithQandPVal=Table[{qValsList[[i]],Sum[pValTab[[flat Places[[i,k]]]],{k,1,Length[flatPlaces[[i]]]}]},{i,1,Length[ flatPlaces]}]; ij=Position[tabWithQandPVal,q]; Sum[tabWithQandPVal[[k,2]],{k,ij[[1,1]],Length[tabWithQ andPVal]}]] predictions={{1,0,0,0},{0,1,1,1},{1,0,0,1},{1,0,1,1},{1,0,0, 1},{1,0,0,0}}; npmCochransQTest[predictions,mthd->approx] Title: Cochran Q Test Test Statistic: 6. Column Totals: {5,1,2,4} Distribution: Chi Square[ 3 ] PValue: 0.11161 npmCochransQTest[predictions,mthd->exact] Title: Cochran Q Test Test Statistic: 6. Column Totals: {5,1,2,4} Distribution: Exact PValue: 0.145833

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬ ‫ ﺣﯾث ﺗدﺧل اﻟﺑﯾﺎﻧﺎت ﺻف ﺻف ﻛﺎﻟﺗﺎﻟﻰ‬predictions ‫اﻟﻘﺎﺋﻣﺔ اﻟﻣﺳﻣﺎه‬ predictions={{1,0,0,0},{0,1,1,1},{1,0,0,1},{1,0,1,1},{1,0,0, 1},{1,0,0,0}};

‫ اﻟﻣﺧرﺟﺎت‬: ‫ﺛﺎﻧﯾﺎ‬ ‫ﻧﺣﺻل ﻋﻠﯾﮭﺎ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ ﻋﻧد اﺳﺗﺧدام ﺗوزﯾﻊ ﻣرﺑﻊ ﻛﺎى ﻣﻊ اﻟﻣﺧرﺟﺎت‬ npmCochransQTest[predictions,mthd->approx]

‫واﻟﻣﺧرج ھو‬ Title: Cochran Q Test Test Statistic: 6. Column Totals: {5,1,2,4} Distribution: Chi Square[ 3 ] ٦٠١

‫‪0.11161‬‬

‫‪PValue:‬‬

‫واﻻﺣﺻﺎء اﻟﻣﻘدر ﺑﺎﺳﺗﺧدام ﺗوزﯾﻊ ﻣرﺑﻊ ﻛﺎى ھو‬ ‫‪6.‬‬

‫‪Test Statistic:‬‬

‫وﻗﯾﻣﺔ ‪ p-value‬ھو‬ ‫‪0.11161‬‬

‫‪PValue:‬‬

‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻛﺑر ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧﻘﺑل ‪. H 0‬‬ ‫ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ وﺑﺎﺳﺗﺧدام اﻟطرﯾﻘﺔ اﻟﻣﺿﺑوطﺔ ﻧﺣﺻل ﻋﻠﻰ اﻻﺣﺻﺎء اﻟﻣﻘدر‬ ‫]‪npmCochransQTest[predictions,mthd->exact‬‬

‫واﻟﻣﺧرج ھو‬ ‫‪Title: Cochran Q Test‬‬ ‫‪Test Statistic: 6.‬‬ ‫}‪Column Totals: {5,1,2,4‬‬ ‫‪Distribution: Exact‬‬ ‫‪PValue: 0.145833‬‬

‫واﻻﺣﺻﺎء اﻟﻣﻘدر ھو‬ ‫‪6.‬‬

‫‪Test Statistic:‬‬

‫وﻗﯾﻣﺔ ‪ p-value‬ھو‬ ‫‪0.145833‬‬

‫‪PValue:‬‬

‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻛﺑر ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧﻘﺑل ‪. H 0‬‬

‫)‪ (١٣-٧‬اﺧﺗﺑﺎرات ﺣول اﻻرﺗﺑﺎط‬ ‫ﻓﻲ ﻛﺛﯾر ﻣن اﻷﺣﯾﺎن ﯾﻛون ﻟدﯾﻧﺎ ﻣﺟﺗﻣﻊ ﻣﺎ وﻧﻛون ﻣﮭﺗﻣﯾن ﺑﻣﺗﻐﯾرﯾن ﻓﻲ ذﻟك اﻟﻣﺟﺗﻣﻊ وﯾﻛون‬ ‫اھﺗﻣﺎﻣﻧﺎ ﺑﻣﻌرﻓﺔ ھل ھﻧﺎك ﻋﻼﻗﺔ ﺑﯾﻧﮭﻣﺎ أم ﻻ‪ ،‬وإن وﺟدت ﻣﺎ ﻧوﻋﮭﺎ‪ ،‬وإذا أردﻧﺎ اﺧﺗﺑﺎر ﺑﻌض‬ ‫اﻟﻔروض اﻟﺗﻲ ﺗدور ﺣول اﻟﻌﻼﻗﺔ ﺑﯾن اﻟﻣﺗﻐﯾرﯾن وﻛﺎﻧت وﺣدة اﻟﻘﯾﺎس ﻟﻠﻣﺗﻐﯾرﯾن ﺑﻔﺗرة ﻋﻠﻰ اﻷﻗل و‬ ‫ﺗوزﯾﻊ اﻟﻣﺟﺗﻣﻊ اﻟﻣﺳﺣوب ﻣﻧﮫ اﻟﻌﯾﻧﺗﯾن ﯾﺗﺑﻊ اﻟﺗوزﯾﻊ اﻟطﺑﯾﻌﻲ اﻟﺛﻧﺎﺋﻲ ﻓﺈﻧﮫ ﯾﻣﻛن ﺣﺳﺎب ﻣﻌﺎﻣل ارﺗﺑﺎط‬ ‫ﺑﯾرﺳون ﻻﺧﺗﺑﺎر اﻟﻔروض اﻟﺗﻲ ﺗدور ﺣول ﻣﻌﺎﻣل اﻷرﺗﺑﺎط‪ ،‬وﻟﻛن إذا ﻟم ﺗﺳﺗوﻓﻰ ھذه اﻟﺷروط ﻓﻼ‬ ‫ﯾﻣﻛن إﺟراء ھذا اﻻﺧﺗﺑﺎر ‪،‬ﻟﻌﻼج ھذه اﻟﻣﺷﻛﻠﺔ ﻧﺟري اﺧﺗﺑﺎرات ﻻﻣﻌﻠﻣﯾﺔ ﺗﻌﺗﻣد ﻋﻠﻰ اﻟرﺗب ﻣﺛل‬ ‫اﺧﺗﺑﺎر ﺳﺑﯾرﻣﺎن أو ﻛﻧدال وﺑذﻟك ﯾﻣﻛن اﻟﺗﻌﺎﻣل ﻣﻊ اﻟﺑﯾﺎﻧﺎت ذات وﺣدة ﻗﯾﺎس أﻗل ﻣن ﻓﺗرة‪،‬ﻛﺄن ﺗﻛون‬ ‫ﺗرﺗﯾﺑﯾﺔ أو أﺳﻣﯾﺔ‪ ،‬وﻣﻊ أن ﻗﯾﻣﺔ ﻣﻌﺎﻣل اﻻرﺗﺑﺎط ﻓﻲ اﻻﺧﺗﺑﺎرﯾن ﺗﺗراوح ﺑﯾن ‪1‬و‪ -1‬ﻓﺈﻧﻧﺎ ﻻ ﻧﺗوﻗﻊ ﻓﻲ‬ ‫ﺟﻣﯾﻊ اﻟﺣﺎﻻت ﺗﺳﺎوي ﻗﯾﻣﺗﯾﮭﻣﺎ ﻟﻧﻔس اﻟﺑﯾﺎﻧﺎت ‪،‬ﻻﺧﺗﻼف اﻻﺳﺎﻟﯾب اﻟﻣﺳﺗﺧدﻣﺔ ﻓﻲ ﺣﺳﺎب ﻛل ﻣﻧﮭﻣﺎ‪.‬‬

‫ﻣﻌﺎﻣل ارﺗﺑﺎط ﺳﺑﯾرﻣﺎن ﻟﻠرﺗب‬ ‫‪The Spearman Rank Correlation Coefficient‬‬ ‫ﺗﻧﺎوﻟﻧ ﺎ ﻣ ن ﻗﺑ ل اﺧﺗﺑ ﺎرات اﻟﻔ روض اﻟﺗ ﻲ ﺗﺧ ص ﻣﻌﺎﻣ ل ارﺗﺑ ﺎط اﻟﻣﺟﺗﻣ ﻊ ‪ ‬ﺗﺣ ت ﻓ رض أن‬ ‫‪ X , Y‬ﻣﺗﻐﯾ رﯾن ﻋﺷ واﺋﯾﯾن ﻟﮭﻣ ﺎ ﺗوزﯾ ﻊ طﺑﯾﻌ ﻲ ﺛﻧ ﺎﺋﻲ ‪ .‬ﻓ ﻲ ﺣﺎﻟ ﺔ ﻋ دم ﺗﺣﻘ ق اﻟﺷ رط اﻟﺳ ﺎﺑق ﻓﺈﻧ ﮫ‬ ‫ﯾﻣﻛﻧﻧﺎ اﺳﺗﺧدام ﻣﻌﺎﻣل ﺳﺑﯾرﻣﺎن ﻛﺈﺣﺻ ﺎء ﻻﺧﺗﺑ ﺎر ﻋ دم وﺟ ود ﻋﻼﻗ ﺔ ) ارﺗﺑ ﺎط( ﺑ ﯾن اﻟﻣﺗﻐﯾ رﯾن ‪X ,‬‬ ‫‪ .Y‬أﯾﺿﺎ ﯾﻣﻛﻧﻧﺎ اﺳﺗﺧدام ﻣﻌﺎﻣل ﺳﺑﯾرﻣﺎن ﻛﻣﻘﯾﺎس وﺻ ﻔﻰ ﻟﻘ وة اﻻرﺗﺑ ﺎط ﺑ ﯾن ﻣﺗﻐﯾ رﯾن ‪ X , Y‬ﻋﻧ دﻣﺎ‬ ‫‪٦٠٢‬‬

‫ﺗﻛون اﻟﺑﯾﺎﻧ ﺎت ﻓ ﻲ اﻟﻌﯾﻧ ﺔ ﻏﯾ ر ﻣﺗ وﻓرة ﻓ ﻲ ﺷ ﻛل ﺑﯾﺎﻧ ﺎت رﻗﻣﯾ ﺔ وﻟﻛ ن ﯾﻣﻛ ن ﺗﻌﯾ ﯾن رﺗ ب ﻟﮭ ﺎ‪ .‬ﻹﺟ راء‬ ‫اﻻﺧﺗﺑﺎر ﻧﺗﺑﻊ اﻵﺗﻲ ‪:‬‬ ‫ﺗﺧﺗﺎر ﻋﯾﻧﺔ ﻋﺷواﺋﯾﺔ ﻣن اﻟﺣﺟ م ‪ n‬ﻣ ن أزواج اﻟﻣﺷ ﺎھدات اﻟرﻗﻣﯾ ﺔ أو اﻟوﺻ ﻔﯾﺔ ‪ .‬ﻛ ل زوج‬ ‫)أ(‬ ‫ﻣن اﻟﻣﺷﺎھدات ﯾﻣﺛل ﻗراءﺗﯾن ﻣﺄﺧوذﺗﯾن ﻋﻠﻰ ﻧﻔس اﻟﻣﻔردة واﻟﻣﺳﻣﺎة وﺣدة اﻻﻗﺗ ران ‪unit of‬‬ ‫‪ . association‬أﯾﺿﺎ ﻗ د ﺗﻣﺛ ل اﻟﺑﯾﺎﻧ ﺎت ﻣﺷ ﺎھدات ﻣ ﺄﺧوذة ﻣ ن ﻣﺟﺗﻣ ﻊ ﺛﻧ ﺎﺋﻲ ‪ .‬ﺳ وف‬ ‫ﻧرﻣز ﻷزواج اﻟﻣﺷﺎھدات ﻛﺎﻟﺗﺎﻟﻲ ) ‪. (x1, y1 ),(x 2 , y 2 ),...,(x n , y n‬‬ ‫)ب( ﻧرﺗب ﻗﯾم اﻟﻣﺷﺎھدات ﻓﻲ اﻟﻌﯾﻧﺔ واﻟﺗﺎﺑﻌﺔ ﻟﻠﻣﺗﻐﯾر ‪ X‬ﺗﺻﺎﻋدﯾﺎ )أو ﺗﻧﺎزﻟﯾ ﺎ( وﺗﻌط ﻲ رﺗﺑ ﺔ ﻟﻛ ل‬ ‫ﻗﯾﻣﺔ ﻣﺷﺎھدة ﺑﺎﻟﻧﺳﺑﺔ ﻟﻛل ﻗﯾم اﻟﻣﺷﺎھدات اﻷﺧرى‪ .‬ﺳوف ﻧرﻣ ز ﻟرﺗﺑ ﺔ اﻟﻣﺷ ﺎھدة رﻗ م ‪xi ، i‬‬ ‫‪،‬ﺑ ﺎﻟرﻣز ) ‪ . r(x i‬ﻋﻧ دﻣﺎ ‪ r(x i )  1‬ﻓﮭ ذا ﯾﻌﻧ ﻰ أن ‪ xi‬ﺗﻣﺛ ل أﻗ ل ﻗﯾﻣ ﺔ ﻣﺷ ﺎھدة ﻣ ن ﻗ ﯾم‬ ‫اﻟﻣﺗﻐﯾر ‪ X‬ﻓﻲ اﻟﻌﯾﻧﺔ ‪.‬‬ ‫)ج( ﻧرﺗب ﻗﯾم اﻟﻣﺷﺎھدات ﻓﻲ اﻟﻌﯾﻧﺔ واﻟﺗﺎﺑﻌﺔ ﻟﻠﻣﺗﻐﯾ ر ‪ Y‬ﺗﺻ ﺎﻋدﯾﺎ ً )أو ﺗﻧﺎزﻟﯾ ﺎ ً( وﺗﻌط ﻰ رﺗﺑ ﺔ ﻟﻛ ل‬ ‫ﻗﯾﻣﺔ ﻣﺷﺎھدة ﺑﺎﻟﻧﺳﺑﺔ ﻟﻛل ﻗﯾم اﻟﻣﺷﺎھدات اﻷﺧرى ‪.‬ﺳوف ﻧرﻣز ﻟرﺗﺑﺔ اﻟﻣﺷ ﺎھدة رﻗ م ‪، yi ، j‬‬ ‫ﺑ ﺎﻟرﻣز ) ‪. r(yi‬ﻋﻧ دﻣﺎ ‪ r(yi )  1‬ﻓﮭ ذا ﯾﻌﻧ ﻰ أن ‪ yi‬ﺗﻣﺛ ل أﻗ ل ﻗﯾﻣ ﺔ ﻣﺷ ﺎھدة ﻣ ن ﻗ ﯾم‬ ‫اﻟﻣﺗﻐﯾر ‪ Y‬ﻓﻲ اﻟﻌﯾﻧﺔ ‪.‬‬ ‫)ح( ﻋﻧد ﺣدوث ﺗداﺧﻼت ﻧﻌطﻰ ﻣﺗوﺳط اﻟرﺗب اﻟﻣﺗﺗﺎﻟﯾﺔ ﺑدﻻ ً ﻣن اﻟرﺗﺑﺔ ﻛﺎﻟﻣﻌﺗﺎد ‪.‬‬ ‫)خ( إذا ﻛﺎﻧت اﻟﺑﯾﺎﻧﺎت وﺻﻔﯾﺔ ﺑﺈﻣﻛﺎﻧﻧﺎ ﺗﺣوﯾﻠﮭﺎ إﻟﻲ رﺗب ‪.‬‬ ‫ﻗﯾﻣﺔ اﻹﺣﺻﺎء اﻟذي ﯾﻌﺗﻣد ﻋﻠﯾﮫ ﻗرارﻧﺎ ھو ﻣﻌﺎﻣل ارﺗﺑﺎط ﺳﺑﯾرﻣﺎن واﻟذي ﯾﺣﺳ ب ﻣ ن اﻟﺻ ﯾﻐﺔ اﻟﺗﺎﻟﯾ ﺔ‬ ‫‪:‬‬ ‫‪6d i2‬‬ ‫‪rs  1 ‬‬ ‫‪,‬‬ ‫)‪n(n 2  1‬‬ ‫ﺣﯾث‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪ d i    r(x i)  r(yi )  .‬‬ ‫ﻟﻛل زوج ﻣن اﻟﻣﺷﺎھدات وﻋﻧدﻣﺎ ﺗﻛون رﺗﺑﺔ ‪ x‬ﻧﻔس رﺗﺑﺔ ‪ ) y‬ارﺗﺑﺎط ﺗﺎم طردي ( ‪ ،‬ﻓ ﺈن ﻛل اﻟﻔ روق‬ ‫‪ di‬ﺳ وف ﺗﺳ ﺎوى ﺻ ﻔر وﻋﻠ ﻰ ذﻟ ك ‪ . rs  1‬إذا ﻛﺎﻧ ت رﺗﺑ ﺔ ﻛ ل ﻣﺗﻐﯾ ر داﺧ ل ﻛ ل زوج ﻣ ن‬ ‫اﻟﻣﺷﺎھدات ﻋﻛس اﻵﺧر‬ ‫) ارﺗﺑﺎط ﺗﺎم ﻋﻛﺳﻲ ( ‪ ،‬أي إذا ﻛﺎن ‪:‬‬ ‫‪[r(x)  1,r(y)  n],[r(x)  2,r(y)  n  1],...,[r(x)  n,r(y)  1].‬‬ ‫وذﻟ ك ﻷزواج اﻟﻣﺷ ﺎھدات اﻟﺗ ﻲ ﻋ ددھﺎ ‪ n‬ﻓ ﺈن ‪. rs  1‬ﻋﻠ ﻰ ﺳ ﺑﯾل اﻟﻣﺛ ﺎل إذا ﻛ ﺎن ﻟ دﯾﻧﺎ أزواج‬ ‫اﻟﻣﺷﺎھدات اﻟﺗﺎﻟﯾﺔ ‪:‬‬ ‫ﻓﺈن اﻟرﺗب ﺗﺻﺑﺢ ‪:‬‬ ‫‪1‬‬

‫‪2‬‬

‫‪3‬‬

‫‪r(x i ) : 4‬‬

‫‪4‬‬

‫‪3‬‬

‫‪2‬‬

‫‪r(yi ) :1‬‬

‫ﺗﻛون ‪ (x i , yi ) : (12,5),(11,6),(10,7),(9,8) :‬ﺳوف ‪  di2‬وﻋﻠﻰ ذﻟك‬

‫‪(3)2  (1)2  (1)2  (3)2  20,‬‬ ‫‪٦٠٣‬‬

‫وﺑﺎﻟﺗﻌوﯾض ﻓﻲ ﻣﻌﺎدﻟﺔ ﺳﺑﯾرﻣﺎن ﻓﺈن ‪:‬‬ ‫‪rs  1  [(6)(20) /(4)(15)  1  2  1.‬‬ ‫ﻣﻌﺎﻣل ارﺗﺑﺎط ﺳﺑﯾرﻣﺎن ﻻ ﯾﻣﻛن أن ﯾزﯾد ﻋن ‪ +1‬وﻻ ﯾﻣﻛ ن أن ﯾﻘل ﻋن ‪ . –1‬ﻓ رض اﻟﻌ دم واﻟﻔ رض‬ ‫اﻟﺑدﯾل ﺳوف ﯾﻛوﻧﺎن ﻋﻠﻰ اﻟﺷﻛل ‪:‬‬ ‫‪ : H 0‬اﻟﻣﺗﻐﯾرﯾن ﻣﺳﺗﻘﻠﯾن ‪.‬‬ ‫‪ : H1‬ﺗوﺟد ﻋﻼﻗﺔ ﺑﯾن اﻟﻣﺗﻐﯾرﯾن ﻓﻲ ﻧﻔس اﻻﺗﺟﺎه أو اﻻﺗﺟﺎه اﻟﻣﻌﺎﻛس ‪.‬‬ ‫ﺑﻔرض أن ‪ H 0‬ﺻﺣﯾﺢ ﻓﺈن ‪ rs‬ﺗﻣﺛل ﻗﯾﻣﺔ ﻟﻺﺣﺻﺎء ‪ R s‬اﻟذي ﻟﮫ ﺗوزﯾﻊ اﺣﺗﻣﺎﻟﻲ ‪ .‬اﻟﻘﯾم اﻟﺣرﺟﺔ‬ ‫‪ rs,* ‬ﻟﻺﺣﺻﺎء ‪ R s‬ﺗﺳﺗﺧرج ﻣن اﻟﺟدول ﻓﻲ ﻣﻠﺣق )‪ (١٣‬ﻟﻌﯾﻧﺎت ﻣن اﻟﺣﺟم ‪ 4‬وﺣﺗﻰ اﻟﺣﺟم ‪30‬‬ ‫ﻋن ﻣﺳﺗوﯾﺎت ﻣﺧﺗﻠﻔﺔ ﻣن اﻟﻣﻌﻧوﯾﺔ ‪ .‬ﻟﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪ ‬ﻓﺈن ﻣﻧطﻘﺔ اﻟرﻓض ‪ R s  rs,  / 2‬أو‬ ‫‪ . R s   rs,  / 2‬إذا وﻗﻌت ‪rs‬‬

‫ﻓﻲ ﻣﻧطﻘﺔ اﻟرﻓض ﻓﺈﻧﻧﺎ ﻧرﻓض ‪ . H 0‬ﻟﻠﻔرض اﻟﺑدﯾل ‪ : H1‬ﺗوﺟد‬

‫ﻋﻼﻗﺔ ﺑﯾن اﻟﻣﺗﻐﯾرﯾن ﻓﻲ ﻧﻔس اﻻﺗﺟﺎه ﻓﺈن ﻣﻧطﻘﺔ اﻟرﻓض ‪ R s  rs, ‬وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ‬ ‫‪ . ‬ﻟﻠﻔرض اﻟﺑدﯾل ‪ : H1‬ﺗوﺟد ﻋﻼﻗﺔ ﺑﯾن اﻟﻣﺗﻐﯾرﯾن ﻓﻲ اﺗﺟﺎه ﻣﻌﺎﻛس ﻓﺈن ﻣﻧطﻘﺔ اﻟرﻓض‬ ‫‪ R s   rs, ‬وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪. ‬‬ ‫اﻟﻘ رارات اﻟﺳ ﺎﺑﻘﺔ ﺗﺳ ﺗﺧدم ﻋﻧ دﻣﺎ ﻻ ﯾﻛ ون ھﻧ ﺎك ﺗ داﺧل أو أن ﯾﻛ ون ﻋ ددھﺎ ﺻ ﻐﯾرا ً ‪ .‬ﻋﻧ دﻣﺎ‬ ‫ﯾﻛ ون ھﻧ ﺎك ﺗ داﺧل و إذا ﻛ ﺎن ﻋ ددھﺎ ﻛﺑﯾ را ً ) اﻟﻌ دد اﻟﺻ ﻐﯾر ﻟﻠﺗ داﺧﻼت ﻻ ﯾ ؤﺛر ﻋﻠ ﻰ ‪ ( rs‬ﻓﯾﺟ ب‬ ‫إﺟ راء ﺗﺻ ﺣﯾﺢ ﻋﻠ ﻰ ‪ rs‬وﻧﺣﺗ ﺎج ﺟ داول ﺧﺎﺻ ﺔ ﻹﺟ راء اﻻﺧﺗﺑ ﺎر ﺳ وف ﻻ ﻧﺗﻌ رض ﻟﮭ ﺎ‪ .‬ﻋﻧ دﻣﺎ‬ ‫ﯾﻛون ﺣﺟم اﻟﻌﯾﻧﺔ ﻛﺑﯾرا ً ) أﻛﺑر ﻣن ‪ (30‬ﻓﺈﻧﻧﺎ ﻻ ﻧﺳﺗطﯾﻊ اﺳﺗﺧدام اﻟﺟداول وﻟﻛن ﺗم إﺛﺑﺎت أن ‪:‬‬

‫‪z  rs / n  1.‬‬ ‫ﻗﯾﻣ ﺔ ﻟﻠﻣﺗﻐﯾ ر اﻟﻌﺷ واﺋﻲ ‪ Z‬واﻟ ذي ﺗﻘرﯾﺑ ﺎ ً ﯾﺗﺑ ﻊ اﻟﺗوزﯾ ﻊ اﻟطﺑﯾﻌ ﻲ اﻟﻘﯾﺎﺳ ﻲ وذﻟ ك ﺑ ﺎﻓﺗراض أن ‪H 0‬‬ ‫ﺻﺣﯾﺢ ‪.‬‬ ‫ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم‬ ‫‪ H 0 :   0‬ﻋﻧ دﻣﺎ ‪ n  10‬ﻓ ﺈن ‪ t  rs n  2 / 1  rs‬ﻗﯾﻣ ﺔ ﻣ ن ﻗ ﯾم ﻣﺗﻐﯾ ر ﻋﺷ واﺋﻰ ﯾﺗﺑ ﻊ‬ ‫ﺗوزﯾﻊ ‪ T‬ﺑدرﺟﺎت ﺣرﯾﺔ ‪n  2‬‬

‫‪٦٠٤‬‬

‫ﻣﺛﺎل)‪(١٨-٧‬‬ ‫ﻟدراﺳﺔ اﻟﻌﻼﻗﺔ ﺑﯾن اﻟﮭﯾﻣوﺟﻠ وﺑﯾن ‪ ) X‬ﻣﻘﺎﺳ ﺎ ً ‪ ( mg/100 ml‬وﻋ دد ﻛ رات اﻟ دم اﻟﺣﻣ راء‬ ‫‪ Y‬ﺑﺎﻟﻣﻠﯾون ﻟﻛل ﻣﻠﻠﯾﻣﺗر ﻣﻛﻌب ‪ ،‬اﺧﺗﯾرت ﻋﯾﻧ ﺔ ﻋﺷ واﺋﯾﺔ ﻣ ن ‪ 12‬ذﻛ ر ﺑ ﺎﻟﻎ ﻣ ن ﻣﺟﺗﻣ ﻊ ﻣ ﺎ‬ ‫وﺗم ﻗﯾﺎس ﺗرﻛﯾزات اﻟﮭﯾﻣوﺟﻠ وﺑﯾن وﻋ دد ﻛرات اﻟ دم اﻟﺣﻣ راء ﻟﻛ ل ﻣﻔ ردة واﻟﺑﯾﺎﻧ ﺎت ﻣﻌط ﺎة‬ ‫ﻓﻲ اﻟﺟدول اﻟﺗﺎﻟﻰ ‪:‬‬ ‫‪d‬‬ ‫‪d2‬‬ ‫اﻟﺷﺧص‬ ‫اﻟﮭﯾﻣوﺟﻠوﺑﯾن‬ ‫ﻛرات اﻟدم اﻟﺣﻣراء‬ ‫‪x‬‬ ‫رﺗب ‪x‬‬ ‫‪y‬‬ ‫رﺗب ‪y‬‬ ‫‪1‬‬ ‫‪15.2‬‬ ‫‪7.5‬‬ ‫‪5.1‬‬ ‫‪9‬‬ ‫‪-1.5‬‬ ‫‪2.25‬‬ ‫‪2‬‬ ‫‪16.4‬‬ ‫‪12‬‬ ‫‪5.4‬‬ ‫‪11‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪14.2‬‬ ‫‪2‬‬ ‫‪4.5‬‬ ‫‪4‬‬ ‫‪-2‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪13.0‬‬ ‫‪1‬‬ ‫‪4.2‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪5‬‬ ‫‪14.5‬‬ ‫‪3‬‬ ‫‪4.3‬‬ ‫‪2.5‬‬ ‫‪0.5‬‬ ‫‪0.25‬‬ ‫‪6‬‬ ‫‪16.1‬‬ ‫‪11‬‬ ‫‪6.1‬‬ ‫‪12‬‬ ‫‪-1‬‬ ‫‪1‬‬ ‫‪7‬‬ ‫‪15.2‬‬ ‫‪7.5‬‬ ‫‪5.2‬‬ ‫‪10‬‬ ‫‪-2.5‬‬ ‫‪6.25‬‬ ‫‪8‬‬ ‫‪14.8‬‬ ‫‪5‬‬ ‫‪4.3‬‬ ‫‪2.5‬‬ ‫‪2.5‬‬ ‫‪6.25‬‬ ‫‪9‬‬ ‫‪15.7‬‬ ‫‪10‬‬ ‫‪4.7‬‬ ‫‪6‬‬ ‫‪4‬‬ ‫‪16‬‬ ‫‪10‬‬ ‫‪14.9‬‬ ‫‪6‬‬ ‫‪4.8‬‬ ‫‪7.5‬‬ ‫‪-1.5‬‬ ‫‪2.25‬‬ ‫‪11‬‬ ‫‪15.6‬‬ ‫‪9‬‬ ‫‪4.6‬‬ ‫‪5‬‬ ‫‪4‬‬ ‫‪16‬‬ ‫‪12‬‬ ‫‪14.7‬‬ ‫‪4‬‬ ‫‪4.8‬‬ ‫‪7.5‬‬ ‫‪-3.5‬‬ ‫‪12.25‬‬ ‫اﻟﻣطﻠوب اﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪ : H 0‬اﻟﻣﺗﻐﯾرﯾن ﻣﺳﺗﻘﻠﯾن ﺿد اﻟﻔرض اﻟﺑدﯾل ‪ : H1‬ﺗوﺟد‬ ‫ﻋﻼﻗﺔ ﺑﯾن اﻟﻣﺗﻐﯾرﯾن ﻓﻲ ﻧﻔس اﻻﺗﺟﺎه أو اﻻﺗﺟﺎه اﻟﻣﻌﺎﻛس وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ‬ ‫‪.   0.05‬‬

‫اﻟﺣــل‪:‬‬ ‫‪  di2  67.5‬وﻋﻠﻰ ذﻟك ﻓﺈن ‪:‬‬

‫‪6di2‬‬ ‫‪rs  1 ‬‬ ‫)‪n(n 2  1‬‬ ‫)‪6(67.5‬‬ ‫‪1‬‬ ‫)‪12(144  1‬‬ ‫‪ 1  0.2360139  0.763986.‬‬

‫‪٦٠٥‬‬

‫( ﻋﻧ د ﻣﺳ ﺗوى ﻣﻌﻧوﯾ ﺔ‬١٣) ‫ واﻟﻣﺳ ﺗﺧرﺟﺔ ﻣ ن اﻟﺟ دول ﻓ ﻲ ﻣﻠﺣ ق‬rs  0.5804  ‫ وﺑﻣ ﺎ أن‬. R s  0.5804 ‫ أو‬R s  0.5804 ‫ ﻣﻧطﻘ ﺔ اﻟ رﻓض‬n  12 ,  0.025 2 . H 0 ‫ ﺗﻘﻊ ﻓﻲ ﻣﻧطﻘﺔ اﻟرﻓض ﻧرﻓض‬rs  0.763986 ‫ وﻓﯾﻣﺎ ﯾﻠﻰ ﺧطوات‬Mathematica ‫ﺳوف ﯾﺗم ﺣل ھذا اﻟﻣﺛﺎل ﺑﺈﺳﺗﺧدام ﺑرﻧﺎﻣﺞ ﺟﺎھز ﻣﻛﺗوب ﺑﻠﻐﺔ‬ . ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت‬ Off[General::spell1] <<Statistics`MultiDescriptiveStatistics` <<Statistics`NormalDistribution` oppbavg={15.2,16.4,14.2,13,14.5,16.1,15.2,14.8,15.7,14.9,15. 6,14.7}; winpct={5.1,5.4,4.5,4.2,4.3,6.1,5.2,4.3,4.7,4.8,4.6,4.8}; SpearmanRankCorrelation[oppbavg,winpct]//N 0.762743 rank[j_,xlist_]:=Module[{}, k=1; flag=0; xsort=Sort[xlist]; While[xlist[[j]]!=xsort[[k]],k=k+1]; m=k; If[m==Length[xlist],flag=1]; If[flag<1,While[xsort[[m]]==xsort[[m+1]],m=m+1]]; num1=m; num2=m-k+1; Sum[val,{val,k,num1}]/num2//N] n=Length[oppbavg] 12 rank1=Table[rank[num,oppbavg],{num,1,n}] {7.5,12.,2.,1.,3.,11.,7.5,5.,10.,6.,9.,4.} rank2=Table[rank[num,winpct],{num,1,n}] {9.,11.,4.,1.,2.5,12.,10.,2.5,6.,7.5,5.,7.5} rexact=Correlation[rank1,rank2] 0.762743 tval=Sqrt[n-2] rexact/Sqrt[1-rexact^2] 3.7297 2(1-CDF[StudentTDistribution[n-2],Abs[tval]]) 0.00391228

: ‫وﻓﯾﻣﺎ ﯾﻠﻰ اﻟﻣدﺧﻼت و اﻟﻣﺧرﺟﺎت ﻟﮭذا اﻟﺑرﻧﺎﻣﺞ‬ ‫ اﻟﻣدﺧﻼت‬:‫اوﻻ‬ y ‫ ﻟﻘﯾم‬winpct ‫ و اﻟﻘﺎﺋﻣﺔ اﻟﻣﺳﻣﺎه‬x ‫ ﻟﻘﯾم‬oppbavg ‫اﻟﻘﺎﺋﻣﺔ اﻟﻣﺳﻣﺎه‬

‫ اﻟﻣﺧرﺟﺎت‬: ‫ﺛﺎﻧﯾﺎ‬ ‫ﻣﻌﺎﻣل ﺳﺑﯾرﻣﺎن ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ SpearmanRankCorrelation[oppbavg,winpct]//N ٦٠٦

‫واﻟﻣﺧرج ھو‬ ‫‪0.762743‬‬

‫وﻣن اﻻﻣر اﻟﺗﺎﻟﻰ ﻧﺣﺻل ﻋﻠﻰ رﺗب ‪x‬‬ ‫]}‪rank1=Table[rank[num,oppbavg],{num,1,n‬‬ ‫واﻟﻤﺨﺮج ھﻮ‬ ‫}‪{7.5,12.,2.,1.,3.,11.,7.5,5.,10.,6.,9.,4.‬‬

‫وﻣن اﻻﻣر اﻟﺗﺎﻟﻰ ﻧﺣﺻل ﻋﻠﻰ رﺗب ‪y‬‬ ‫]}‪rank2=Table[rank[num,winpct],{num,1,n‬‬ ‫واﻟﻤﺨﺮج ھﻮ‬ ‫}‪{9.,11.,4.,1.,2.5,12.,10.,2.5,6.,7.5,5.,7.5‬‬

‫ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪ H 0 :   0‬ﻧﺳﺗﺧدم اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫)]]‪2(1-CDF[StudentTDistribution[n-2],Abs[tval‬‬

‫واﻟﻣﺧرج ھو‬ ‫‪0.00391228‬‬

‫واﻟذى ﯾﻣﺛل ﻗﯾﻣﺔ ‪p-value‬‬ ‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻗل ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧرﻓض ‪. H0‬‬ ‫ﻣﺛﺎل) ‪(١٩-٧‬‬

‫ﯾﻌطﻰ اﻟﺟدول اﻟﺗﺎﻟﻰ ﺗﻘدﯾرات ‪ 10‬طﻼب ﻓﻲ ﻛل ﻣن اﻹﺣﺻﺎء واﻟرﯾﺎﺿﯾﺎت ‪.‬‬ ‫ﺟﯾد‬

‫ﺟﯾد‬

‫ﻣﻣﺗﺎز‬

‫ﺟﯾد‬

‫ﻣﻣﺗﺎز‬

‫ﺟﯾد‬ ‫ﺟدا‬

‫ﻣﻣﺗﺎز‬

‫ﻣﻘﺑول‬

‫ﺟﯾد‬ ‫ﺟدا‬

‫ﺟﯾد‬

‫ﻣﻘﺑول‬

‫ﺟﯾد‬

‫ﺟﯾد‬ ‫ﺟدا‬

‫ﺟﯾد‬

‫ﺟﯾد ﺗﻘدﯾرات‬ ‫اﻟرﯾﺎﺿﯾﺎت‬

‫ﺟﯾد ﺟدا‬ ‫ﺟﯾد‬

‫ﺟﯾد‬ ‫ﺟدا ً‬

‫ﻣﻘﺑول‬

‫ﺗﻘدﯾرات‬ ‫اﻹﺣﺻﺎء‬

‫أﺧﺗﺑر ﻓرض اﻟﻌدم ‪ : H0‬اﻟﻣﺗﻐﯾرﯾن ﻣﺳﺗﻘﻠﯾن ‪.‬‬ ‫ﺿد اﻟﻔرض اﻟﺑدﯾل ‪:‬‬ ‫‪ : H1‬ﺗوﺟد ﻋﻼﻗﺔ ﺑﯾن اﻟﻣﺗﻐﯾرﯾن ﻓﻲ ﻧﻔس اﻻﺗﺟﺎه ‪.‬‬ ‫وذﻟك ﻋﻧد ﻣﺳﺗوى ﻣﻌﻧوﯾﺔ ‪.   0.05‬‬

‫اﻟﺣــل‪:‬‬ ‫ﺳ وف ﯾ ﺗم ﺣ ل ھ ذا اﻟﻣﺛ ﺎل ﺑﺈﺳ ﺗﺧدام ﻧﻔ س اﻟﺑرﻧ ﺎﻣﺞ اﻟﺧ ﺎص ﺑﺎﻟﻣﺛ ﺎل اﻟﺳ ﺎﺑق وﺳ وف ﻧﻛﺗﻔ ﻰ ھﻧ ﺎ‬ ‫ﺑﺗوﺿ ﯾﺢ اﻟﻣ دﺧﻼت واﻟﺧرﺟ ﺎت‪ .‬وﺑﻣ ﺎ ان اﻟﺑﯾﺎﻧ ﺎت وﺻ ﻔﯾﺔ ﻓﺳ وف ﻧﺣوﻟﮭ ﺎ اﻟ ﻰ ﺑﯾﺎﻧ ﺎت رﻗﻣﯾ ﺔ ﺣﯾ ث‬ ‫ﯾﻌط ﻰ ﻋﻠ ﻰ ﺳ ﺑﯾل اﻟﻣﺛ ﺎل اﻟ رﻗم واﺣ د اﻟ ﻰ ﻣﻘﺑ ول واﻟ رﻗم اﺛﻧ ﯾن اﻟ ﻰ ﺟﯾ د وھﻛ ذا‪ .‬وﻓﯾﻣ ﺎ ﯾﻠ ﻰ ﺧط وات‬ ‫اﻟﺑرﻧﺎﻣﺞ واﻟﻣﺧرﺟﺎت ‪.‬‬ ‫]‪Off[General::spell1‬‬ ‫‪٦٠٧‬‬

<<Statistics`MultiDescriptiveStatistics` <<Statistics`NormalDistribution` oppbavg={3,2,1,2,3,4,2,4,2,2}; winpct={2,2,1,3,2,3,1,4,3,3}; SpearmanRankCorrelation[oppbavg,winpct]//N 0.508678 rank[j_,xlist_]:=Module[{}, k=1; flag=0; xsort=Sort[xlist]; While[xlist[[j]]!=xsort[[k]],k=k+1]; m=k; If[m==Length[xlist],flag=1]; If[flag<1,While[xsort[[m]]==xsort[[m+1]],m=m+1]]; num1=m; num2=m-k+1; Sum[val,{val,k,num1}]/num2//N] n=Length[oppbavg] 10 rank1=Table[rank[num,oppbavg],{num,1,n}] {7.5,4.,1.,4.,7.5,9.5,4.,9.5,4.,4.} rank2=Table[rank[num,winpct],{num,1,n}] {4.,4.,1.5,7.5,4.,7.5,1.5,10.,7.5,7.5} rexact=Correlation[rank1,rank2] 0.508678 tval=Sqrt[n-2] rexact/Sqrt[1-rexact^2] 1.67111 2(1-CDF[StudentTDistribution[n-2],Abs[tval]]) 0.133244

‫ اﻟﻣﺧرﺟﺎت‬: ‫ﺛﺎﻧﯾﺎ‬ ‫ﻣﻌﺎﻣل ﺳﺑﯾرﻣﺎن ﻧﺣﺻل ﻋﻠﯾﮫ ﻣن اﻻﻣر اﻟﺗﺎﻟﻰ‬ SpearmanRankCorrelation[oppbavg,winpct]//N

‫واﻟﻣﺧرج ھو‬ 0.508678

x ‫وﻣن اﻻﻣر اﻟﺗﺎﻟﻰ ﻧﺣﺻل ﻋﻠﻰ رﺗب‬ rank1=Table[rank[num,oppbavg],{num,1,n}] ‫واﻟﻤﺨﺮج ھﻮ‬ {7.5,4.,1.,4.,7.5,9.5,4.,9.5,4.,4.}

y ‫وﻣن اﻻﻣر اﻟﺗﺎﻟﻰ ﻧﺣﺻل ﻋﻠﻰ رﺗب‬ ٦٠٨

‫]}‪rank2=Table[rank[num,winpct],{num,1,n‬‬ ‫واﻟﻤﺨﺮج ھﻮ‬ ‫}‪{4.,4.,1.5,7.5,4.,7.5,1.5,10.,7.5,7.5‬‬

‫ﻻﺧﺗﺑﺎر ﻓرض اﻟﻌدم ‪ H 0 :   0‬ﻧﺳﺗﺧدم اﻻﻣر اﻟﺗﺎﻟﻰ‬ ‫)]]‪2(1-CDF[StudentTDistribution[n-2],Abs[tval‬‬

‫واﻟﻣﺧرج ھو‬ ‫‪0.133244‬‬

‫واﻟذى ﯾﻣﺛل ﻗﯾﻣﺔ ‪p-value‬‬ ‫وذﻟك ﻻﺧﺗﺑﺎر ﻣن ﺟﺎﻧﺑﯾن وﺑﻣﺎ ان اﻟﻘﯾﻣﺔ اﻛﺑر ﻣن ‪ 0.05‬ﻓﺈﻧﻧﺎ ﻧﻘﺑل ‪. H 0‬‬

‫‪٦٠٩‬‬

٦١٠

‫اﳌﺮاﺟـﻊ‬ ‫‪REFERENCES‬‬ ‫أوﻻً ‪ :‬اﻟﻤﺮاﺟﻊ اﻟﻌﺮﺑﻴﺔ‬ ‫‪ - ١‬أﺣﻣ د ﻋﺑ ﺎدة ﺳ رﺣﺎن ‪ ، (١٩٦٨) ،‬ﻣﻘدﻣ ﺔ ﻓ ﻰ ط رق اﻟﺗﺣﻠﯾ ل اﻹﺣﺻ ﺎﺋﻲ ‪ ،‬ﻣﻌﮭ د اﻟدراﺳ ﺎت‬ ‫واﻟﺑﺣوث اﻹﺣﺻﺎﺋﯾﺔ – ﺟﺎﻣﻌﺔ اﻟﻘﺎھرة‪.‬‬ ‫‪ - ٢‬أﺣﻣ د ﻋﺑ ﺎدة ﺳ رﺣﺎن ‪ ، (١٩٨٣) ،‬ﺗﺻ ﻣﯾم اﻟﺗﺟ ﺎرب وﺗﺣﻠﯾﻠﮭ ﺎ ‪ ،‬دار اﻟﻛﺗ ب اﻟﺟﺎﻣﻌﯾ ﺔ –‬ ‫اﻟﻘﺎھرة ‪ ،‬ﺟﻣﮭورﯾﺔ ﻣﺻر اﻟﻌرﺑﯾﺔ ‪.‬‬ ‫‪ - ٣‬أﺣﻣد ﻓؤاد ﻏﺎﻟب وآﺧرون ‪ ، (١٩٩٣) ،‬اﻟرﯾﺎﺿﺔ ﻟدارﺳ ﻲ اﻟﻌﻠ وم اﻟﺣﯾوﯾ ﺔ – ﺗرﺟﻣ ﺔ ﻟﻛﺗ ﺎب‬ ‫ﺟ ﺎدﯾش س‪ .‬أرﺑ ﺎ وروﺑ ﯾن و‪.‬ﻻردﻧ ر – اﻟ دار اﻟدوﻟﯾ ﺔ ﻟﻠﻧﺷ ر واﻟﺗوزﯾ ﻊ – اﻟﻘ ﺎھرة –‬ ‫ﺟﻣﮭورﯾﺔ ﻣﺻر اﻟﻌرﺑﯾﺔ ‪.‬‬ ‫‪ - ٤‬أﻧ ﯾس إﺳ ﻣﺎﻋﯾل ﻛﻧﺟ و ‪ ، (١٩٩٣) ،‬اﻹﺣﺻ ﺎء واﻹﺣﺗﻣ ﺎل – ﺟﺎﻣﻌ ﺔ اﻟﻣﻠ ك ﺳ ﻌود – ﻋﻣﺎﻧ ﮫ‬ ‫ﺷؤن اﻟﻣﻛﺗﺑﺎت‪.‬‬ ‫‪ - ٥‬ﺑدرﯾ ﺔ ﺷ وﻗﻰ ﻋﺑ د اﻟوھ ﺎب وﻣﺣﻣ د ﻛﺎﻣ ل اﻟﺷ رﺑﯾﻧﻰ ‪ ، (١٩٨٤) ،‬اﻟﻣﺑ ﺎدئ اﻷوﻟﯾ ﺔ ﻓ ﻰ‬ ‫اﻹﺣﺻﺎء – ﺗرﺟﻣﺔ ﻟﻛﺗﺎب ﺑول ج‪ .‬ھوﯾل – اﻟطﺑﻌﺔ اﻟراﺑﻌﺔ – دار ﺟون واﯾﻠﻰ وأﺑﻧﺎﺋﮫ‪.‬‬ ‫‪ - ٦‬ﺛروت ﻣﺣﻣ د ﻋﺑ د اﻟﻣ ﻧﻌم ‪ ، (٢٠٠٤) ،‬ﺗﺻ ﻣﯾم وﺗﺣﻠﯾ ل اﻟﺗﺟ ﺎرب – ﻣﻛﺗﺑ ﺔ اﻻﻧﺟﻠ و اﻟﻣﺻ رﯾﺔ‬ ‫– اﻟﻘﺎھرة‪.‬‬ ‫‪ - ٧‬ﺛروت ﻣﺣﻣد ﻋﺑد اﻟﻣﻧﻌم ‪ ، (٢٠٠٥) ،‬اﻻﻧﺣدار – ﻣﻛﺗﺑﺔ اﻻﻧﺟﻠو اﻟﻣﺻرﯾﺔ – اﻟﻘﺎھرة‪.‬‬ ‫‪ - ٨‬ﺛ روت ﻣﺣﻣ د ﻋﺑ د اﻟﻣ ﻧﻌم ‪ ، (٢٠١١) ،‬ﻣ دﺧل ﺣ دﯾث ﻟﻼﺣﺻ ﺎء واﻻﺣﺗﻣ ﺎﻻت– اﻟطﺑﻌ ﺔ‬ ‫اﻟراﺑﻌﺔ – ﻣﻛﺗﺑﺔ اﻟﻌﺑﯾﻛﺎن – اﻟدﻣﺎم – اﻟﻣﻣﻠﻛﺔ اﻟﻌرﺑﯾﺔ اﻟﺳﻌودﯾﺔ‪.‬‬ ‫‪ - ٩‬ﺟﻼل ﻣﺻطﻔﻰ اﻟﺻﯾﺎد وﻣﺣﻣ د اﻟدﺳ وﻗﻰ ﺣﺑﯾ ب ‪ ، (١٩٩٠) ،‬ﻣﻘدﻣ ﺔ ﻓ ﻰ اﻟط رق اﻹﺣﺻ ﺎﺋﯾﺔ‬ ‫– اﻟطﺑﻌﺔ اﻟﺛﺎﻧﯾﺔ – ﺗﮭﺎﻣﺔ – ﺟدة – اﻟﻣﻣﻠﻛﺔ اﻟﻌرﺑﯾﺔ اﻟﺳﻌودﯾﺔ‪.‬‬ ‫‪ - ١٠‬ذﻛرﯾ ﺎ اﻟﺷ رﺑﯾﻧﻰ ‪ ، (١٩٩٥) ،‬اﻹﺣﺻ ﺎء وﺗﺻ ﻣﯾم اﻟﺗﺟ ﺎرب ﻓ ﻰ اﻟﺑﺣ وث اﻟﻧﻔﺳ ﯾﺔ واﻟﺗرﺑوﯾ ﺔ‬ ‫واﻹﺟﺗﻣﺎﻋﯾﺔ – ﻣﻛﺗﺑﺔ اﻷﻧﺟﻠو اﻟﻣﺻرﯾﺔ – اﻟﻘﺎھرة‪.‬‬ ‫‪ - ١١‬راﻓ ت رﯾ ﺎض رزق ﷲ ‪ ، (٢٠٠٠) ،‬ﻣﺎﺛﯾﻣﺎﺗﯾﻛ ﺎ – اﻟرﯾﺎﺿ ﯾﺎت ﺑﺎﺳ ﺗﺧدام اﻟﻛﻣﺑﯾ وﺗر –‬ ‫اﻟﻣﻛﺗﺑﺔ اﻻﻛﺎدﯾﻣﯾﺔ– اﻟﻘﺎھرة‪.‬‬ ‫‪٦١١‬‬

‫‪ - ١٢‬رﺑﯾ ﻊ ذﻛ ر ﻋ ﺎﻣر ‪ ، (١٩٨٩) ،‬ﺗﺣﻠﯾ ل اﻻﻧﺣ دار – أﺳ ﺎﻟﯾﺑﺔ وﺗطﺑﯾﻘﺎﺗ ﮫ اﻟﻌﻠﻣﯾ ﺔ ﺑﺎﺳ ﺗﺧدام‬ ‫اﻟﺑرﻧﺎﻣﺞ اﻟﺟﺎھز ‪ - SPSS/PC‬ﻣﻌﮭد اﻟدراﺳﺎت واﻟﺑﺣوث اﻹﺣﺻﺎﺋﯾﺔ – ﺟﺎﻣﻌﺔ اﻟﻘﺎھرة‪.‬‬ ‫‪ - ١٣‬ﺳ ﻌدﯾﺔ ﺣ ﺎﻓظ ﻣﻧﺗﺻ ر ‪ ، (١٩٨٢) ،‬ﻣﻠﺧﺻ ﺎت ﺷ وم – ﻧظرﯾ ﺎت وﻣﺳ ﺎﺋل ﻓ ﻲ اﻹﺣﺻ ﺎء‬ ‫واﻻﻗﺗﺻﺎد اﻟﻘﯾﺎﺳﻲ – ﺗرﺟﻣﺔ ﻟﻛﺗﺎب دوﻣﯾﻧﯾك ﺳﺎﻟﻔﺎﺗور – دار ﻣﺎﻛﺟروھﯾل – ﻧﯾوﯾورك‪.‬‬ ‫‪ - ١٤‬ﺳﻣﯾر ﻛﺎﻣل ﻋﺎﺷور وﺳ ﺎﻣﯾﺔ ﺳ ﺎﻟم أﺑ و اﻟﻔﺗ وح ‪ ، (١٩٩٠) ،‬ﻣﻘدﻣ ﺔ ﻓ ﻰ اﻹﺣﺻ ﺎء اﻟﺗﺣﻠﯾﻠ ﻰ‬ ‫ ﻣﻌﮭد اﻟدراﺳﺎت واﻟﺑﺣوث اﻹﺣﺻﺎﺋﯾﺔ – ﺟﺎﻣﻌﺔ اﻟﻘﺎھرة‪.‬‬‫‪ - ١٥‬ﺳﻣﯾر ﻛﺎﻣل ﻋﺎﺷور وﺳ ﺎﻣﯾﺔ ﺳ ﺎﻟم أﺑ و اﻟﻔﺗ وح ‪ ، (١٩٩٠) ،‬ﻣﻘدﻣ ﺔ ﻓ ﻰ اﻹﺣﺻ ﺎء اﻟوﺻ ﻔﻰ‬ ‫ ﻣﻌﮭد اﻟدراﺳﺎت واﻟﺑﺣوث اﻹﺣﺻﺎﺋﯾﺔ – ﺟﺎﻣﻌﺔ اﻟﻘﺎھرة‪.‬‬‫‪ - ١٦‬ﺳ ﻣﯾر ﻛﺎﻣ ل ﻋﺎﺷ ور وﺳ ﺎﻣﯾﺔ ﺳ ﺎﻟم أﺑ و اﻟﻔﺗ وح ‪ ، (١٩٩٥) ،‬اﻻﺧﺗﺑ ﺎرات اﻟﻼﻣﻌﻠﻣﯾ ﺔ ‪-‬‬ ‫ﻣﻌﮭد اﻟدراﺳﺎت واﻟﺑﺣوث اﻹﺣﺻﺎﺋﯾﺔ – ﺟﺎﻣﻌﺔ اﻟﻘﺎھرة‪.‬‬ ‫‪ - ١٧‬ﻋدﻧﺎن ﺑن ﻣﺎﺟد ﻋﺑد اﻟرﺣﻣن ﺑ رى وﻣﺣﻣ ود ﻣﺣﻣ د ﺑ راھﯾم ھﻧﯾ دى وأﻧ ور أﺣﻣ د ﻣﺣﻣ د ﻋﺑ د‬ ‫ﷲ ‪ ، (١٩٩١) ،‬ﻣﺑﺎدئ اﻹﺣﺻﺎء واﻻﺣﺗﻣﺎﻻت – ﻋﻣﺎده ﺷ ؤون اﻟﻣﻛﺗﺑ ﺎت – ﺟﺎﻣﻌ ﺔ اﻟﻣﻠك‬ ‫ﺳﻌود – اﻟﻣﻣﻠﻛﺔ اﻟﻌرﺑﯾﺔ اﻟﺳﻌودﯾﺔ‪.‬‬ ‫‪ - ١٨‬ﻋﻔ ﺎف اﻟ دش ‪ ، (١٩٩٤) ،‬اﻹﺣﺻ ﺎء اﻟﺗطﺑﯾﻘ ﻰ ﻟﻠﺗﺟ ﺎرﯾﯾن – اﻟطﺑﻌ ﺔ اﻟﺛﺎﻧﯾ ﺔ – ﺟﺎﻣﻌ ﺔ‬ ‫ﺣﻠوان – اﻟﻘﺎھرة‪.‬‬ ‫‪ - ١٩‬ﻣﺣﻣ د ﺻ ﺑﺣﻰ أﺑ و ﺻ ﺎﻟﺢ وﻋ دﻧﺎن ﻣﺣﻣ د ﻋ وض ‪ ، (٠١٩٨٣ ،‬ﻣﻘدﻣ ﺔ ﻓ ﻰ اﻹﺣﺻ ﺎء –‬ ‫اﻟطﺑﻌﺔ اﻟراﺑﻌﺔ – دار ﺟون واﯾﻠﻰ وأﺑﻧﺎﺋﺔ – ﻧﯾوﯾورك‪.‬‬ ‫‪ - ٢٠‬ﻣﺣﻣد ﻣﺣﻣ د اﻟط ﺎھر اﻹﻣ ﺎم )‪ ، (١٩٩٤‬ﺗﺻ ﻣﯾم وﺗﺣﻠﯾ ل اﻟﺗﺟ ﺎرب – دار اﻟﻣ رﯾﺦ – اﻟﻣﻣﻠﻛ ﺔ‬ ‫اﻟﻌرﺑﯾﺔ اﻟﺳﻌودﯾﺔ‪.‬‬

‫ﺛﺎﻧﻴﺎ ‪ :‬اﻟﻤﺮاﺟﻊ اﻷﺟﻨﺒﻴﺔ‬ ‫‪1- Abell, M. L. et.al. (1999) Statistics with Mathematica, Academic‬‬ ‫‪Press, New York.‬‬ ‫‪2- Abell, M. L. et.al. (1992) The Mathematica Handbook, Academic‬‬ ‫‪Press, New York.‬‬ ‫‪3- Bain, L. J. (1992) Introduction to Probability and Mathematical‬‬ ‫‪Statistics, Second Edition, Duxbury Press - An Imprint of‬‬ ‫‪Wadsworth Publishing Company Belmont, California.‬‬ ‫‪٦١٢‬‬

4- Cangelosi, V. E.; Taylor, P. H. and Rice, P. F. (1979) Basic statistics - A Real World Approach, Second Edition, West Publishing Company, New York. 5- Cochran, W. G. (1963) Sampling Techniques, Second Edition, New York : John Willey & Sons, Inc. 6- Daniel, W. W. (1978) Applied Nonparametric Statistic, Houghton Mifflin Company, London. 7- Devore, J. L. (1995) Probability and Statistics for Engineering and the Sciences, Fourth Edition, Duxburg Press-An International Themson Publishing Company, London. 8- Draper, N. R. and Smith, H. (1981) Applied Regression Analysis, Second Edition, John Wiley & Sons Inc., U.S.A. . 9- Frank, H. and Althoen, S. C. (1997) Statistics- Concepts and Applications, Low Price Edition, Cambridge University Press.

10- Hamburg, M. (1979) Basic Statistics : A Modern Approach, Second Edition, Harcourt Brace Jovanovich, Inc., New York. 11- Mendenhall, W. (1975) Introduction to Probability and Statistics, Company, Inc. Belmont, California Fourth Edition, Duxburg Press, A Division of Wadsworth Publishing 12- Neter, J.; Wasserman, W. and Whitmere, G. A. (1993) Applied Statistics, Fourth Edition, ALLYN AND BACON, London. 13- Owen, F. and Jones, R. (1994) Statistics, Fourth Edition, Pitman Publishing, London. 14- Schelfer, W. (1979) Statistics for the Bidogical Sciences, Second Edition, Addison-Wesly Publishing Company. Inc. Philippines.

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15- Yates, F. (1934) Contingency Tables Involving Small Numbers and the 2 13-Test, J. Roy. Statist. Soc., 1,217-235. 16- Walpole, R. E. (1982) Introduction to Statistics, Macmillan Publishing Co. Inc. New York. 17- Weisberg, S. (1980), Applied Linear Regression, John Wiley & Sons Inc., New York, U.S.A.. 18- Winer, B. J. Brown, D. R. and Michels, K. M.(1991) Statistical Experimental Design, Third Edition, McGraw-Hill, Inc., New York. 19- Yamane, T. (1967) Elementary Sampling Theory, Prentice-Hall, Inc., Englewood Cliffs, N. J.

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٦١٥

‫اﻟﻤﻼﺣﻖ‬ ‫ﻣﻠﺤﻖ ) ‪ ( ١‬ﺟﺪول اﻟﻤﺴﺎﺣﺎت ﺗﺤﺖ اﻟﻤﻨﺤﻨﻰ اﻟﻄﺒﯿﻌﻲ اﻟﻘﯿﺎﺳﻲ )‪. P(0  Z  z‬‬ ‫ﻣﻠﺤﻖ ) ‪ ( ٢‬ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ‪ t ‬ﻟﺘﻮزﯾﻊ ‪. t‬‬ ‫ﻣﻠﺤﻖ ) ‪ ( ٣‬ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ‪  2‬ﻟﺘﻮزﯾﻊ ‪.  2‬‬ ‫ﻣﻠﺤﻖ ) ‪ ( ٤‬ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ) ‪ f  (1,  2‬ﻟﺘﻮزﯾﻊ ‪ F‬ﻋﻨﺪ ) ‪. (  0.05‬‬ ‫ﻣﻠﺤﻖ ) ‪ (٥‬ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ) ‪ f  (1,  2‬ﻟﺘﻮزﯾﻊ ‪ F‬ﻋﻨﺪ ) ‪. (  0.01‬‬ ‫ﻣﻠﺤﻖ ) ‪ ( ٦‬ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ )‪ q  (p , ‬ﻟﻨﯿﻮﻣﻦ ‪.‬‬ ‫ﻣﻠﺤﻖ ) ‪ ( ٧‬ﺟﺪول ﺣﺴﺎب‬

‫‪r‬‬ ‫)‪ b ( x ; n , p‬‬ ‫‪x 0‬‬

‫ﻟﻤﺘﻐﯿﺮ ﻋﺸﻮاﺋﻲ ﯾﺘﺒﻊ ﺗﻮزﯾﻊ ذي اﻟﺤﺪﯾﻦ ‪.‬‬

‫ﻣﻠﺤﻖ ) ‪ ( ٨‬ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ )‪ d ( n , ), d ( n , ‬ﻻﺧﺘﺒﺎر إﺷﺎرة اﻟﺮﺗﺐ ‪.‬‬ ‫ﻣﻠﺤﻖ ) ‪ ( ٩‬ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ‪ r1‬اﻟﺴﻔﻠﻲ ﻻﺧﺘﺒﺎر اﻟﺪورات ‪.‬‬ ‫ﻣﻠﺤﻖ ) ‪ ( ١٠‬ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ‪ r2‬اﻟﻌﻠﯿﺎ ﻻﺧﺘﺒﺎر اﻟﺪورات ‪.‬‬ ‫ﻣﻠﺣق ) ‪ ( ١١‬ﺟدول اﻟﻘﯾم اﻟﺣرﺟﺔ ﻻﺧﺗﺑﺎر ‪Mann-Whitney-Wilcoxon‬‬ ‫ﻣﻠﺤﻖ ) ‪ ( ١٢‬ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ﻻﺧﺘﺒﺎر ‪. Kruskal – Wallis‬‬

‫ﻣﻠﺤﻖ ) ‪ ( ١٣‬ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ‬

‫*‬ ‫‪rs , ‬‬

‫ﻻﺧﺘﺒﺎر ﺳﺒﯿﺮﻣﺎن ‪.‬‬

‫‪٦١٦‬‬

‫ﻣﻠﺤﻖ )‪(١‬‬

‫ﺟدول اﻟﻣﺳﺎﺣﺎت ﺗﺣت اﻟﻣﻧﺣﻧﻰ اﻟطﺑﯾﻌﻲ اﻟﻘﯾﺎﺳﻲ‬ ‫)‪P(0<Z<z‬‬ ‫‪.09‬‬ ‫‪.0359‬‬ ‫‪.0753‬‬ ‫‪.1141‬‬ ‫‪.1517‬‬ ‫‪.1879‬‬ ‫‪.2224‬‬ ‫‪.2549‬‬ ‫‪.2852‬‬ ‫‪.3133‬‬ ‫‪.3389‬‬ ‫‪.3621‬‬ ‫‪.3830‬‬ ‫‪.4015‬‬ ‫‪.4177‬‬ ‫‪.4319‬‬ ‫‪.4441‬‬ ‫‪.4545‬‬ ‫‪.4633‬‬ ‫‪.4706‬‬ ‫‪.4767‬‬ ‫‪.4817‬‬ ‫‪.4857‬‬ ‫‪.4890‬‬ ‫‪.4916‬‬ ‫‪.4936‬‬ ‫‪.4952‬‬ ‫‪.4964‬‬ ‫‪.4974‬‬ ‫‪.4981‬‬ ‫‪.4986‬‬ ‫‪.4990‬‬

‫‪.08‬‬ ‫‪.0319‬‬ ‫‪.0714‬‬ ‫‪.1103‬‬ ‫‪.1480‬‬ ‫‪.1844‬‬ ‫‪.2190‬‬ ‫‪.2517‬‬ ‫‪.2823‬‬ ‫‪.3106‬‬ ‫‪.3365‬‬ ‫‪.3599‬‬ ‫‪.3810‬‬ ‫‪.3997‬‬ ‫‪.4162‬‬ ‫‪.4306‬‬ ‫‪.4429‬‬ ‫‪.4535‬‬ ‫‪.4625‬‬ ‫‪.4699‬‬ ‫‪.4761‬‬ ‫‪.4812‬‬ ‫‪.4854‬‬ ‫‪.4887‬‬ ‫‪.4913‬‬ ‫‪.4934‬‬ ‫‪.4951‬‬ ‫‪.4963‬‬ ‫‪.4973‬‬ ‫‪.4980‬‬ ‫‪.4986‬‬ ‫‪.4990‬‬

‫‪.07‬‬ ‫‪.0279‬‬ ‫‪.0675‬‬ ‫‪.1064‬‬ ‫‪.1443‬‬ ‫‪.1808‬‬ ‫‪.2157‬‬ ‫‪.2486‬‬ ‫‪.2794‬‬ ‫‪.3078‬‬ ‫‪.3340‬‬ ‫‪.3577‬‬ ‫‪.3790‬‬ ‫‪.3980‬‬ ‫‪.4147‬‬ ‫‪.4292‬‬ ‫‪.4418‬‬ ‫‪.4525‬‬ ‫‪.4616‬‬ ‫‪.4693‬‬ ‫‪.4756‬‬ ‫‪.4808‬‬ ‫‪.4850‬‬ ‫‪.4884‬‬ ‫‪.4911‬‬ ‫‪.4932‬‬ ‫‪.4949‬‬ ‫‪.4962‬‬ ‫‪.4972‬‬ ‫‪.4979‬‬ ‫‪.4985‬‬ ‫‪.4989‬‬

‫‪.06‬‬ ‫‪.0239‬‬ ‫‪.0636‬‬ ‫‪.1026‬‬ ‫‪.1406‬‬ ‫‪.1772‬‬ ‫‪.2123‬‬ ‫‪.2454‬‬ ‫‪.2764‬‬ ‫‪.3051‬‬ ‫‪.3315‬‬ ‫‪.3554‬‬ ‫‪.3770‬‬ ‫‪.3962‬‬ ‫‪.4131‬‬ ‫‪.4279‬‬ ‫‪.4406‬‬ ‫‪.4515‬‬ ‫‪.4608‬‬ ‫‪.4686‬‬ ‫‪.4750‬‬ ‫‪.4803‬‬ ‫‪.4846‬‬ ‫‪.4881‬‬ ‫‪.4909‬‬ ‫‪.4931‬‬ ‫‪.4948‬‬ ‫‪.4961‬‬ ‫‪.4971‬‬ ‫‪.4979‬‬ ‫‪.4985‬‬ ‫‪.4989‬‬

‫‪.05‬‬ ‫‪.0199‬‬ ‫‪.0596‬‬ ‫‪.0987‬‬ ‫‪.1368‬‬ ‫‪.1736‬‬ ‫‪.2088‬‬ ‫‪.2422‬‬ ‫‪.2734‬‬ ‫‪.3023‬‬ ‫‪.3289‬‬ ‫‪.3531‬‬ ‫‪.3749‬‬ ‫‪.3944‬‬ ‫‪.4115‬‬ ‫‪.4265‬‬ ‫‪.4394‬‬ ‫‪.4505‬‬ ‫‪.4599‬‬ ‫‪.4678‬‬ ‫‪.4744‬‬ ‫‪.4798‬‬ ‫‪.4842‬‬ ‫‪.4878‬‬ ‫‪.4906‬‬ ‫‪.4929‬‬ ‫‪.4946‬‬ ‫‪.4960‬‬ ‫‪.4970‬‬ ‫‪.4978‬‬ ‫‪.4984‬‬ ‫‪.4989‬‬

‫‪.04‬‬ ‫‪.0160‬‬ ‫‪.0557‬‬ ‫‪.0948‬‬ ‫‪.1331‬‬ ‫‪.1700‬‬ ‫‪.2054‬‬ ‫‪.2389‬‬ ‫‪.2704‬‬ ‫‪.2995‬‬ ‫‪.3264‬‬ ‫‪.3508‬‬ ‫‪.3729‬‬ ‫‪.3925‬‬ ‫‪.4099‬‬ ‫‪.4251‬‬ ‫‪.4382‬‬ ‫‪.4495‬‬ ‫‪.4591‬‬ ‫‪.4671‬‬ ‫‪.4738‬‬ ‫‪.4793‬‬ ‫‪.4838‬‬ ‫‪.4875‬‬ ‫‪.4904‬‬ ‫‪.4927‬‬ ‫‪.4945‬‬ ‫‪.4959‬‬ ‫‪.4969‬‬ ‫‪.4977‬‬ ‫‪.4984‬‬ ‫‪.4988‬‬

‫اﻟﻤﺼﺪر ‪ :‬ﻋﻦ ])‪[Daniel (1978‬‬

‫‪٦١٧‬‬

‫‪.03‬‬ ‫‪.0120‬‬ ‫‪.0517‬‬ ‫‪.0910‬‬ ‫‪.1293‬‬ ‫‪.1664‬‬ ‫‪.2019‬‬ ‫‪.2357‬‬ ‫‪.2673‬‬ ‫‪.2967‬‬ ‫‪.3238‬‬ ‫‪.3485‬‬ ‫‪.3708‬‬ ‫‪.3907‬‬ ‫‪.4082‬‬ ‫‪.4236‬‬ ‫‪.4370‬‬ ‫‪.4484‬‬ ‫‪.4582‬‬ ‫‪.4664‬‬ ‫‪.4732‬‬ ‫‪.4788‬‬ ‫‪.4834‬‬ ‫‪.4871‬‬ ‫‪.4901‬‬ ‫‪.4925‬‬ ‫‪.4943‬‬ ‫‪.4957‬‬ ‫‪.4968‬‬ ‫‪.4977‬‬ ‫‪.4983‬‬ ‫‪.4988‬‬

‫‪.02‬‬ ‫‪.0080‬‬ ‫‪.0478‬‬ ‫‪.0871‬‬ ‫‪.1255‬‬ ‫‪.1628‬‬ ‫‪.1985‬‬ ‫‪.2324‬‬ ‫‪.2642‬‬ ‫‪.2939‬‬ ‫‪.3212‬‬ ‫‪.3461‬‬ ‫‪.3686‬‬ ‫‪.3888‬‬ ‫‪.4066‬‬ ‫‪.4222‬‬ ‫‪.4357‬‬ ‫‪.4474‬‬ ‫‪.4573‬‬ ‫‪.4656‬‬ ‫‪.4726‬‬ ‫‪.4783‬‬ ‫‪.4830‬‬ ‫‪.4868‬‬ ‫‪.4898‬‬ ‫‪.4922‬‬ ‫‪.4941‬‬ ‫‪.4956‬‬ ‫‪.4967‬‬ ‫‪.4976‬‬ ‫‪.4982‬‬ ‫‪.4987‬‬

‫‪.01‬‬ ‫‪.0040‬‬ ‫‪.0438‬‬ ‫‪.0832‬‬ ‫‪.1217‬‬ ‫‪.1591‬‬ ‫‪.1950‬‬ ‫‪.2291‬‬ ‫‪.2611‬‬ ‫‪.2910‬‬ ‫‪.3186‬‬ ‫‪.3438‬‬ ‫‪.3665‬‬ ‫‪.3869‬‬ ‫‪.4049‬‬ ‫‪.4207‬‬ ‫‪.4345‬‬ ‫‪.4463‬‬ ‫‪.4564‬‬ ‫‪.4649‬‬ ‫‪.4719‬‬ ‫‪.4778‬‬ ‫‪.4826‬‬ ‫‪.4864‬‬ ‫‪.4896‬‬ ‫‪.4920‬‬ ‫‪.4940‬‬ ‫‪.4955‬‬ ‫‪.4966‬‬ ‫‪.4975‬‬ ‫‪.4982‬‬ ‫‪.4987‬‬

‫‪.00‬‬ ‫‪.0000‬‬ ‫‪.0398‬‬ ‫‪.079‬‬ ‫‪.1179‬‬ ‫‪.1554‬‬ ‫‪.1915‬‬ ‫‪.2257‬‬ ‫‪.2580‬‬ ‫‪.2881‬‬ ‫‪.3159‬‬ ‫‪.3413‬‬ ‫‪.3643‬‬ ‫‪.3849‬‬ ‫‪.4032‬‬ ‫‪.4192‬‬ ‫‪.4332‬‬ ‫‪.4452‬‬ ‫‪.4554‬‬ ‫‪.4641‬‬ ‫‪.4713‬‬ ‫‪.4772‬‬ ‫‪.4821‬‬ ‫‪.4861‬‬ ‫‪.4893‬‬ ‫‪.4918‬‬ ‫‪.4938‬‬ ‫‪.4953‬‬ ‫‪.4965‬‬ ‫‪.4974‬‬ ‫‪.4981‬‬ ‫‪.4987‬‬

‫‪Z‬‬ ‫‪0.0‬‬ ‫‪0.1‬‬ ‫‪0.2‬‬ ‫‪0.3‬‬ ‫‪0.4‬‬ ‫‪0.5‬‬ ‫‪0.6‬‬ ‫‪0.7‬‬ ‫‪0.8‬‬ ‫‪0.9‬‬ ‫‪1.0‬‬ ‫‪1.1‬‬ ‫‪1.2‬‬ ‫‪1.3‬‬ ‫‪1.4‬‬ ‫‪1.5‬‬ ‫‪1.6‬‬ ‫‪1.7‬‬ ‫‪1.8‬‬ ‫‪1.9‬‬ ‫‪2.0‬‬ ‫‪2.1‬‬ ‫‪2.2‬‬ ‫‪2.3‬‬ ‫‪2.4‬‬ ‫‪2.5‬‬ ‫‪2.6‬‬ ‫‪2.7‬‬ ‫‪2.8‬‬ ‫‪2.9‬‬ ‫‪3.0‬‬

‫ﻣﻠﺤﻖ )‪(٢‬‬ ‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ‪ t ‬ﻟﺘﻮزﯾﻊ‬

‫‪t‬‬

‫‪‬‬ ‫‪.0005‬‬ ‫‪636.62‬‬ ‫‪31.598‬‬ ‫‪12.924‬‬ ‫‪8.610‬‬ ‫‪6.869‬‬ ‫‪5.959‬‬ ‫‪5.408‬‬ ‫‪5.041‬‬ ‫‪4.781‬‬ ‫‪4.587‬‬ ‫‪4.437‬‬ ‫‪4.318‬‬ ‫‪4.221‬‬ ‫‪4.140‬‬ ‫‪4.073‬‬ ‫‪4.015‬‬ ‫‪3.965‬‬ ‫‪3.922‬‬ ‫‪3.883‬‬ ‫‪3.850‬‬ ‫‪3.819‬‬ ‫‪3.792‬‬ ‫‪3.767‬‬ ‫‪3.745‬‬ ‫‪3.725‬‬ ‫‪3.707‬‬ ‫‪3.690‬‬ ‫‪3.674‬‬ ‫‪3.659‬‬ ‫‪3.646‬‬ ‫‪3.551‬‬ ‫‪3.460‬‬ ‫‪3.373‬‬ ‫‪3.291‬‬

‫اﻟﻤﺼﺪر‬

‫‪.001‬‬ ‫‪318.31‬‬ ‫‪22.326‬‬ ‫‪10.213‬‬ ‫‪7.173‬‬ ‫‪5.893‬‬ ‫‪5.208‬‬ ‫‪4.785‬‬ ‫‪4.501‬‬ ‫‪4.297‬‬ ‫‪4.144‬‬ ‫‪4.025‬‬ ‫‪3.930‬‬ ‫‪3.852‬‬ ‫‪3.787‬‬ ‫‪3.733‬‬ ‫‪3.686‬‬ ‫‪3.646‬‬ ‫‪3.610‬‬ ‫‪3.579‬‬ ‫‪3.552‬‬ ‫‪3.527‬‬ ‫‪3.505‬‬ ‫‪3.485‬‬ ‫‪3.467‬‬ ‫‪3.450‬‬ ‫‪3.435‬‬ ‫‪3.421‬‬ ‫‪3.408‬‬ ‫‪3.396‬‬ ‫‪3.385‬‬ ‫‪3.307‬‬ ‫‪3.232‬‬ ‫‪3.160‬‬ ‫‪3.090‬‬

‫‪.005‬‬ ‫‪63.657‬‬ ‫‪9.925‬‬ ‫‪5.841‬‬ ‫‪4.604‬‬ ‫‪4.032‬‬ ‫‪3.707‬‬ ‫‪3.499‬‬ ‫‪3.355‬‬ ‫‪3.250‬‬ ‫‪3.169‬‬ ‫‪3.106‬‬ ‫‪3.055‬‬ ‫‪3.012‬‬ ‫‪2.977‬‬ ‫‪2.947‬‬ ‫‪2.921‬‬ ‫‪2.898‬‬ ‫‪2.878‬‬ ‫‪2.861‬‬ ‫‪2.845‬‬ ‫‪2.831‬‬ ‫‪2.819‬‬ ‫‪2.807‬‬ ‫‪2.797‬‬ ‫‪2.787‬‬ ‫‪2.779‬‬ ‫‪2.771‬‬ ‫‪2.763‬‬ ‫‪2.756‬‬ ‫‪2.750‬‬ ‫‪2.704‬‬ ‫‪2.660‬‬ ‫‪2.617‬‬ ‫‪2.576‬‬

‫‪.01‬‬ ‫‪31.821‬‬ ‫‪6.965‬‬ ‫‪4.541‬‬ ‫‪3.747‬‬ ‫‪3.365‬‬ ‫‪3.143‬‬ ‫‪2.998‬‬ ‫‪2.896‬‬ ‫‪2.821‬‬ ‫‪2.764‬‬ ‫‪2.718‬‬ ‫‪2.681‬‬ ‫‪2.650‬‬ ‫‪2.624‬‬ ‫‪2.602‬‬ ‫‪2.583‬‬ ‫‪2.567‬‬ ‫‪2.552‬‬ ‫‪2.539‬‬ ‫‪2.528‬‬ ‫‪2.518‬‬ ‫‪2.508‬‬ ‫‪2.500‬‬ ‫‪2.492‬‬ ‫‪2.485‬‬ ‫‪2.479‬‬ ‫‪2.473‬‬ ‫‪2.467‬‬ ‫‪2.462‬‬ ‫‪2.457‬‬ ‫‪2.423‬‬ ‫‪2.390‬‬ ‫‪2.358‬‬ ‫‪2.326‬‬

‫‪ :‬ﻋﻦ ])‪[Devore (1995‬‬

‫‪٦١٨‬‬

‫‪.025‬‬ ‫‪12.706‬‬ ‫‪4.303‬‬ ‫‪3.182‬‬ ‫‪2.776‬‬ ‫‪2.571‬‬ ‫‪2.447‬‬ ‫‪2.365‬‬ ‫‪2.306‬‬ ‫‪2.262‬‬ ‫‪2.228‬‬ ‫‪2.201‬‬ ‫‪2.179‬‬ ‫‪2.160‬‬ ‫‪2.145‬‬ ‫‪2.131‬‬ ‫‪2.120‬‬ ‫‪2.110‬‬ ‫‪2.101‬‬ ‫‪2.093‬‬ ‫‪2.086‬‬ ‫‪2.080‬‬ ‫‪2.074‬‬ ‫‪2.069‬‬ ‫‪2.064‬‬ ‫‪2.060‬‬ ‫‪2.056‬‬ ‫‪2.052‬‬ ‫‪2.048‬‬ ‫‪2.045‬‬ ‫‪2.042‬‬ ‫‪2.021‬‬ ‫‪2.000‬‬ ‫‪1.980‬‬ ‫‪1.960‬‬

‫‪.05‬‬ ‫‪6.314‬‬ ‫‪2.920‬‬ ‫‪2.353‬‬ ‫‪2.132‬‬ ‫‪2.015‬‬ ‫‪1.943‬‬ ‫‪1.895‬‬ ‫‪1.860‬‬ ‫‪1.833‬‬ ‫‪1.812‬‬ ‫‪1.796‬‬ ‫‪1.782‬‬ ‫‪1.771‬‬ ‫‪1.761‬‬ ‫‪1.753‬‬ ‫‪1.746‬‬ ‫‪1.740‬‬ ‫‪1.734‬‬ ‫‪1.729‬‬ ‫‪1.725‬‬ ‫‪1.721‬‬ ‫‪1.717‬‬ ‫‪1.714‬‬ ‫‪1.711‬‬ ‫‪1.708‬‬ ‫‪1.706‬‬ ‫‪1.703‬‬ ‫‪1.701‬‬ ‫‪1.699‬‬ ‫‪1.697‬‬ ‫‪1.684‬‬ ‫‪1.671‬‬ ‫‪1.658‬‬ ‫‪1.645‬‬

‫‪.10‬‬ ‫‪3.078‬‬ ‫‪1.886‬‬ ‫‪1.638‬‬ ‫‪1.533‬‬ ‫‪1.476‬‬ ‫‪1.440‬‬ ‫‪1.415‬‬ ‫‪1.397‬‬ ‫‪1.383‬‬ ‫‪1.372‬‬ ‫‪1.363‬‬ ‫‪1.356‬‬ ‫‪1.350‬‬ ‫‪1.345‬‬ ‫‪1.341‬‬ ‫‪1.337‬‬ ‫‪1.333‬‬ ‫‪1.330‬‬ ‫‪1.328‬‬ ‫‪1.325‬‬ ‫‪1.323‬‬ ‫‪1.321‬‬ ‫‪1.319‬‬ ‫‪1.318‬‬ ‫‪1.316‬‬ ‫‪1.315‬‬ ‫‪1.314‬‬ ‫‪1.313‬‬ ‫‪1.311‬‬ ‫‪1.310‬‬ ‫‪1.303‬‬ ‫‪1.296‬‬ ‫‪1.289‬‬ ‫‪1.282‬‬

‫‪‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪13‬‬ ‫‪14‬‬ ‫‪15‬‬ ‫‪16‬‬ ‫‪17‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪23‬‬ ‫‪24‬‬ ‫‪25‬‬ ‫‪26‬‬ ‫‪27‬‬ ‫‪28‬‬ ‫‪29‬‬ ‫‪30‬‬ ‫‪40‬‬ ‫‪60‬‬ ‫‪120‬‬ ‫‪‬‬

‫ﻣﻠﺤﻖ )‪(٣‬‬ ‫‪2‬‬

‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ‪‬‬

‫ﻟﺘﻮزﯾﻊ ‪ 2‬‬ ‫‪‬‬

‫‪.005‬‬

‫‪.01‬‬

‫‪.025‬‬

‫‪.05‬‬

‫‪.10‬‬

‫‪.90‬‬

‫‪.95‬‬

‫‪.975‬‬

‫‪.99‬‬

‫‪.995‬‬

‫‪‬‬

‫‪7.882‬‬ ‫‪10.59‬‬ ‫‪12.83‬‬ ‫‪14.86‬‬ ‫‪16.74‬‬ ‫‪18.54‬‬ ‫‪20.27‬‬ ‫‪21.95‬‬ ‫‪23.58‬‬ ‫‪25.18‬‬ ‫‪26.75‬‬ ‫‪28.30‬‬ ‫‪29.81‬‬ ‫‪31.31‬‬ ‫‪32.79‬‬ ‫‪34.26‬‬ ‫‪35.71‬‬ ‫‪37.15‬‬ ‫‪38.58‬‬ ‫‪39.99‬‬ ‫‪41.39‬‬ ‫‪42.79‬‬ ‫‪44.17‬‬ ‫‪45.55‬‬ ‫‪46.92‬‬ ‫‪48.29‬‬ ‫‪49.64‬‬ ‫‪50.99‬‬ ‫‪52.33‬‬ ‫‪53.67‬‬ ‫‪55.00‬‬ ‫‪56.32‬‬ ‫‪57.64‬‬ ‫‪58.96‬‬ ‫‪60.27‬‬ ‫‪61.58‬‬ ‫‪62.88‬‬ ‫‪64.18‬‬ ‫‪65.47‬‬ ‫‪66.76‬‬

‫‪6.637‬‬ ‫‪9.210‬‬ ‫‪11.34‬‬ ‫‪13.27‬‬ ‫‪15.08‬‬ ‫‪16.81‬‬ ‫‪18.47‬‬ ‫‪20.09‬‬ ‫‪21.66‬‬ ‫‪23.20‬‬ ‫‪24.72‬‬ ‫‪26.21‬‬ ‫‪27.68‬‬ ‫‪29.14‬‬ ‫‪30.57‬‬ ‫‪32.00‬‬ ‫‪33.40‬‬ ‫‪34.80‬‬ ‫‪36.19‬‬ ‫‪37.56‬‬ ‫‪38.93‬‬ ‫‪40.28‬‬ ‫‪41.63‬‬ ‫‪42.98‬‬ ‫‪44.31‬‬ ‫‪45.64‬‬ ‫‪46.96‬‬ ‫‪48.27‬‬ ‫‪49.58‬‬ ‫‪50.89‬‬ ‫‪52.19‬‬ ‫‪53.48‬‬ ‫‪54.77‬‬ ‫‪56.06‬‬ ‫‪57.34‬‬ ‫‪58.61‬‬ ‫‪59.89‬‬ ‫‪61.16‬‬ ‫‪62.42‬‬ ‫‪63.69‬‬

‫‪5.025‬‬ ‫‪7.378‬‬ ‫‪9.348‬‬ ‫‪11.14‬‬ ‫‪12.83‬‬ ‫‪14.44‬‬ ‫‪16.01‬‬ ‫‪17.53‬‬ ‫‪19.02‬‬ ‫‪20.48‬‬ ‫‪21.92‬‬ ‫‪23.33‬‬ ‫‪24.73‬‬ ‫‪26.11‬‬ ‫‪27.48‬‬ ‫‪28.84‬‬ ‫‪30.19‬‬ ‫‪31.52‬‬ ‫‪32.85‬‬ ‫‪34.17‬‬ ‫‪35.47‬‬ ‫‪36.78‬‬ ‫‪38.07‬‬ ‫‪39.36‬‬ ‫‪40.64‬‬ ‫‪41.92‬‬ ‫‪43.19‬‬ ‫‪44.46‬‬ ‫‪45.77‬‬ ‫‪46.97‬‬ ‫‪48.23‬‬ ‫‪49.48‬‬ ‫‪50.72‬‬ ‫‪51.96‬‬ ‫‪53.20‬‬ ‫‪54.43‬‬ ‫‪55.66‬‬ ‫‪56.89‬‬ ‫‪58.11‬‬ ‫‪59.34‬‬

‫‪3.843‬‬ ‫‪5.992‬‬ ‫‪7.815‬‬ ‫‪9.488‬‬ ‫‪11.07‬‬ ‫‪12.59‬‬ ‫‪14.06‬‬ ‫‪15.50‬‬ ‫‪16.91‬‬ ‫‪18.30‬‬ ‫‪19.67‬‬ ‫‪21.02‬‬ ‫‪22.36‬‬ ‫‪23.68‬‬ ‫‪24.99‬‬ ‫‪26.29‬‬ ‫‪27.58‬‬ ‫‪28.86‬‬ ‫‪30.14‬‬ ‫‪31.41‬‬ ‫‪32.67‬‬ ‫‪33.92‬‬ ‫‪35.17‬‬ ‫‪36.41‬‬ ‫‪37.65‬‬ ‫‪38.88‬‬ ‫‪40.11‬‬ ‫‪41.33‬‬ ‫‪42.55‬‬ ‫‪43.77‬‬ ‫‪44.98‬‬ ‫‪46.19‬‬ ‫‪47.40‬‬ ‫‪48.60‬‬ ‫‪49.80‬‬ ‫‪50.99‬‬ ‫‪52.19‬‬ ‫‪53.38‬‬ ‫‪54.57‬‬ ‫‪55.75‬‬

‫‪2.706‬‬ ‫‪4.605‬‬ ‫‪6.251‬‬ ‫‪7.779‬‬ ‫‪9.236‬‬ ‫‪10.64‬‬ ‫‪12.01‬‬ ‫‪13.36‬‬ ‫‪14.68‬‬ ‫‪15.98‬‬ ‫‪17.27‬‬ ‫‪18.54‬‬ ‫‪19.81‬‬ ‫‪21.06‬‬ ‫‪22.30‬‬ ‫‪23.54‬‬ ‫‪24.76‬‬ ‫‪25.98‬‬ ‫‪27.20‬‬ ‫‪28.41‬‬ ‫‪29.61‬‬ ‫‪30.81‬‬ ‫‪32.00‬‬ ‫‪33.19‬‬ ‫‪34.38‬‬ ‫‪35.56‬‬ ‫‪36.74‬‬ ‫‪37.91‬‬ ‫‪39.08‬‬ ‫‪40.25‬‬ ‫‪41.42‬‬ ‫‪42.58‬‬ ‫‪43.74‬‬ ‫‪44.90‬‬ ‫‪46.05‬‬ ‫‪47.21‬‬ ‫‪48.36‬‬ ‫‪49.51‬‬ ‫‪50.66‬‬ ‫‪51.80‬‬

‫‪0.016‬‬ ‫‪0.211‬‬ ‫‪0.584‬‬ ‫‪1.064‬‬ ‫‪1.610‬‬ ‫‪2.204‬‬ ‫‪2.833‬‬ ‫‪3.490‬‬ ‫‪4.168‬‬ ‫‪4.865‬‬ ‫‪5.578‬‬ ‫‪6.304‬‬ ‫‪7.041‬‬ ‫‪7.790‬‬ ‫‪8.547‬‬ ‫‪9.312‬‬ ‫‪10.08‬‬ ‫‪10.86‬‬ ‫‪11.65‬‬ ‫‪12.44‬‬ ‫‪13.24‬‬ ‫‪14.04‬‬ ‫‪14.84‬‬ ‫‪15.65‬‬ ‫‪16.47‬‬ ‫‪17.29‬‬ ‫‪18.11‬‬ ‫‪18.93‬‬ ‫‪19.76‬‬ ‫‪20.59‬‬ ‫‪21.43‬‬ ‫‪22.27‬‬ ‫‪23.11‬‬ ‫‪23.95‬‬ ‫‪24.79‬‬ ‫‪25.64‬‬ ‫‪26.49‬‬ ‫‪27.34‬‬ ‫‪28.19‬‬ ‫‪29.05‬‬

‫‪0.004‬‬ ‫‪0.103‬‬ ‫‪0.352‬‬ ‫‪0.711‬‬ ‫‪1.145‬‬ ‫‪1.635‬‬ ‫‪2.167‬‬ ‫‪2.733‬‬ ‫‪3.325‬‬ ‫‪3.940‬‬ ‫‪4.575‬‬ ‫‪5.226‬‬ ‫‪5.892‬‬ ‫‪6.571‬‬ ‫‪7.261‬‬ ‫‪7.962‬‬ ‫‪8.682‬‬ ‫‪9.390‬‬ ‫‪10.11‬‬ ‫‪10.85‬‬ ‫‪11.59‬‬ ‫‪12.33‬‬ ‫‪13.09‬‬ ‫‪13.84‬‬ ‫‪14.61‬‬ ‫‪15.37‬‬ ‫‪16.15‬‬ ‫‪16.92‬‬ ‫‪17.70‬‬ ‫‪18.49‬‬ ‫‪19.28‬‬ ‫‪20.07‬‬ ‫‪20.86‬‬ ‫‪21.66‬‬ ‫‪22.46‬‬ ‫‪23.26‬‬ ‫‪24.07‬‬ ‫‪24.88‬‬ ‫‪25.69‬‬ ‫‪26.50‬‬

‫‪0.001‬‬ ‫‪0.051‬‬ ‫‪0.216‬‬ ‫‪0.484‬‬ ‫‪0.831‬‬ ‫‪1.237‬‬ ‫‪1.690‬‬ ‫‪2.180‬‬ ‫‪2.700‬‬ ‫‪3.247‬‬ ‫‪3.816‬‬ ‫‪4.404‬‬ ‫‪5.009‬‬ ‫‪5.629‬‬ ‫‪6.262‬‬ ‫‪6.908‬‬ ‫‪7.564‬‬ ‫‪8.231‬‬ ‫‪8.906‬‬ ‫‪9.591‬‬ ‫‪10.28‬‬ ‫‪10.98‬‬ ‫‪11.68‬‬ ‫‪12.40‬‬ ‫‪13.12‬‬ ‫‪13.84‬‬ ‫‪14.57‬‬ ‫‪15.30‬‬ ‫‪16.14‬‬ ‫‪16.79‬‬ ‫‪17.53‬‬ ‫‪18.29‬‬ ‫‪19.04‬‬ ‫‪19.80‬‬ ‫‪20.56‬‬ ‫‪21.33‬‬ ‫‪22.10‬‬ ‫‪22.87‬‬ ‫‪23.65‬‬ ‫‪24.43‬‬

‫‪0.000‬‬ ‫‪0.020‬‬ ‫‪0.115‬‬ ‫‪0.297‬‬ ‫‪0.554‬‬ ‫‪0.872‬‬ ‫‪1.239‬‬ ‫‪1.646‬‬ ‫‪2.088‬‬ ‫‪2.558‬‬ ‫‪3.053‬‬ ‫‪3.571‬‬ ‫‪4.107‬‬ ‫‪4.660‬‬ ‫‪5.229‬‬ ‫‪5.812‬‬ ‫‪6.407‬‬ ‫‪7.015‬‬ ‫‪7.632‬‬ ‫‪8.260‬‬ ‫‪8.897‬‬ ‫‪9.542‬‬ ‫‪10.19‬‬ ‫‪10.85‬‬ ‫‪11.52‬‬ ‫‪12.19‬‬ ‫‪12.87‬‬ ‫‪13.56‬‬ ‫‪14.25‬‬ ‫‪14.95‬‬ ‫‪15.65‬‬ ‫‪16.36‬‬ ‫‪17.07‬‬ ‫‪17.78‬‬ ‫‪18.50‬‬ ‫‪19.23‬‬ ‫‪19.96‬‬ ‫‪20.69‬‬ ‫‪21.42‬‬ ‫‪22.16‬‬

‫‪0.000‬‬ ‫‪0.010‬‬ ‫‪0.072‬‬ ‫‪0.207‬‬ ‫‪0.412‬‬ ‫‪0.676‬‬ ‫‪0.989‬‬ ‫‪1.344‬‬ ‫‪1.735‬‬ ‫‪2.156‬‬ ‫‪2.603‬‬ ‫‪3.074‬‬ ‫‪3.565‬‬ ‫‪4.075‬‬ ‫‪4.600‬‬ ‫‪5.142‬‬ ‫‪5.697‬‬ ‫‪6.265‬‬ ‫‪6.843‬‬ ‫‪7.434‬‬ ‫‪8.033‬‬ ‫‪8.643‬‬ ‫‪9.260‬‬ ‫‪9.886‬‬ ‫‪10.51‬‬ ‫‪11.16‬‬ ‫‪11.80‬‬ ‫‪12.46‬‬ ‫‪13.12‬‬ ‫‪13.78‬‬ ‫‪14.45‬‬ ‫‪15.13‬‬ ‫‪15.81‬‬ ‫‪16.50‬‬ ‫‪17.19‬‬ ‫‪17.88‬‬ ‫‪18.58‬‬ ‫‪19.28‬‬ ‫‪19.99‬‬ ‫‪20.70‬‬

‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪13‬‬ ‫‪14‬‬ ‫‪15‬‬ ‫‪16‬‬ ‫‪17‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪23‬‬ ‫‪24‬‬ ‫‪25‬‬ ‫‪26‬‬ ‫‪27‬‬ ‫‪28‬‬ ‫‪29‬‬ ‫‪30‬‬ ‫‪31‬‬ ‫‪32‬‬ ‫‪33‬‬ ‫‪34‬‬ ‫‪35‬‬ ‫‪36‬‬ ‫‪37‬‬ ‫‪38‬‬ ‫‪39‬‬ ‫‪40‬‬

‫اﻟﻤﺼﺪر ‪ :‬ﻋﻦ ])‪[Devore(1995‬‬

‫‪٦١٩‬‬

‫ﻣﻠﺤﻖ )‪(٤‬‬ ‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ) ‪ f  (1,  2‬ﻟﺘﻮزﯾﻊ ‪ F‬ﻋﻨﺪ‬

‫)‪(  0.05‬‬ ‫‪1‬‬

‫‪‬‬

‫‪120‬‬

‫‪60‬‬

‫‪40‬‬

‫‪30‬‬

‫‪24‬‬

‫‪20‬‬

‫‪15‬‬

‫‪12‬‬

‫‪10‬‬

‫‪9‬‬

‫‪8‬‬

‫‪7‬‬

‫‪236. 238. 240. 241. 243. 245. 248. 249. 250. 251. 252. 253.3 254.‬‬ ‫‪8 19.3‬‬ ‫‪9 19.3‬‬ ‫‪5 19.4‬‬ ‫‪9 19.4‬‬ ‫‪9 19.4‬‬ ‫‪9 19.4‬‬ ‫‪0 19.4‬‬ ‫‪1 19.4‬‬ ‫‪1 19.4‬‬ ‫‪1 19.4‬‬ ‫‪2 19.49 19.5‬‬ ‫‪3‬‬ ‫‪19.3‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪0‬‬ ‫‪8.89 8.85 8.81 8.79 8.74 8.70 8.66 8.64 8.62 8.59 8.57 8.55 8.53‬‬

‫‪6‬‬

‫‪5‬‬

‫‪4‬‬

‫‪3‬‬

‫‪2‬‬

‫‪1‬‬

‫‪161. 199. 215. 224. 230. 234.‬‬ ‫‪4 19.0‬‬ ‫‪5 19.1‬‬ ‫‪7 19.2‬‬ ‫‪6 19.3‬‬ ‫‪2 19.3‬‬ ‫‪0‬‬ ‫‪18.5‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪6‬‬ ‫‪5‬‬ ‫‪0‬‬ ‫‪3‬‬ ‫‪10.1 9..55 9.28 9.12 9.01 8.94‬‬ ‫‪3 6.94 6.59 6.39 6.26 6.16‬‬ ‫‪7.71‬‬

‫‪‬‬ ‫‪2‬‬

‫‪‬‬

‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬

‫‪6.09 6.04 6.00 5.96 5.91 5.86 5.80 5.77 5.75 5.72 5.69 5.66 5.63‬‬ ‫‪6.‬‬ ‫‪6.61 5.79 5.41 5.19 5.05 4.95 4.88 4.82 4.77 4.74 4.68 4.62 4.56 4.53 4.50 4.46 4.43 4.40 4.36‬‬

‫‪5‬‬

‫‪5.99 5.14 4.76 4.53 4.39 4.28 4.21 4.15 4.10 4.06 4.00 3.94 3.87 3.84 3.81 3.77 3.74 3.70 3.67‬‬

‫‪6‬‬

‫‪5.59 4.74 4.35 4.12 3.97 3.87 3.79 3.73 3.68 3.64 3.57 3.51 3.44 3.41 3.38 3.34 3.30 3.27 3.23‬‬

‫‪7‬‬

‫‪5.32 4.46 4.07 3.84 3.69 3.58 3.50 3.44 3.39 3.35 3.28 3.22 3.15 3.12 3.08 3.04 3.01 2.97 2.93‬‬

‫‪8‬‬

‫‪5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 3.18 3.14 3.07 3.01 2.94 2.90 2.86 2.83 2.79 2.75 2.71‬‬

‫‪9‬‬

‫‪4.96 4.10 3.71 3.48 3.33 3.22 3.14 3.07 3.02 2.98 2.91 2.85 2.77 2.74 2.70 2.66 2.62 2.58 2.54‬‬

‫‪10‬‬

‫‪4.84 3.98 3.59 3.36 3.20 3.09 3.01 2.95 2.90 2.85 2.79 2.72 2.65 2.61 2.57 2.53 2.49 2.45 2.40‬‬

‫‪11‬‬

‫‪4.75 3.89 3.49 3.26 3.11 3.00 2.91 2.85 2.80 2.75 2.69 2.62 2.54 2.51 2.47 2.43 2.38 2.34 2.30‬‬

‫‪12‬‬

‫‪4.67 3.81 3.41 3.18 3.03 2.92 2.83 2.77 2.71 2.67 2.60 2.53 2.46 2.42 2.38 2.34 2.30 2.25 2.21‬‬

‫‪13‬‬

‫‪4.60 3.74 3.34 3.11 2.96 2.85 2.76 2.70 2.65 2.60 2.53 2.46 2.39 2.35 2.31 2.27 2.22 2.18 2.13‬‬

‫‪14‬‬

‫‪4.54 3.68 3.29 3.06 2.90 2.79 2.71 2.64 2.59 2.54 2.48 2.40 2.33 2.29 2.25 2.20 2.16 2.11 2.07‬‬

‫‪15‬‬

‫‪4.49 3.63 3.24 3.01 2.85 2.74 2.66 2.59 2.54 2.49 2.42 2.35 2.28 2.24 2.19 2.15 2.11 2.06 2.07‬‬

‫‪16‬‬

‫‪4.45 3.59 3.20 2.96 2.81 2.70 2.61 2.55 2.49 2.45 2.38 2.31 2.23 2.19 2.15 2.10 2.06 2.01 1.96‬‬

‫‪17‬‬

‫‪4.41 3.55 3.16 2.93 2.77 2.66 2.58 2.51 2.46 2.41 2.34 2.27 2.19 2.15 2.11 2.06 2.02 1.97 1.92‬‬

‫‪18‬‬

‫‪4.38 3.52 3.13 2.90 2.74 2.63 2.54 2.48 2.42 2.38 2.31 2.23 2.16 2.11 2.07 2.03 1.98 1.93 1.88‬‬

‫‪19‬‬

‫‪4.35 3.49 3.10 2.87 2.71 2.60 2.51 2.45 2.39 2.35 2.28 2.20 2.12 2.08 2.04 1.99 1.95 1.90 1.84‬‬

‫‪20‬‬

‫‪4.32 3.47 3.07 2.84 2.68 2.57 2.49 2.42 2.37 2.32 2.25 2.18 2.10 2.05 2.01 1.96 1.92 1.87 1.81‬‬

‫‪21‬‬

‫‪4.30 3.44 3.05 2.82 2.66 2.55 2.46 2.40 2.34 2.30 2.23 2.15 2.07 2.03 1.98 1.94 1.89 1.84 1.78‬‬

‫‪22‬‬

‫‪4.28 3.42 3.03 2.80 2.64 2.53 2.44 2.37 2.32 2.27 2.20 2.13 2.05 2.01 1.96 1.91 1.86 1.81 1.76‬‬

‫‪23‬‬

‫‪4.26 3.40 3.01 2.78 2.62 2.51 2.42 2.36 2.30 2.25 2.18 2.11 2.03 1.98 1.94 1.89 1.84 1.79 1.73‬‬

‫‪24‬‬

‫‪2.24 3.39 2.99 2.76 2.60 2.49 2.40 2.34 2.28 2.24 2.16 2.09 2.01 1.96 1.92 1.87 1.82 1.77 1.71‬‬

‫‪25‬‬

‫‪4.23 3.37 2.98 2.74 2.59 2.47 2.39 2.32 2.27 2.22 2.15 2.07 1.99 1.95 1.90 1.58 1.80 1.75 1.69‬‬

‫‪26‬‬

‫‪4.21 3.35 2.96 2.73 2.57 2.46 2.37 2.31 2.25 2.20 2.13 2.06 1.97 1.93 1.88 1.84 1.79 1.73 1.67‬‬

‫‪27‬‬

‫‪4.20 3.34 2.95 2.71 2.56 2.45 2.36 2.29 2.24 2.19 2.12 2.04 1.96 1.91 1.87 1.82 1.77 1.71 1.65‬‬

‫‪28‬‬

‫‪4.18 3.33 2.93 2.70 2.55 2.43 2.35 2.28 2.22 2.18 2.10 2.03 1.94 1.90 1.85 1.81 1.75 1.70 1.64‬‬

‫‪29‬‬

‫‪4.17 3.32 2.92 2.69 2.53 2.42 2.33 2.27 2.21 2.16 2.09 2.01 1.93 1.89 1.84 1.79 1.74 1.68 1.62‬‬

‫‪30‬‬

‫‪4.08 3.23 2.48 2.61 2.45 2.34 2.25 2.18 2.12 2.08 2.00 1.92 1.84 1.79 1.74 1.69 1.64 1.58 1.51‬‬

‫‪40‬‬

‫‪4.00 3.15 2.76 2.53 2.37 2.25 2.17 2.10 2.04 1.99 1.92 1.84 1.75 1.70 1.65 1.59 1.53 1.47 1.39‬‬

‫‪60‬‬

‫‪3.92 3.07 2.68 2.45 2.29 2.17 2.09 2.02 1.96 1.91 1.83 1.75 1.66 1.61 1.55 1.50 1.43 1.35 1.25‬‬

‫‪120‬‬

‫‪3.84 3.84 2.60 2.37 2.21 2.10 2.01 1.94 1.88 1.83 1.75 1.67 1.57 1.52 1.46 1.39 1.32 1.22 1.00‬‬

‫‪‬‬

‫اﻟﻤﺼﺪر ‪ :‬ﻋﻦ ])‪[Devore (1995‬‬

‫‪٦٢٠‬‬

‫ﻣﻠﺤﻖ )‪(٥‬‬ ‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ) ‪ f  (1,  2‬ﻟﺘﻮزﯾﻊ‬

‫‪ F‬ﻋﻨﺪ ) ‪(   0.01‬‬ ‫‪1‬‬

‫‪‬‬

‫‪120‬‬

‫‪60‬‬

‫‪40‬‬

‫‪30‬‬

‫‪24‬‬

‫‪20‬‬

‫‪15‬‬

‫‪12‬‬

‫‪10‬‬

‫‪9‬‬

‫‪8‬‬

‫‪7‬‬

‫‪6‬‬

‫‪5‬‬

‫‪4‬‬

‫‪3‬‬

‫‪2‬‬

‫‪1‬‬

‫‪‬‬ ‫‪2‬‬

‫‪‬‬

‫‪4052 5000 5403 5625 5764 5859 5928 5981 6022 6056 6106 6157 6209 6235 6261 6287 6313 6339 6366‬‬

‫‪1‬‬

‫‪98.50 99.00 99.17 99.25 99.30 99.33 99.36 99.37 99.39 99.40 99.42 99.43 99.45 99.46 99.47 99.47 99.48 99.49 99.50‬‬

‫‪2‬‬

‫‪34.12 30.82 29.46 28.71 28.24 27.91 27.67 27.49 27.35 27.23 27.05 26.87 26.69 26.60 26.50 26.41 26.32 26.22 26.13‬‬

‫‪3‬‬

‫‪21.20 18.00 16.69 15.98 15.52 15.21 14.98 14.80 14.66 14.55 14.37 14.20 14.02 13.93 13.84 13.75 13.65 13.56 13.46‬‬

‫‪4‬‬

‫‪9.02‬‬

‫‪9.11‬‬

‫‪9.20‬‬

‫‪9.29‬‬

‫‪9.38‬‬

‫‪9.47‬‬

‫‪9.55‬‬

‫‪9.72‬‬

‫‪16.26 13.27 12.06 11.39 10.97 10.67 10.46 10.29 10.16 10.05 9.89‬‬

‫‪5‬‬

‫‪6.88‬‬

‫‪6.97‬‬

‫‪7.06‬‬

‫‪7.14‬‬

‫‪7.23‬‬

‫‪7.31‬‬

‫‪7.40‬‬

‫‪7.56‬‬

‫‪7.72‬‬

‫‪7.87‬‬

‫‪7.98‬‬

‫‪8.10‬‬

‫‪8.26‬‬

‫‪8.47‬‬

‫‪8.75‬‬

‫‪9.15‬‬

‫‪13.57 10.92 9.78‬‬

‫‪6‬‬

‫‪5.65‬‬

‫‪5.74‬‬

‫‪5.82‬‬

‫‪5.91‬‬

‫‪5.99‬‬

‫‪6.07‬‬

‫‪6.16‬‬

‫‪6.31‬‬

‫‪6.47‬‬

‫‪6.62‬‬

‫‪6.72‬‬

‫‪6.84‬‬

‫‪6.99‬‬

‫‪7.19‬‬

‫‪7.46‬‬

‫‪7.85‬‬

‫‪8.45‬‬

‫‪12.25 9.55‬‬

‫‪7‬‬

‫‪4.86‬‬

‫‪4.95‬‬

‫‪5.03‬‬

‫‪5.12‬‬

‫‪5.20‬‬

‫‪5.28‬‬

‫‪5.36‬‬

‫‪5.52‬‬

‫‪5.67‬‬

‫‪5.81‬‬

‫‪5.91‬‬

‫‪6.03‬‬

‫‪6.18‬‬

‫‪6.37‬‬

‫‪6.63‬‬

‫‪7.01‬‬

‫‪7.59‬‬

‫‪11.26 8.65‬‬

‫‪8‬‬

‫‪4.31‬‬

‫‪4.40‬‬

‫‪4.48‬‬

‫‪4.57‬‬

‫‪4.65‬‬

‫‪4.73‬‬

‫‪4.81‬‬

‫‪4.96‬‬

‫‪5.11‬‬

‫‪5.26‬‬

‫‪5.35‬‬

‫‪5.47‬‬

‫‪5.61‬‬

‫‪5.80‬‬

‫‪6.06‬‬

‫‪6.42‬‬

‫‪6.99‬‬

‫‪10.56 8.02‬‬

‫‪9‬‬

‫‪3.91‬‬

‫‪4.00‬‬

‫‪4.08‬‬

‫‪4.17‬‬

‫‪4.25‬‬

‫‪4.33‬‬

‫‪4.41‬‬

‫‪4.56‬‬

‫‪4.71‬‬

‫‪4.85‬‬

‫‪4.94‬‬

‫‪5.06‬‬

‫‪5.20‬‬

‫‪5.39‬‬

‫‪5.64‬‬

‫‪5.99‬‬

‫‪6.55‬‬

‫‪10.04 7.56‬‬

‫‪10‬‬

‫‪3.60‬‬

‫‪3.69‬‬

‫‪3.78‬‬

‫‪3.86‬‬

‫‪3.94‬‬

‫‪4.02‬‬

‫‪4.10‬‬

‫‪4.25‬‬

‫‪4.40‬‬

‫‪4.54‬‬

‫‪4.63‬‬

‫‪4.74‬‬

‫‪4.89‬‬

‫‪5.07‬‬

‫‪5.32‬‬

‫‪5.67‬‬

‫‪6.22‬‬

‫‪7.21‬‬

‫‪9.65‬‬

‫‪11‬‬

‫‪3.45 3.369‬‬ ‫‪.07‬‬ ‫‪3.25 3.17‬‬

‫‪3.54‬‬

‫‪3.62‬‬

‫‪3.70‬‬

‫‪3.78‬‬

‫‪3.86‬‬

‫‪4.01‬‬

‫‪4.16‬‬

‫‪4.30‬‬

‫‪4.39‬‬

‫‪4.50‬‬

‫‪4.64‬‬

‫‪4.82‬‬

‫‪5.06‬‬

‫‪5.41‬‬

‫‪5.95‬‬

‫‪6.93‬‬

‫‪9.33‬‬

‫‪12‬‬

‫‪3.34‬‬

‫‪3.43‬‬

‫‪3.51‬‬

‫‪3.59‬‬

‫‪3.66‬‬

‫‪3.82‬‬

‫‪3.96‬‬

‫‪4.10‬‬

‫‪4.19‬‬

‫‪4.30‬‬

‫‪4.41‬‬

‫‪4.62‬‬

‫‪4.86‬‬

‫‪5.21‬‬

‫‪5.74‬‬

‫‪6.70‬‬

‫‪9.07‬‬

‫‪13‬‬

‫‪3.00‬‬

‫‪3.09‬‬

‫‪3.18‬‬

‫‪3.27‬‬

‫‪3.35‬‬

‫‪3.43‬‬

‫‪3.51‬‬

‫‪3.66‬‬

‫‪3.80‬‬

‫‪3.94‬‬

‫‪4.03‬‬

‫‪4.14‬‬

‫‪4.28‬‬

‫‪4.46‬‬

‫‪4.69‬‬

‫‪5.04‬‬

‫‪5.56‬‬

‫‪6.51‬‬

‫‪8.86‬‬

‫‪14‬‬

‫‪2.87‬‬

‫‪2.96‬‬

‫‪3.05‬‬

‫‪3.13‬‬

‫‪3.21‬‬

‫‪3.29‬‬

‫‪3.37‬‬

‫‪3.52‬‬

‫‪3.67‬‬

‫‪3.80‬‬

‫‪3.89‬‬

‫‪4.00‬‬

‫‪4.14‬‬

‫‪4.32‬‬

‫‪4.56‬‬

‫‪4.89‬‬

‫‪5.42‬‬

‫‪6.36‬‬

‫‪8.68‬‬

‫‪15‬‬

‫‪2.75‬‬

‫‪2.84‬‬

‫‪2.93‬‬

‫‪3.02‬‬

‫‪3.10‬‬

‫‪3.18‬‬

‫‪3.26‬‬

‫‪3.41‬‬

‫‪3.55‬‬

‫‪3.69‬‬

‫‪3.78‬‬

‫‪3.89‬‬

‫‪4.03‬‬

‫‪4.20‬‬

‫‪4.44‬‬

‫‪4.77‬‬

‫‪5.29‬‬

‫‪6.23‬‬

‫‪8.53‬‬

‫‪16‬‬

‫‪2.65‬‬

‫‪2.75‬‬

‫‪2.83‬‬

‫‪2.92‬‬

‫‪3.00‬‬

‫‪3.08‬‬

‫‪3.16‬‬

‫‪3.31‬‬

‫‪3.46‬‬

‫‪3.59‬‬

‫‪3.68‬‬

‫‪3.79‬‬

‫‪3.93‬‬

‫‪4.10‬‬

‫‪4.34‬‬

‫‪4.67‬‬

‫‪5.18‬‬

‫‪6.11‬‬

‫‪8.40‬‬

‫‪17‬‬

‫‪2.57‬‬

‫‪2.66‬‬

‫‪2.75‬‬

‫‪2.84‬‬

‫‪2.92‬‬

‫‪3.00‬‬

‫‪3.08‬‬

‫‪3.23‬‬

‫‪3.37‬‬

‫‪3.51‬‬

‫‪3.60‬‬

‫‪3.71‬‬

‫‪3.84‬‬

‫‪4.01‬‬

‫‪4.25‬‬

‫‪4.58‬‬

‫‪5.09‬‬

‫‪6.01‬‬

‫‪8.29‬‬

‫‪18‬‬

‫‪2.49‬‬

‫‪2.58‬‬

‫‪2.67‬‬

‫‪2.76‬‬

‫‪2.84‬‬

‫‪2.92‬‬

‫‪3.00‬‬

‫‪3.15‬‬

‫‪3.30‬‬

‫‪3.43‬‬

‫‪3.52‬‬

‫‪3.63‬‬

‫‪3.77‬‬

‫‪3.94‬‬

‫‪4.17‬‬

‫‪4.50‬‬

‫‪5.01‬‬

‫‪5.93‬‬

‫‪8.18‬‬

‫‪19‬‬

‫‪2.42‬‬

‫‪2.52‬‬

‫‪2.61‬‬

‫‪2.69‬‬

‫‪2.78‬‬

‫‪2.86‬‬

‫‪2.94‬‬

‫‪3.09‬‬

‫‪3.23‬‬

‫‪3.37‬‬

‫‪3.46‬‬

‫‪3.56‬‬

‫‪3.70‬‬

‫‪3.87‬‬

‫‪4.10‬‬

‫‪4.43‬‬

‫‪4.94‬‬

‫‪5.85‬‬

‫‪8.10‬‬

‫‪20‬‬

‫‪2.36‬‬

‫‪2.46‬‬

‫‪2.55‬‬

‫‪2.64‬‬

‫‪2.72‬‬

‫‪2.80‬‬

‫‪2.88‬‬

‫‪3.03‬‬

‫‪3.17‬‬

‫‪3.31‬‬

‫‪3.40‬‬

‫‪3.51‬‬

‫‪3.64‬‬

‫‪3.81‬‬

‫‪4.04‬‬

‫‪4.37‬‬

‫‪4.87‬‬

‫‪5.78‬‬

‫‪8.02‬‬

‫‪21‬‬

‫‪2.31‬‬

‫‪2.40‬‬

‫‪2.50‬‬

‫‪2.58‬‬

‫‪2.67‬‬

‫‪2.75‬‬

‫‪2.83‬‬

‫‪2.98‬‬

‫‪3.12‬‬

‫‪3.26‬‬

‫‪3.35‬‬

‫‪3.45‬‬

‫‪3.59‬‬

‫‪3.76‬‬

‫‪3.99‬‬

‫‪4.31‬‬

‫‪4.82‬‬

‫‪5.72‬‬

‫‪7.95‬‬

‫‪22‬‬

‫‪2.26‬‬

‫‪2.35‬‬

‫‪2.45‬‬

‫‪2.54‬‬

‫‪2.62‬‬

‫‪2.70‬‬

‫‪2.78‬‬

‫‪2.93‬‬

‫‪3.07‬‬

‫‪3.21‬‬

‫‪3.30‬‬

‫‪3.41‬‬

‫‪3.54‬‬

‫‪3.71‬‬

‫‪3.94‬‬

‫‪4.26‬‬

‫‪4.76‬‬

‫‪5.66‬‬

‫‪7.88‬‬

‫‪23‬‬

‫‪2.21‬‬

‫‪2.31‬‬

‫‪2.40‬‬

‫‪2.49‬‬

‫‪2.58‬‬

‫‪2.66‬‬

‫‪2.74‬‬

‫‪3.03 2..89‬‬

‫‪3.17‬‬

‫‪3.26‬‬

‫‪3.36‬‬

‫‪3.50‬‬

‫‪3.67‬‬

‫‪3.90‬‬

‫‪4.22‬‬

‫‪4.72‬‬

‫‪5.61‬‬

‫‪7.82‬‬

‫‪24‬‬

‫‪2.17‬‬

‫‪2.27‬‬

‫‪2.36‬‬

‫‪2.45‬‬

‫‪2.54‬‬

‫‪2.62‬‬

‫‪2.70‬‬

‫‪2.85‬‬

‫‪2.99‬‬

‫‪3.13‬‬

‫‪3.22‬‬

‫‪3.32‬‬

‫‪3.46‬‬

‫‪3.63‬‬

‫‪3.85‬‬

‫‪4.18‬‬

‫‪4.68‬‬

‫‪5.57‬‬

‫‪7.77‬‬

‫‪25‬‬

‫‪2.13‬‬

‫‪2.23‬‬

‫‪2.33‬‬

‫‪2.42‬‬

‫‪2.50‬‬

‫‪2.58‬‬

‫‪2.66‬‬

‫‪2.81‬‬

‫‪2.96‬‬

‫‪3.09‬‬

‫‪3.18‬‬

‫‪3.29‬‬

‫‪3.42‬‬

‫‪3.59‬‬

‫‪3.82‬‬

‫‪4.14‬‬

‫‪4.64‬‬

‫‪5.53‬‬

‫‪7.72‬‬

‫‪26‬‬

‫‪2.10‬‬

‫‪2.20‬‬

‫‪2.29‬‬

‫‪2.38‬‬

‫‪2.47‬‬

‫‪2.55‬‬

‫‪2.63‬‬

‫‪2.78‬‬

‫‪2.93‬‬

‫‪3.06‬‬

‫‪3.15‬‬

‫‪3.26‬‬

‫‪3.39‬‬

‫‪3.56‬‬

‫‪3.78‬‬

‫‪4.11‬‬

‫‪4.60‬‬

‫‪5.49‬‬

‫‪7.86‬‬

‫‪27‬‬

‫‪2.06‬‬

‫‪2.17‬‬

‫‪2.26‬‬

‫‪2.35‬‬

‫‪2.44‬‬

‫‪2.52‬‬

‫‪2.60‬‬

‫‪2.75‬‬

‫‪2.90‬‬

‫‪3.03‬‬

‫‪3.12‬‬

‫‪3.23‬‬

‫‪3.36‬‬

‫‪3.53‬‬

‫‪3.75‬‬

‫‪4.07‬‬

‫‪4.57‬‬

‫‪5.45‬‬

‫‪7.64‬‬

‫‪28‬‬

‫‪2.03‬‬

‫‪2.14‬‬

‫‪2.23‬‬

‫‪2.33‬‬

‫‪2.41‬‬

‫‪2.49‬‬

‫‪2.57‬‬

‫‪2.73‬‬

‫‪2.87‬‬

‫‪3.00‬‬

‫‪3.09‬‬

‫‪3.20‬‬

‫‪3.33‬‬

‫‪3.50‬‬

‫‪3.73‬‬

‫‪4.04‬‬

‫‪4.54‬‬

‫‪5.42‬‬

‫‪7.60‬‬

‫‪29‬‬

‫‪2.01‬‬

‫‪2.11‬‬

‫‪2.21‬‬

‫‪2.30‬‬

‫‪2.39‬‬

‫‪2.47‬‬

‫‪2.55‬‬

‫‪2.70‬‬

‫‪2.84‬‬

‫‪2.98‬‬

‫‪3.07‬‬

‫‪3.17‬‬

‫‪3.30‬‬

‫‪3.47‬‬

‫‪3.70‬‬

‫‪4.02‬‬

‫‪4.51‬‬

‫‪5.39‬‬

‫‪7.56‬‬

‫‪30‬‬

‫‪1.80‬‬

‫‪1.92‬‬

‫‪2.02‬‬

‫‪2.11‬‬

‫‪2.20‬‬

‫‪2.29‬‬

‫‪2.37‬‬

‫‪2.52‬‬

‫‪2.66‬‬

‫‪2.80‬‬

‫‪3.89‬‬

‫‪2.99‬‬

‫‪3.12‬‬

‫‪3.29‬‬

‫‪3.51‬‬

‫‪3.83‬‬

‫‪4.31‬‬

‫‪5.18‬‬

‫‪7.31‬‬

‫‪40‬‬

‫‪1.60‬‬

‫‪1.73‬‬

‫‪1.84‬‬

‫‪1.94‬‬

‫‪2.03‬‬

‫‪2.12‬‬

‫‪2.20‬‬

‫‪2.35‬‬

‫‪2.50‬‬

‫‪2.63‬‬

‫‪3.72‬‬

‫‪2.82‬‬

‫‪2.95‬‬

‫‪3.12‬‬

‫‪3.34‬‬

‫‪3.65‬‬

‫‪4.13‬‬

‫‪4.98‬‬

‫‪7.08‬‬

‫‪60‬‬

‫‪1.38‬‬

‫‪1.53‬‬

‫‪1.66‬‬

‫‪1.76‬‬

‫‪1.86‬‬

‫‪1.95‬‬

‫‪2.03‬‬

‫‪2.19‬‬

‫‪2.34‬‬

‫‪2.47‬‬

‫‪3.56‬‬

‫‪2.66‬‬

‫‪2.79‬‬

‫‪2.96‬‬

‫‪3.17‬‬

‫‪3.48‬‬

‫‪3.95‬‬

‫‪4.79‬‬

‫‪6.85‬‬

‫‪120‬‬

‫‪1.00‬‬

‫‪1.32‬‬

‫‪1.47‬‬

‫‪1.59‬‬

‫‪1.70‬‬

‫‪1.79‬‬

‫‪1.88‬‬

‫‪2.04‬‬

‫‪2.18‬‬

‫‪2.32‬‬

‫‪3.41‬‬

‫‪2.51‬‬

‫‪2.64‬‬

‫‪2.80‬‬

‫‪3.02‬‬

‫‪3.32‬‬

‫‪3.78‬‬

‫‪4.61‬‬

‫‪6.63‬‬

‫‪‬‬

‫‪٦٢١‬‬

‫ﻣﻠﺤﻖ )‪(٦‬‬ ‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ )‪ q  (p , ‬ﻟﻨﯿﻮﻣﻦ‬ ‫‪M‬‬

‫‪‬‬

‫‪11‬‬

‫‪10‬‬

‫‪9‬‬

‫‪8‬‬

‫‪7‬‬

‫‪6‬‬

‫‪5‬‬

‫‪4‬‬

‫‪3‬‬

‫‪2‬‬

‫‪7.17‬‬

‫‪6.99‬‬

‫‪6.80‬‬

‫‪6.58‬‬

‫‪6.33‬‬

‫‪6.03‬‬

‫‪5.67‬‬

‫‪5.22‬‬

‫‪4.60‬‬

‫‪3.64‬‬

‫‪.05‬‬

‫‪10.48‬‬

‫‪10.24‬‬

‫‪9.97‬‬

‫‪9.67‬‬

‫‪9.32‬‬

‫‪8.91‬‬

‫‪8.42‬‬

‫‪7.80‬‬

‫‪6.98‬‬

‫‪5.70‬‬

‫‪.01‬‬

‫‪6.65‬‬

‫‪6.49‬‬

‫‪6.32‬‬

‫‪6.12‬‬

‫‪5.90‬‬

‫‪5.63‬‬

‫‪5.30‬‬

‫‪4.90‬‬

‫‪4.34‬‬

‫‪3.46‬‬

‫‪.05‬‬

‫‪9.30‬‬

‫‪9.10‬‬

‫‪8.87‬‬

‫‪8.61‬‬

‫‪8.32‬‬

‫‪7.97‬‬

‫‪7.56‬‬

‫‪7.03‬‬

‫‪6.33‬‬

‫‪5.24‬‬

‫‪.01‬‬

‫‪6.30‬‬

‫‪6.16‬‬

‫‪6.00‬‬

‫‪5.82‬‬

‫‪5.61‬‬

‫‪5.36‬‬

‫‪5.06‬‬

‫‪4.68‬‬

‫‪4.16‬‬

‫‪3.34‬‬

‫‪.05‬‬

‫‪8.55‬‬

‫‪8.37‬‬

‫‪8.17‬‬

‫‪6.94‬‬

‫‪7.68‬‬

‫‪7.37‬‬

‫‪7.01‬‬

‫‪6.54‬‬

‫‪5.92‬‬

‫‪4.95‬‬

‫‪.01‬‬

‫‪6.05‬‬

‫‪5.92‬‬

‫‪5.77‬‬

‫‪5.60‬‬

‫‪5.40‬‬

‫‪5.17‬‬

‫‪4.89‬‬

‫‪4.53‬‬

‫‪4.04‬‬

‫‪3.26‬‬

‫‪.05‬‬

‫‪8.03‬‬

‫‪7.86‬‬

‫‪7.68‬‬

‫‪7.47‬‬

‫‪7.24‬‬

‫‪6.96‬‬

‫‪6.62‬‬

‫‪6.20‬‬

‫‪5.64‬‬

‫‪4.75‬‬

‫‪.01‬‬

‫‪5.87‬‬

‫‪5.74‬‬

‫‪5.59‬‬

‫‪5.43‬‬

‫‪5.24‬‬

‫‪5.02‬‬

‫‪4.76‬‬

‫‪4.41‬‬

‫‪3.95‬‬

‫‪3.20‬‬

‫‪.05‬‬

‫‪7.65‬‬

‫‪7.49‬‬

‫‪7.33‬‬

‫‪7.13‬‬

‫‪6.91‬‬

‫‪6.66‬‬

‫‪6.35‬‬

‫‪5.96‬‬

‫‪5.43‬‬

‫‪4.60‬‬

‫‪.01‬‬

‫‪5.72‬‬

‫‪5.60‬‬

‫‪5.46‬‬

‫‪5.30‬‬

‫‪5.12‬‬

‫‪4.91‬‬

‫‪4.65‬‬

‫‪4.33‬‬

‫‪3.88‬‬

‫‪3.15‬‬

‫‪.05‬‬

‫‪7.36‬‬

‫‪7.21‬‬

‫‪7.05‬‬

‫‪6.87‬‬

‫‪6.67‬‬

‫‪6.43‬‬

‫‪6.14‬‬

‫‪5.77‬‬

‫‪5.27‬‬

‫‪4.48‬‬

‫‪.01‬‬

‫‪5.61‬‬

‫‪5.49‬‬

‫‪5.35‬‬

‫‪5.20‬‬

‫‪5.03‬‬

‫‪4.82‬‬

‫‪4.57‬‬

‫‪4.26‬‬

‫‪3.82‬‬

‫‪3.11‬‬

‫‪.05‬‬

‫‪7.13‬‬

‫‪6.99‬‬

‫‪6.84‬‬

‫‪6.67‬‬

‫‪6.48‬‬

‫‪6.25‬‬

‫‪5.97‬‬

‫‪5.62‬‬

‫‪5.15‬‬

‫‪4.39‬‬

‫‪.01‬‬

‫‪5.51‬‬

‫‪5.39‬‬

‫‪5.27‬‬

‫‪5.12‬‬

‫‪4.95‬‬

‫‪4.75‬‬

‫‪4.51‬‬

‫‪4.20‬‬

‫‪3.77‬‬

‫‪3.08‬‬

‫‪.05‬‬

‫‪6.94‬‬

‫‪6.81‬‬

‫‪6.67‬‬

‫‪6.51‬‬

‫‪6.32‬‬

‫‪6.10‬‬

‫‪5.84‬‬

‫‪5.50‬‬

‫‪5.05‬‬

‫‪4.32‬‬

‫‪.01‬‬

‫‪5.43‬‬

‫‪5.32‬‬

‫‪5.19‬‬

‫‪5.05‬‬

‫‪4.88‬‬

‫‪4.69‬‬

‫‪4.45‬‬

‫‪4.15‬‬

‫‪3.73‬‬

‫‪3.06‬‬

‫‪.05‬‬

‫‪6.79‬‬

‫‪6.67‬‬

‫‪6.53‬‬

‫‪6.37‬‬

‫‪6.19‬‬

‫‪5.98‬‬

‫‪5.73‬‬

‫‪5.40‬‬

‫‪4.96‬‬

‫‪4.26‬‬

‫‪.01‬‬

‫‪5.36‬‬

‫‪5.25‬‬

‫‪5.13‬‬

‫‪4.99‬‬

‫‪4.83‬‬

‫‪4.64‬‬

‫‪4.41‬‬

‫‪4.11‬‬

‫‪3.70‬‬

‫‪3.03‬‬

‫‪.05‬‬

‫‪6.66‬‬

‫‪6.54‬‬

‫‪6.41‬‬

‫‪6.26‬‬

‫‪6.08‬‬

‫‪5.88‬‬

‫‪5.63‬‬

‫‪5.32‬‬

‫‪4.89‬‬

‫‪4.21‬‬

‫‪.01‬‬

‫‪5.31‬‬

‫‪5.20‬‬

‫‪5.08‬‬

‫‪4.94‬‬

‫‪4.78‬‬

‫‪4.59‬‬

‫‪4.37‬‬

‫‪4.08‬‬

‫‪3.67‬‬

‫‪3.01‬‬

‫‪.05‬‬

‫‪6.55‬‬

‫‪6.44‬‬

‫‪6.31‬‬

‫‪6.16‬‬

‫‪5.99‬‬

‫‪5.80‬‬

‫‪5.56‬‬

‫‪5.25‬‬

‫‪4.84‬‬

‫‪4.17‬‬

‫‪.01‬‬

‫‪5.26‬‬

‫‪5.15‬‬

‫‪5.03‬‬

‫‪4.90‬‬

‫‪4.74‬‬

‫‪4.56‬‬

‫‪4.33‬‬

‫‪4.05‬‬

‫‪3.65‬‬

‫‪3.00‬‬

‫‪.05‬‬

‫‪6.46‬‬

‫‪6.35‬‬

‫‪6.22‬‬

‫‪6.08‬‬

‫‪5.92‬‬

‫‪5.72‬‬

‫‪5.49‬‬

‫‪5.19‬‬

‫‪4.79‬‬

‫‪4.13‬‬

‫‪.01‬‬

‫‪5.21‬‬

‫‪5.11‬‬

‫‪4.99‬‬

‫‪4.86‬‬

‫‪4.70‬‬

‫‪4.52‬‬

‫‪4.30‬‬

‫‪4.02‬‬

‫‪3.63‬‬

‫‪2.98‬‬

‫‪.05‬‬

‫‪6.38‬‬

‫‪6.27‬‬

‫‪6.15‬‬

‫‪6.01‬‬

‫‪5.85‬‬

‫‪5.66‬‬

‫‪5.43‬‬

‫‪5.14‬‬

‫‪4.74‬‬

‫‪4.10‬‬

‫‪.01‬‬

‫‪5.17‬‬

‫‪5.07‬‬

‫‪4.96‬‬

‫‪4.82‬‬

‫‪4.67‬‬

‫‪4.49‬‬

‫‪4.28‬‬

‫‪4.00‬‬

‫‪3.61‬‬

‫‪2.97‬‬

‫‪.05‬‬

‫‪6.31‬‬

‫‪6.20‬‬

‫‪6.08‬‬

‫‪5.94‬‬

‫‪5.79‬‬

‫‪5.60‬‬

‫‪5.38‬‬

‫‪5.09‬‬

‫‪4.70‬‬

‫‪4.07‬‬

‫‪.01‬‬

‫‪5.14‬‬

‫‪5.04‬‬

‫‪4.92‬‬

‫‪4.79‬‬

‫‪4.65‬‬

‫‪4.47‬‬

‫‪4.25‬‬

‫‪3.98‬‬

‫‪3.59‬‬

‫‪2.96‬‬

‫‪.05‬‬

‫‪6.25‬‬

‫‪6.14‬‬

‫‪6.02‬‬

‫‪5.89‬‬

‫‪5.73‬‬

‫‪5.55‬‬

‫‪5.33‬‬

‫‪5.05‬‬

‫‪4.67‬‬

‫‪4.05‬‬

‫‪.01‬‬

‫‪5.11‬‬

‫‪5.01‬‬

‫‪4.90‬‬

‫‪4.77‬‬

‫‪4.62‬‬

‫‪4.45‬‬

‫‪4.23‬‬

‫‪3.96‬‬

‫‪3.58‬‬

‫‪2.95‬‬

‫‪.05‬‬

‫‪6.19‬‬

‫‪6.09‬‬

‫‪5.97‬‬

‫‪5.84‬‬

‫‪5.69‬‬

‫‪5.51‬‬

‫‪5.29‬‬

‫‪5.02‬‬

‫‪4.64‬‬

‫‪4.02‬‬

‫‪.01‬‬

‫‪5.01‬‬

‫‪4.92‬‬

‫‪4.81‬‬

‫‪4.68‬‬

‫‪4.54‬‬

‫‪4.37‬‬

‫‪4.17‬‬

‫‪3.90‬‬

‫‪3.53‬‬

‫‪2.92‬‬

‫‪.05‬‬

‫‪6.02‬‬

‫‪5.92‬‬

‫‪5.81‬‬

‫‪5.69‬‬

‫‪5.54‬‬

‫‪5.37‬‬

‫‪5.17‬‬

‫‪4.91‬‬

‫‪4.55‬‬

‫‪3.96‬‬

‫‪.01‬‬

‫‪4.92‬‬

‫‪4.82‬‬

‫‪4.72‬‬

‫‪4.60‬‬

‫‪4.46‬‬

‫‪4.30‬‬

‫‪4.10‬‬

‫‪3.85‬‬

‫‪3.49‬‬

‫‪2.89‬‬

‫‪.05‬‬

‫‪5.85‬‬

‫‪5.76‬‬

‫‪5.65‬‬

‫‪5.54‬‬

‫‪5.40‬‬

‫‪5.24‬‬

‫‪5.05‬‬

‫‪4.80‬‬

‫‪4.45‬‬

‫‪3.89‬‬

‫‪.01‬‬

‫‪4.82‬‬

‫‪4.73‬‬

‫‪4.63‬‬

‫‪4.52‬‬

‫‪4.39‬‬

‫‪4.23‬‬

‫‪4.04‬‬

‫‪3.79‬‬

‫‪3.44‬‬

‫‪2.86‬‬

‫‪.05‬‬

‫‪5.69‬‬

‫‪5.60‬‬

‫‪5.50‬‬

‫‪5.39‬‬

‫‪5.26‬‬

‫‪5.11‬‬

‫‪4.93‬‬

‫‪4.70‬‬

‫‪4.37‬‬

‫‪3.82‬‬

‫‪.01‬‬

‫‪4.73‬‬

‫‪4.65‬‬

‫‪4.55‬‬

‫‪4.44‬‬

‫‪4.31‬‬

‫‪4.16‬‬

‫‪3.98‬‬

‫‪3.74‬‬

‫‪3.40‬‬

‫‪2.83‬‬

‫‪.05‬‬

‫‪5.53‬‬

‫‪5.45‬‬

‫‪5.36‬‬

‫‪5.25‬‬

‫‪5.13‬‬

‫‪4.99‬‬

‫‪4.82‬‬

‫‪4.59‬‬

‫‪4.28‬‬

‫‪3.76‬‬

‫‪.01‬‬

‫‪4.64‬‬

‫‪4.56‬‬

‫‪4.47‬‬

‫‪4.36‬‬

‫‪4.24‬‬

‫‪4.10‬‬

‫‪3.92‬‬

‫‪3.68‬‬

‫‪3.36‬‬

‫‪2.80‬‬

‫‪.05‬‬

‫‪٦٢٢‬‬

‫‪‬‬ ‫‪5‬‬

‫‪6‬‬

‫‪7‬‬

‫‪8‬‬

‫‪9‬‬

‫‪10‬‬

‫‪11‬‬

‫‪12‬‬

‫‪13‬‬

‫‪14‬‬

‫‪15‬‬

‫‪16‬‬

‫‪17‬‬

‫‪18‬‬

‫‪19‬‬

‫‪20‬‬

‫‪24‬‬

‫‪30‬‬

‫‪40‬‬

‫‪60‬‬

‫‪120‬‬

.01

3.70

4.20

4.50

4.71

4.87

5.01

5.12

5.21

5.30

5.37

.05

2.77

3.31

3.63

3.86

4.03

4.17

4.29

4.39

4.47

4.55

.01

3.64

4.12

4.40

4.60

4.76

4.88

4.99

5.08

5.16

5.23

٦٢٣

‫ﺗﺎﺑﻊ ﻣﻠﺤﻖ )‪(٦‬‬ ‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ )‪ q  (p , ‬ﻟﻨﯿﻮﻣﻦ‬ ‫‪m‬‬

‫‪‬‬ ‫‪5‬‬ ‫‪6‬‬

‫‪7‬‬

‫‪8‬‬

‫‪9‬‬

‫‪10‬‬

‫‪11‬‬

‫‪12‬‬

‫‪13‬‬

‫‪14‬‬

‫‪15‬‬

‫‪16‬‬

‫‪17‬‬

‫‪18‬‬

‫‪19‬‬

‫‪20‬‬

‫‪24‬‬

‫‪30‬‬

‫‪40‬‬

‫‪60‬‬

‫‪120‬‬

‫‪‬‬ ‫‪‬‬

‫‪20‬‬

‫‪19‬‬

‫‪18‬‬

‫‪17‬‬

‫‪16‬‬

‫‪15‬‬

‫‪14‬‬

‫‪13‬‬

‫‪12‬‬

‫‪.05‬‬

‫‪8.21‬‬

‫‪8.12‬‬

‫‪8.03‬‬

‫‪7.93‬‬

‫‪7.83‬‬

‫‪7.72‬‬

‫‪7.60‬‬

‫‪7.47‬‬

‫‪7.32‬‬

‫‪.01‬‬

‫‪11.93‬‬

‫‪11.81‬‬

‫‪11.68‬‬

‫‪11.55‬‬

‫‪11.40‬‬

‫‪11.24‬‬

‫‪11.08‬‬

‫‪10.89‬‬

‫‪10.70‬‬

‫‪.05‬‬

‫‪7.59‬‬

‫‪7.51‬‬

‫‪7.43‬‬

‫‪7.34‬‬

‫‪7.24‬‬

‫‪7.14‬‬

‫‪7.03‬‬

‫‪6.92‬‬

‫‪6.79‬‬

‫‪.01‬‬

‫‪10.54‬‬

‫‪10.43‬‬

‫‪10.32‬‬

‫‪10.21‬‬

‫‪10.08‬‬

‫‪9.95‬‬

‫‪9.81‬‬

‫‪9.65‬‬

‫‪9.48‬‬

‫‪.05‬‬

‫‪7.17‬‬

‫‪7.10‬‬

‫‪7.02‬‬

‫‪6.94‬‬

‫‪6.85‬‬

‫‪6.76‬‬

‫‪6.66‬‬

‫‪6.55‬‬

‫‪6.43‬‬

‫‪.01‬‬

‫‪9.65‬‬

‫‪9.55‬‬

‫‪9.46‬‬

‫‪9.35‬‬

‫‪9.24‬‬

‫‪9.12‬‬

‫‪9.00‬‬

‫‪8.86‬‬

‫‪8.71‬‬

‫‪.05‬‬

‫‪6.87‬‬

‫‪6.80‬‬

‫‪6.73‬‬

‫‪6.65‬‬

‫‪6.57‬‬

‫‪6.48‬‬

‫‪6.39‬‬

‫‪6.29‬‬

‫‪6.18‬‬

‫‪.01‬‬

‫‪9.03‬‬

‫‪8.94‬‬

‫‪8.85‬‬

‫‪8.76‬‬

‫‪8.66‬‬

‫‪8.55‬‬

‫‪8.44‬‬

‫‪8.31‬‬

‫‪8.18‬‬

‫‪.05‬‬

‫‪6.64‬‬

‫‪6.58‬‬

‫‪6.51‬‬

‫‪6.44‬‬

‫‪6.36‬‬

‫‪6.28‬‬

‫‪6.19‬‬

‫‪6.09‬‬

‫‪5.98‬‬

‫‪.01‬‬

‫‪8.57‬‬

‫‪8.49‬‬

‫‪8.41‬‬

‫‪8.33‬‬

‫‪8.23‬‬

‫‪8.13‬‬

‫‪8.03‬‬

‫‪7.91‬‬

‫‪7.78‬‬

‫‪.05‬‬

‫‪6.47‬‬

‫‪6.40‬‬

‫‪6.34‬‬

‫‪6.27‬‬

‫‪6.19‬‬

‫‪6.11‬‬

‫‪6.03‬‬

‫‪5.93‬‬

‫‪5.83‬‬

‫‪.01‬‬

‫‪8.23‬‬

‫‪8.15‬‬

‫‪8.08‬‬

‫‪7.99‬‬

‫‪7.91‬‬

‫‪7.81‬‬

‫‪7.71‬‬

‫‪7.60‬‬

‫‪7.49‬‬

‫‪.05‬‬

‫‪6.33‬‬

‫‪6.27‬‬

‫‪6.20‬‬

‫‪6.13‬‬

‫‪6.06‬‬

‫‪5.98‬‬

‫‪5.90‬‬

‫‪5.81‬‬

‫‪5.71‬‬

‫‪.01‬‬

‫‪7.95‬‬

‫‪7.88‬‬

‫‪7.81‬‬

‫‪7.73‬‬

‫‪7.65‬‬

‫‪7.56‬‬

‫‪7.46‬‬

‫‪7.36‬‬

‫‪7.25‬‬

‫‪.05‬‬

‫‪6.21‬‬

‫‪6.15‬‬

‫‪6.09‬‬

‫‪6.02‬‬

‫‪5.95‬‬

‫‪5.88‬‬

‫‪5.80‬‬

‫‪5.71‬‬

‫‪5.61‬‬

‫‪.01‬‬

‫‪7.73‬‬

‫‪7.66‬‬

‫‪7.59‬‬

‫‪7.52‬‬

‫‪7.44‬‬

‫‪7.36‬‬

‫‪7.26‬‬

‫‪7.17‬‬

‫‪7.06‬‬

‫‪.05‬‬

‫‪6.11‬‬

‫‪6.05‬‬

‫‪5.99‬‬

‫‪5.93‬‬

‫‪5.86‬‬

‫‪5.79‬‬

‫‪5.71‬‬

‫‪5.63‬‬

‫‪5.53‬‬

‫‪.01‬‬

‫‪7.55‬‬

‫‪7.48‬‬

‫‪7.42‬‬

‫‪7.35‬‬

‫‪7.27‬‬

‫‪7.19‬‬

‫‪7.10‬‬

‫‪7.01‬‬

‫‪6.90‬‬

‫‪.05‬‬

‫‪6.03‬‬

‫‪5.97‬‬

‫‪5.91‬‬

‫‪5.85‬‬

‫‪5.79‬‬

‫‪5.71‬‬

‫‪5.64‬‬

‫‪5.55‬‬

‫‪5.46‬‬

‫‪.01‬‬

‫‪7.39‬‬

‫‪7.33‬‬

‫‪7.27‬‬

‫‪7.20‬‬

‫‪7.13‬‬

‫‪7.05‬‬

‫‪6.96‬‬

‫‪6.87‬‬

‫‪6.77‬‬

‫‪.05‬‬

‫‪5.96‬‬

‫‪5.90‬‬

‫‪5.85‬‬

‫‪5.78‬‬

‫‪5.72‬‬

‫‪5.65‬‬

‫‪5.57‬‬

‫‪5.49‬‬

‫‪5.40‬‬

‫‪.01‬‬

‫‪7.26‬‬

‫‪7.20‬‬

‫‪7.14‬‬

‫‪7.07‬‬

‫‪7.00‬‬

‫‪6.93‬‬

‫‪6.84‬‬

‫‪6.76‬‬

‫‪6.66‬‬

‫‪.05‬‬

‫‪5.90‬‬

‫‪5.84‬‬

‫‪5.79‬‬

‫‪5.73‬‬

‫‪5.66‬‬

‫‪5.59‬‬

‫‪5.52‬‬

‫‪5.44‬‬

‫‪5.35‬‬

‫‪.01‬‬

‫‪7.15‬‬

‫‪7.09‬‬

‫‪7.03‬‬

‫‪6.97‬‬

‫‪6.90‬‬

‫‪6.82‬‬

‫‪6.74‬‬

‫‪6.66‬‬

‫‪6.56‬‬

‫‪.05‬‬

‫‪5.84‬‬

‫‪5.79‬‬

‫‪5.73‬‬

‫‪5.67‬‬

‫‪5.61‬‬

‫‪5.54‬‬

‫‪5.47‬‬

‫‪5.39‬‬

‫‪5.31‬‬

‫‪.01‬‬

‫‪7.05‬‬

‫‪7.00‬‬

‫‪6.94‬‬

‫‪6.87‬‬

‫‪6.81‬‬

‫‪6.73‬‬

‫‪6.66‬‬

‫‪6.57‬‬

‫‪6.48‬‬

‫‪.05‬‬

‫‪5.79‬‬

‫‪5.74‬‬

‫‪5.69‬‬

‫‪5.63‬‬

‫‪5.57‬‬

‫‪5.50‬‬

‫‪5.43‬‬

‫‪5.35‬‬

‫‪5.27‬‬

‫‪.01‬‬

‫‪6.97‬‬

‫‪6.91‬‬

‫‪6.85‬‬

‫‪6.79‬‬

‫‪6.73‬‬

‫‪6.65‬‬

‫‪6.58‬‬

‫‪6.50‬‬

‫‪6.41‬‬

‫‪.05‬‬

‫‪5.75‬‬

‫‪5.70‬‬

‫‪5.65‬‬

‫‪5.59‬‬

‫‪5.53‬‬

‫‪5.46‬‬

‫‪5.39‬‬

‫‪5.31‬‬

‫‪5.23‬‬

‫‪.01‬‬

‫‪6.89‬‬

‫‪6.84‬‬

‫‪6.78‬‬

‫‪6.72‬‬

‫‪6.65‬‬

‫‪6.58‬‬

‫‪6.51‬‬

‫‪6.43‬‬

‫‪6.34‬‬

‫‪.05‬‬

‫‪5.71‬‬

‫‪5.66‬‬

‫‪5.61‬‬

‫‪5.55‬‬

‫‪5.49‬‬

‫‪5.43‬‬

‫‪5.36‬‬

‫‪5.28‬‬

‫‪5.20‬‬

‫‪.01‬‬

‫‪6.82‬‬

‫‪6.77‬‬

‫‪6.71‬‬

‫‪6.65‬‬

‫‪6.59‬‬

‫‪6.52‬‬

‫‪6.45‬‬

‫‪6.37‬‬

‫‪6.28‬‬

‫‪.05‬‬

‫‪5.59‬‬

‫‪5.55‬‬

‫‪5.49‬‬

‫‪5.44‬‬

‫‪5.38‬‬

‫‪5.32‬‬

‫‪5.25‬‬

‫‪5.18‬‬

‫‪5.10‬‬

‫‪.01‬‬

‫‪6.61‬‬

‫‪6.56‬‬

‫‪6.51‬‬

‫‪6.45‬‬

‫‪6.39‬‬

‫‪6.33‬‬

‫‪6.26‬‬

‫‪6.19‬‬

‫‪6.11‬‬

‫‪.05‬‬

‫‪5.47‬‬

‫‪5.43‬‬

‫‪5.38‬‬

‫‪5.33‬‬

‫‪5.27‬‬

‫‪5.21‬‬

‫‪5.15‬‬

‫‪5.08‬‬

‫‪5.00‬‬

‫‪.01‬‬

‫‪6.41‬‬

‫‪6.36‬‬

‫‪6.31‬‬

‫‪6.26‬‬

‫‪6.20‬‬

‫‪6.14‬‬

‫‪6.08‬‬

‫‪6.01‬‬

‫‪5.93‬‬

‫‪.05‬‬

‫‪5.36‬‬

‫‪5.31‬‬

‫‪5.27‬‬

‫‪5.22‬‬

‫‪5.16‬‬

‫‪5.11‬‬

‫‪5.04‬‬

‫‪4.98‬‬

‫‪4.90‬‬

‫‪.01‬‬

‫‪6.21‬‬

‫‪6.16‬‬

‫‪6.12‬‬

‫‪6.07‬‬

‫‪6.02‬‬

‫‪5.96‬‬

‫‪5.90‬‬

‫‪5.83‬‬

‫‪5.76‬‬

‫‪.05‬‬

‫‪5.24‬‬

‫‪5.20‬‬

‫‪5.15‬‬

‫‪5.11‬‬

‫‪5.06‬‬

‫‪5.00‬‬

‫‪4.94‬‬

‫‪4.88‬‬

‫‪4.81‬‬

‫‪.01‬‬

‫‪6.01‬‬

‫‪5.97‬‬

‫‪5.93‬‬

‫‪5.89‬‬

‫‪5.84‬‬

‫‪5.78‬‬

‫‪5.73‬‬

‫‪5.67‬‬

‫‪5.60‬‬

‫‪.05‬‬

‫‪5.13‬‬

‫‪5.09‬‬

‫‪5.04‬‬

‫‪5.00‬‬

‫‪4.95‬‬

‫‪4.90‬‬

‫‪4.84‬‬

‫‪4.78‬‬

‫‪4.71‬‬

‫‪٦٢٤‬‬

5.44

5.50

5.56

5.61

5.66

5.71

5.75

5.79

5.83

.01

4.62

4.68

5.74

4.80

4.85

4.89

4.93

4.97

5.01

.05

5.29

5.35

5.40

5.45

5.49

5.54

5.57

5.61

5.65

.01

٦٢٥



‫ﻣﻠﺤﻖ )‪(٧‬‬ ‫‪x‬‬ ‫)‪ b( k ; n , p‬‬ ‫‪k 0‬‬

‫ﻟﻤﺘﻐﯿﺮ ﻋﺸﻮاﺋﻲ ﯾﺘﺒﻊ ﺗﻮزﯾﻊ ذي اﻟﺤﺪﯾﻦ‬

‫‪0.99‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.001‬‬ ‫‪.049‬‬

‫‪0.95‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.001‬‬ ‫‪.023‬‬ ‫‪.226‬‬

‫‪0.90‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.009‬‬ ‫‪.081‬‬ ‫‪.410‬‬

‫‪0.80‬‬ ‫‪.000‬‬ ‫‪.007‬‬ ‫‪.058‬‬ ‫‪.263‬‬ ‫‪.672‬‬

‫‪0.75‬‬ ‫‪.001‬‬ ‫‪.016‬‬ ‫‪.104‬‬ ‫‪.367‬‬ ‫‪.763‬‬

‫‪0.70‬‬ ‫‪.002‬‬ ‫‪.031‬‬ ‫‪.163‬‬ ‫‪.472‬‬ ‫‪.832‬‬

‫‪0.60‬‬ ‫‪.010‬‬ ‫‪.087‬‬ ‫‪.317‬‬ ‫‪.663‬‬ ‫‪.922‬‬

‫‪0.50‬‬ ‫‪.031‬‬ ‫‪.188‬‬ ‫‪.500‬‬ ‫‪.812‬‬ ‫‪.969‬‬

‫‪0.40‬‬ ‫‪.078‬‬ ‫‪.337‬‬ ‫‪.683‬‬ ‫‪.913‬‬ ‫‪.990‬‬

‫‪0.30‬‬ ‫‪.168‬‬ ‫‪.528‬‬ ‫‪.837‬‬ ‫‪.969‬‬ ‫‪.998‬‬

‫‪0.25‬‬ ‫‪.237‬‬ ‫‪.633‬‬ ‫‪.896‬‬ ‫‪.984‬‬ ‫‪.999‬‬

‫‪0.20‬‬ ‫‪.328‬‬ ‫‪.737‬‬ ‫‪.942‬‬ ‫‪.993‬‬ ‫‪1.00‬‬

‫‪0.1‬‬ ‫‪.590‬‬ ‫‪.919‬‬ ‫‪.991‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0.05‬‬ ‫‪.774‬‬ ‫‪.977‬‬ ‫‪.999‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0.99‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.004‬‬ ‫‪.096‬‬

‫‪0.95‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.001‬‬ ‫‪.012‬‬ ‫‪.086‬‬ ‫‪.401‬‬

‫‪0.90‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.002‬‬ ‫‪.013‬‬ ‫‪.070‬‬ ‫‪.264‬‬ ‫‪.651‬‬

‫‪0.80‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.001‬‬ ‫‪.006‬‬ ‫‪.033‬‬ ‫‪.121‬‬ ‫‪.322‬‬ ‫‪.624‬‬ ‫‪.893‬‬

‫‪0.75‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.004‬‬ ‫‪.020‬‬ ‫‪.078‬‬ ‫‪.224‬‬ ‫‪.474‬‬ ‫‪.756‬‬ ‫‪.944‬‬

‫‪0.70‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.002‬‬ ‫‪.011‬‬ ‫‪.047‬‬ ‫‪.150‬‬ ‫‪.350‬‬ ‫‪.617‬‬ ‫‪.851‬‬ ‫‪.972‬‬

‫‪0.60‬‬ ‫‪.000‬‬ ‫‪.002‬‬ ‫‪.012‬‬ ‫‪.055‬‬ ‫‪.166‬‬ ‫‪.367‬‬ ‫‪.618‬‬ ‫‪.833‬‬ ‫‪.954‬‬ ‫‪.994‬‬

‫‪0.50‬‬ ‫‪.001‬‬ ‫‪.011‬‬ ‫‪.055‬‬ ‫‪.172‬‬ ‫‪.377‬‬ ‫‪.623‬‬ ‫‪.828‬‬ ‫‪.945‬‬ ‫‪.989‬‬ ‫‪.999‬‬

‫‪0.40‬‬ ‫‪.006‬‬ ‫‪.046‬‬ ‫‪.167‬‬ ‫‪.382‬‬ ‫‪.633‬‬ ‫‪.834‬‬ ‫‪.945‬‬ ‫‪.988‬‬ ‫‪.998‬‬ ‫‪1.00‬‬

‫‪0.30‬‬ ‫‪.028‬‬ ‫‪.149‬‬ ‫‪.383‬‬ ‫‪.650‬‬ ‫‪.850‬‬ ‫‪.953‬‬ ‫‪.989‬‬ ‫‪.998‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0.25‬‬ ‫‪.056‬‬ ‫‪.244‬‬ ‫‪.526‬‬ ‫‪.776‬‬ ‫‪.922‬‬ ‫‪.980‬‬ ‫‪.996‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0.20‬‬ ‫‪.107‬‬ ‫‪.376‬‬ ‫‪.678‬‬ ‫‪.879‬‬ ‫‪.967‬‬ ‫‪.994‬‬ ‫‪.999‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0.10‬‬ ‫‪.349‬‬ ‫‪.736‬‬ ‫‪.930‬‬ ‫‪.987‬‬ ‫‪.998‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0.05‬‬ ‫‪.599‬‬ ‫‪.914‬‬ ‫‪.988‬‬ ‫‪.999‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫ﺟﺪول ﺣﺴﺎب‬

‫‪n=5‬‬ ‫‪p‬‬ ‫‪0.01‬‬ ‫‪.951‬‬ ‫‪.999‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬

‫‪x‬‬

‫‪n=10‬‬ ‫‪p‬‬ ‫‪0.01‬‬ ‫‪.904‬‬ ‫‪.996‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪9‬‬

‫‪x‬‬

‫ﺗﺎﺑﻊ ﻣﻠﺤﻖ )‪(٧‬‬ ‫ﺟﺪول ﺣﺴﺎب‬

‫‪x‬‬ ‫)‪ b( k ; n , p‬‬ ‫‪k 0‬‬

‫ﻟﻤﺘﻐﯿﺮ ﻋﺸﻮاﺋﻲ ﯾﺘﺒﻊ ﺗﻮزﯾﻊ ذي اﻟﺤﺪﯾﻦ‬

‫‪0.60‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.002‬‬ ‫‪.009‬‬ ‫‪.034‬‬ ‫‪.095‬‬ ‫‪.213‬‬ ‫‪.390‬‬ ‫‪.597‬‬ ‫‪.783‬‬ ‫‪.909‬‬ ‫‪.973‬‬ ‫‪.995‬‬ ‫‪1.00‬‬

‫‪0.05‬‬ ‫‪.463‬‬ ‫‪.829‬‬ ‫‪.964‬‬ ‫‪.995‬‬ ‫‪.999‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪n=15‬‬ ‫‪0.99‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.010‬‬ ‫‪.140‬‬

‫‪0.95‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.001‬‬ ‫‪.005‬‬ ‫‪.036‬‬ ‫‪.171‬‬ ‫‪.537‬‬

‫‪0.90‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.002‬‬ ‫‪.013‬‬ ‫‪.056‬‬ ‫‪.184‬‬ ‫‪.451‬‬ ‫‪.794‬‬

‫‪0.80‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.001‬‬ ‫‪.004‬‬ ‫‪.018‬‬ ‫‪.061‬‬ ‫‪.164‬‬ ‫‪.352‬‬ ‫‪.602‬‬ ‫‪.833‬‬ ‫‪.965‬‬

‫‪0.75‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.001‬‬ ‫‪.004‬‬ ‫‪.017‬‬ ‫‪.057‬‬ ‫‪.148‬‬ ‫‪.314‬‬ ‫‪.539‬‬ ‫‪.764‬‬ ‫‪.920‬‬ ‫‪.987‬‬

‫‪0.70‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.001‬‬ ‫‪.004‬‬ ‫‪.015‬‬ ‫‪.050‬‬ ‫‪.131‬‬ ‫‪.278‬‬ ‫‪.485‬‬ ‫‪.703‬‬ ‫‪.873‬‬ ‫‪.965‬‬ ‫‪.995‬‬

‫‪0.50‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.004‬‬ ‫‪.018‬‬ ‫‪.059‬‬ ‫‪.151‬‬ ‫‪.304‬‬ ‫‪.500‬‬ ‫‪.696‬‬ ‫‪.849‬‬ ‫‪.941‬‬ ‫‪.982‬‬ ‫‪.996‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫اﻟﻤﺼﺪر ‪ :‬ﻋﻦ ])‪[Devore(1995‬‬ ‫‪٦٢٦‬‬

‫‪0.40‬‬ ‫‪.000‬‬ ‫‪.005‬‬ ‫‪.027‬‬ ‫‪.091‬‬ ‫‪.217‬‬ ‫‪.403‬‬ ‫‪.610‬‬ ‫‪.787‬‬ ‫‪.905‬‬ ‫‪.966‬‬ ‫‪.991‬‬ ‫‪.998‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0.30‬‬ ‫‪.005‬‬ ‫‪.035‬‬ ‫‪.127‬‬ ‫‪.297‬‬ ‫‪.515‬‬ ‫‪.722‬‬ ‫‪.869‬‬ ‫‪.950‬‬ ‫‪.985‬‬ ‫‪.996‬‬ ‫‪.999‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0.25‬‬ ‫‪.013‬‬ ‫‪.080‬‬ ‫‪.236‬‬ ‫‪.461‬‬ ‫‪.686‬‬ ‫‪.852‬‬ ‫‪.943‬‬ ‫‪.983‬‬ ‫‪.996‬‬ ‫‪.999‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0.20‬‬ ‫‪.035‬‬ ‫‪.167‬‬ ‫‪.398‬‬ ‫‪.648‬‬ ‫‪.836‬‬ ‫‪.939‬‬ ‫‪.982‬‬ ‫‪.996‬‬ ‫‪.999‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0.10‬‬ ‫‪.206‬‬ ‫‪.549‬‬ ‫‪.816‬‬ ‫‪.944‬‬ ‫‪.987‬‬ ‫‪.998‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪p‬‬ ‫‪0.01‬‬ ‫‪.860‬‬ ‫‪.990‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪13‬‬ ‫‪14‬‬

‫‪x‬‬

‫ﺗﺎﺑﻊ ﻣﻠﺤﻖ )‪(٧‬‬ ‫‪x‬‬ ‫)‪ b( k ; n , p‬‬ ‫‪k 0‬‬

‫ﻟﻤﺘﻐﯿﺮ ﻋﺸﻮاﺋﻲ ﯾﺘﺒﻊ ﺗﻮزﯾﻊ ذي اﻟﺤﺪﯾﻦ‬

‫‪0.99‬‬

‫‪0.95‬‬

‫‪0.90‬‬

‫‪0.80‬‬

‫‪0.75‬‬

‫‪0.70‬‬

‫‪0.60‬‬

‫‪0.50‬‬

‫‪0.40‬‬

‫‪0.30‬‬

‫‪0.25‬‬

‫‪0.20‬‬

‫‪0.10‬‬

‫‪0.05‬‬

‫‪0.01‬‬

‫‪.000‬‬

‫‪.001‬‬

‫‪.003‬‬

‫‪.012‬‬

‫‪.122‬‬

‫‪.358‬‬

‫‪.818‬‬

‫‪0‬‬

‫‪.000‬‬

‫‪.001‬‬

‫‪.008‬‬

‫‪.024‬‬

‫‪.069‬‬

‫‪.392‬‬

‫‪.736‬‬

‫‪.983‬‬

‫‪1‬‬

‫‪.000‬‬

‫‪.004‬‬

‫‪.035‬‬

‫‪.091‬‬

‫‪.206‬‬

‫‪.677‬‬

‫‪.925‬‬

‫‪.999‬‬

‫‪2‬‬

‫‪.000‬‬

‫‪.001‬‬

‫‪.016‬‬

‫‪.107‬‬

‫‪.225‬‬

‫‪.411‬‬

‫‪.867‬‬

‫‪.984‬‬

‫‪1.00‬‬

‫‪3‬‬

‫‪.000‬‬

‫‪.006‬‬

‫‪.051‬‬

‫‪.238‬‬

‫‪.415‬‬

‫‪.630‬‬

‫‪.957‬‬

‫‪.997‬‬

‫‪1.00‬‬

‫‪4‬‬

‫‪.000‬‬

‫‪.002‬‬

‫‪.021‬‬

‫‪.126‬‬

‫‪.416‬‬

‫‪.617‬‬

‫‪.804‬‬

‫‪.989‬‬

‫‪1.00‬‬

‫‪5‬‬

‫‪.000‬‬

‫‪.006‬‬

‫‪.058‬‬

‫‪.250‬‬

‫‪.608‬‬

‫‪.786‬‬

‫‪.913‬‬

‫‪.998‬‬

‫‪1.00‬‬

‫‪6‬‬

‫‪.000‬‬

‫‪.001‬‬

‫‪.021‬‬

‫‪.132‬‬

‫‪.416‬‬

‫‪.772‬‬

‫‪.898‬‬

‫‪.968‬‬

‫‪1.00‬‬

‫‪7‬‬

‫‪.000‬‬

‫‪.001‬‬

‫‪.005‬‬

‫‪.057‬‬

‫‪.252‬‬

‫‪.596‬‬

‫‪.887‬‬

‫‪.959‬‬

‫‪.990‬‬

‫‪1.00‬‬

‫‪8‬‬

‫‪.000‬‬

‫‪.001‬‬

‫‪.004‬‬

‫‪.017‬‬

‫‪.128‬‬

‫‪.412‬‬

‫‪.755‬‬

‫‪.952‬‬

‫‪.986‬‬

‫‪.997‬‬

‫‪1.00‬‬

‫‪9‬‬

‫‪.000‬‬

‫‪.003‬‬

‫‪.014‬‬

‫‪.048‬‬

‫‪.245‬‬

‫‪.588‬‬

‫‪.872‬‬

‫‪.983‬‬

‫‪.996‬‬

‫‪.999‬‬

‫‪1.00‬‬

‫‪10‬‬

‫‪.000‬‬

‫‪.010‬‬

‫‪.041‬‬

‫‪.113‬‬

‫‪.404‬‬

‫‪.748‬‬

‫‪.943‬‬

‫‪.995‬‬

‫‪.999‬‬

‫‪1.00‬‬

‫‪11‬‬

‫‪.000‬‬

‫‪.032‬‬

‫‪.102‬‬

‫‪.228‬‬

‫‪.584‬‬

‫‪.868‬‬

‫‪.979‬‬

‫‪.999‬‬

‫‪1.00‬‬

‫‪12‬‬

‫‪.000‬‬

‫‪.002‬‬

‫‪.087‬‬

‫‪.214‬‬

‫‪.392‬‬

‫‪.750‬‬

‫‪.942‬‬

‫‪.994‬‬

‫‪1.00‬‬

‫‪13‬‬

‫‪.000‬‬

‫‪.011‬‬

‫‪.196‬‬

‫‪.383‬‬

‫‪.584‬‬

‫‪.874‬‬

‫‪.979‬‬

‫‪.998‬‬

‫‪1.00‬‬

‫ﺟﺪول ﺣﺴﺎب‬

‫‪n=20‬‬ ‫‪p‬‬

‫‪14‬‬ ‫‪1.00‬‬

‫‪.000‬‬

‫‪.003‬‬

‫‪.043‬‬

‫‪.370‬‬

‫‪.585‬‬

‫‪.762‬‬

‫‪.949‬‬

‫‪.994‬‬

‫‪1.00‬‬

‫‪15‬‬

‫‪.000‬‬

‫‪.016‬‬

‫‪.133‬‬

‫‪.589‬‬

‫‪.775‬‬

‫‪.893‬‬

‫‪.984‬‬

‫‪.999‬‬

‫‪1.00‬‬

‫‪16‬‬

‫‪.001‬‬

‫‪.075‬‬

‫‪.323‬‬

‫‪.794‬‬

‫‪.909‬‬

‫‪.965‬‬

‫‪1.00‬‬

‫‪17‬‬

‫‪.996‬‬ ‫‪.017‬‬

‫‪.264‬‬

‫‪.608‬‬

‫‪.931‬‬

‫‪.976‬‬

‫‪.992‬‬

‫‪.999‬‬

‫‪1.00‬‬

‫‪18‬‬

‫‪.182‬‬

‫‪.642‬‬

‫‪.878‬‬

‫‪.988‬‬

‫‪.997‬‬

‫‪.999‬‬

‫‪1.00‬‬

‫‪19‬‬

‫‪٦٢٧‬‬

‫‪x‬‬

‫ﺗﺎﺑﻊ ﻣﻠﺤﻖ )‪(٧‬‬

‫ﺟﺪول ﺣﺴﺎب‬

‫‪x‬‬ ‫)‪ b( k ; n , p‬‬ ‫‪k 0‬‬

‫ﻟﻤﺘﻐﯿﺮ ﻋﺸﻮاﺋﻲ ﯾﺘﺒﻊ ﺗﻮزﯾﻊ ذي‬

‫اﻟﺤﺪﯾﻦ‬ ‫‪n=25‬‬ ‫‪0.99‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬

‫‪0.95‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬

‫‪0.90‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬

‫‪0.80‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬

‫‪0.75‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬

‫‪0.70‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬

‫‪0.60‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.00‬‬ ‫‪.000‬‬

‫‪0.50‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬

‫‪0.40‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.002‬‬ ‫‪.009‬‬

‫‪0.30‬‬ ‫‪.000‬‬ ‫‪.002‬‬ ‫‪.009‬‬ ‫‪.033‬‬ ‫‪.090‬‬

‫‪0.25‬‬ ‫‪.001‬‬ ‫‪.007‬‬ ‫‪.032‬‬ ‫‪.096‬‬ ‫‪.214‬‬

‫‪0.20‬‬ ‫‪.004‬‬ ‫‪.027‬‬ ‫‪.098‬‬ ‫‪.234‬‬ ‫‪.421‬‬

‫‪0.10‬‬ ‫‪.072‬‬ ‫‪.271‬‬ ‫‪.537‬‬ ‫‪.764‬‬ ‫‪.902‬‬

‫‪0.05‬‬ ‫‪.277‬‬ ‫‪.642‬‬ ‫‪.873‬‬ ‫‪.966‬‬ ‫‪.993‬‬

‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬

‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.002‬‬ ‫‪.006‬‬

‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.001‬‬ ‫‪.003‬‬ ‫‪.020‬‬ ‫‪.030‬‬

‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.002‬‬ ‫‪.006‬‬ ‫‪.017‬‬ ‫‪.044‬‬ ‫‪.098‬‬

‫‪.000‬‬ ‫‪.000‬‬ ‫‪.001‬‬ ‫‪.004‬‬ ‫‪.013‬‬ ‫‪.034‬‬ ‫‪.078‬‬ ‫‪.154‬‬ ‫‪.268‬‬ ‫‪.414‬‬

‫‪.002‬‬ ‫‪.007‬‬ ‫‪.022‬‬ ‫‪.054‬‬ ‫‪.115‬‬ ‫‪.212‬‬ ‫‪.345‬‬ ‫‪.500‬‬ ‫‪.655‬‬ ‫‪.788‬‬

‫‪.029‬‬ ‫‪.074‬‬ ‫‪.154‬‬ ‫‪.274‬‬ ‫‪.425‬‬ ‫‪.586‬‬ ‫‪.732‬‬ ‫‪.846‬‬ ‫‪.922‬‬ ‫‪.966‬‬

‫‪.193‬‬ ‫‪.341‬‬ ‫‪.512‬‬ ‫‪.677‬‬ ‫‪.811‬‬ ‫‪.902‬‬ ‫‪.956‬‬ ‫‪.983‬‬ ‫‪.994‬‬ ‫‪.998‬‬

‫‪.378‬‬ ‫‪.561‬‬ ‫‪.727‬‬ ‫‪.851‬‬ ‫‪.929‬‬ ‫‪.970‬‬ ‫‪.980‬‬ ‫‪.997‬‬ ‫‪.999‬‬ ‫‪1.00‬‬

‫‪.617‬‬ ‫‪.780‬‬ ‫‪.891‬‬ ‫‪.953‬‬ ‫‪.983‬‬ ‫‪.994‬‬ ‫‪.998‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪.967‬‬ ‫‪.991‬‬ ‫‪.998‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪.999‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬

‫‪.000‬‬ ‫‪.000‬‬ ‫‪.002‬‬

‫‪.017‬‬ ‫‪.047‬‬ ‫‪.109‬‬

‫‪.071‬‬ ‫‪.149‬‬ ‫‪.273‬‬

‫‪.189‬‬ ‫‪.323‬‬ ‫‪.488‬‬

‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.000‬‬ ‫‪.002‬‬ ‫‪.026‬‬ ‫‪.222‬‬

‫‪.000‬‬ ‫‪.001‬‬ ‫‪.007‬‬ ‫‪.034‬‬ ‫‪.127‬‬ ‫‪.358‬‬ ‫‪.723‬‬

‫‪.009‬‬ ‫‪.033‬‬ ‫‪.098‬‬ ‫‪.236‬‬ ‫‪.463‬‬ ‫‪.729‬‬ ‫‪.928‬‬

‫‪.220‬‬ ‫‪.383‬‬ ‫‪.579‬‬ ‫‪.766‬‬ ‫‪.902‬‬ ‫‪.973‬‬ ‫‪.996‬‬

‫‪.439‬‬ ‫‪.622‬‬ ‫‪.786‬‬ ‫‪.904‬‬ ‫‪.968‬‬ ‫‪.993‬‬ ‫‪.999‬‬

‫‪.659‬‬ ‫‪.807‬‬ ‫‪.910‬‬ ‫‪.967‬‬ ‫‪.991‬‬ ‫‪.998‬‬ ‫‪1.00‬‬

‫‪p‬‬ ‫‪0.01‬‬ ‫‪.778‬‬ ‫‪.974‬‬ ‫‪.998‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬

‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪13‬‬ ‫‪14‬‬

‫‪.575‬‬ ‫‪.726‬‬

‫‪.885‬‬ ‫‪.946‬‬ ‫‪.978‬‬

‫‪.987‬‬ ‫‪.996‬‬ ‫‪.999‬‬

‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪15‬‬ ‫‪16‬‬ ‫‪17‬‬

‫‪.926‬‬ ‫‪.971‬‬ ‫‪.991‬‬ ‫‪.998‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪.993‬‬ ‫‪.998‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬ ‫‪1.00‬‬

‫‪18‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪23‬‬ ‫‪24‬‬

‫‪1.00‬‬

‫‪.846‬‬

‫‪٦٢٨‬‬

‫‪x‬‬

(٨) ‫ﻣﻠﺤﻖ‬ ‫ ﻻﺧﺘﺒﺎر إﺷﺎرة اﻟﺮﺗﺐ‬d ( n ,  ), d ( n ,  ) ‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ‬ n 3 4 5 6

d 1 1 1 2 1 2 3 4 1 2 4 5 1 2 4 5 6 7 2 3 6 7 9 10 4 5 9 10 11 12 6 7

Confidence coefficient .750 .875 .938 .875 .969 .937 .906 .844 .984 .969 .922 .891 .992 .984 .961 .945 .922 .891 .992 .988 .961 .945 .902 .871 .990 .986 .951 .936 .916 .895 .990 .986 ٦٢٩

  .250 .125 .062 .125 .031 .063 .094 .156 .016 .031 .078 .109 .008 .016 .039 .055 .078 .109 .008 .012 .039 .055 .098 .129 .010 .014 .049 .064 .084 .105 .010 .014

  .125 .063 .031 .063 .016 .031 .047 .078 .008 .016 .039 .055 .004 .008 .020 .027 .039 .055 .004 .006 .020 .027 .049 .065 .005 .007 .024 .032 .042 .053 .005 .007

‫‪.021‬‬ ‫‪.027‬‬ ‫‪.042‬‬ ‫‪.051‬‬ ‫‪.005‬‬ ‫‪.006‬‬ ‫‪.021‬‬ ‫‪.026‬‬ ‫‪.046‬‬ ‫‪.055‬‬ ‫‪.004‬‬ ‫‪.005‬‬ ‫‪.024‬‬ ‫‪.029‬‬ ‫‪.047‬‬ ‫‪.055‬‬

‫‪.042‬‬ ‫‪.054‬‬ ‫‪.083‬‬ ‫‪.102‬‬ ‫‪.009‬‬ ‫‪.012‬‬ ‫‪.042‬‬ ‫‪.052‬‬ ‫‪.092‬‬ ‫‪.110‬‬ ‫‪.008‬‬ ‫‪.010‬‬ ‫‪.048‬‬ ‫‪.057‬‬ ‫‪.094‬‬ ‫‪.110‬‬

‫‪.958‬‬ ‫‪.946‬‬ ‫‪.917‬‬ ‫‪.898‬‬ ‫‪.991‬‬ ‫‪.988‬‬ ‫‪.958‬‬ ‫‪.948‬‬ ‫‪.908‬‬ ‫‪.890‬‬ ‫‪.992‬‬ ‫‪.990‬‬ ‫‪.952‬‬ ‫‪.943‬‬ ‫‪.906‬‬ ‫‪.890‬‬

‫اﻟﻤﺼﺪر ‪ :‬ﻋﻦ ])‪[Daniel (1978‬‬

‫‪٦٣٠‬‬

‫‪11‬‬ ‫‪12‬‬ ‫‪14‬‬ ‫‪15‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪14‬‬ ‫‪15‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪22‬‬ ‫‪23‬‬

‫‪12‬‬

‫‪13‬‬

‫ ﻻﺧﺘﺒﺎر إﺷﺎرة اﻟﺮﺗﺐ‬d ( n ,  ), d ( n ,  ) ‫( ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ‬٨) ‫ ﻣﻠﺤﻖ‬: ‫ﺗﺎﺑﻊ‬ n 14

d 13 14 22 23 26 27 16 17 26 27 31 32 20 21 30 31 36 37 24 25 35 36 42 43 28 29 41 42 48 49 33 34 47 48 54 55 38 39 53 54 61 62 43 44 59 60 68 69

Confidence coefficient .991 .989 .951 .942 .909 .896 .992 .990 .952 .945 .905 .893 .991 .989 .956 .949 .907 .895 .991 .989 .955 .949 .902 .891 .991 .990 .952 .946 .901 .892 .991 .989 .951 .945 .904 .896 .991 .989 .952 .947 .903 .895 .991 .990 .954 .950 .904 .897

  .009 .011 .049 .058 .091 .104 .008 .010 .048 .055 .095 .107 .009 .011 .044 .051 .093 .105 .009 .011 .045 .051 .098 .109 .009 .010 .048 .054 .099 .108 .009 .011 .049 .055 .096 .104 .009 .011 .048 .053 .097 .105 .009 .010 .046 .050 .096 .103

  .004 .005 .025 .029 .045 .052 .004 .005 .024 .028 .047 .054 .005 .006 .022 .025 .047 .052 .005 .006 .022 .025 .049 .054 .005 .005 .024 .027 .049 .054 .005 .005 .025 .027 .048 .052 .005 .005 .024 .027 .049 .053 .005 .005 .023 .025 .048 .052

[Daniel (1978)] ‫ ﻋﻦ‬: ‫اﻟﻤﺼﺪر‬ ٦٣١

(٨) ‫ ﻣﻠﺤﻖ‬: ‫ﺗﺎﺑﻊ‬ ‫ ﻻﺧﺘﺒﺎر اﻹﺷﺎرة‬d ( n , ), d ( n , ) ‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ‬ n 22

d 49 50 66 67 76 77 55 56 74 75 84 85 62 63 82 83 92 93 69 70 90 91 101 102

Confidence coefficient .991 .990 .954 .950 .902 .895 .991 .990 .952 .948 .902 .895 .990 .989 .951 .947 .905 .899 .990 .989 .952 .948 .904 .899

  .009 .010 .046 .050 .098 .105 .009 .010 .048 .052 .098 .105 .010 .011 .049 .053 .095 .101 .010 .011 .048 .052 .096 .101

  .005 .005 .023 .025 .049 .053 .005 .005 .024 .026 .049 .052 .005 .005 .025 .026 .048 .051 .005 .005 .024 .026 .048 .051

[Daniel (1978)] ‫ ﻋﻦ‬: ‫اﻟﻤﺼﺪر‬

٦٣٢

‫ﻣﻠﺤﻖ )‪(٩‬‬ ‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ‪ r1‬اﻟﺴﻔﻠﻲ ﻻﺧﺘﺒﺎر اﻟﺪورات‬ ‫‪20‬‬

‫‪19‬‬

‫‪18‬‬

‫‪17‬‬

‫‪16‬‬

‫‪15‬‬

‫‪14‬‬

‫‪13‬‬

‫‪12‬‬

‫‪11‬‬

‫‪10‬‬

‫‪9‬‬

‫‪8‬‬

‫‪7‬‬

‫‪6‬‬

‫‪5‬‬

‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪12‬‬ ‫‪13‬‬ ‫‪13‬‬ ‫‪13‬‬ ‫‪14‬‬

‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪12‬‬ ‫‪13‬‬ ‫‪13‬‬ ‫‪13‬‬

‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪12‬‬ ‫‪13‬‬ ‫‪13‬‬

‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪11‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪12‬‬ ‫‪13‬‬

‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪11‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪12‬‬

‫‪2‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪11‬‬ ‫‪11‬‬ ‫‪12‬‬

‫‪2‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪10‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪11‬‬

‫‪2‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪10‬‬ ‫‪10‬‬ ‫‪10‬‬

‫‪2‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪10‬‬

‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪9‬‬

‫‪2‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪9‬‬

‫‪2‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪8‬‬ ‫‪8‬‬

‫‪2‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪7‬‬

‫‪2‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪6‬‬

‫‪2‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪6‬‬

‫‪2‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬

‫‪4‬‬

‫‪3‬‬

‫‪2‬‬

‫‪n2‬‬ ‫‪n1‬‬

‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬

‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬

‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬

‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪13‬‬ ‫‪14‬‬ ‫‪15‬‬ ‫‪16‬‬ ‫‪17‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪20‬‬

‫اﻟﻤﺼﺪر ‪ :‬ﻋﻦ ])‪[Daniel (1978‬‬

‫ﻣﻠﺤﻖ )‪(١٠‬‬ ‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ‪ r2‬اﻟﻌﻠﯿﺎ ﻻﺧﺘﺒﺎر اﻟﺪورات‬ ‫‪20‬‬

‫‪19‬‬

‫‪18‬‬

‫‪17‬‬

‫‪16‬‬

‫‪15‬‬

‫‪14‬‬

‫‪13‬‬

‫‪1‬‬ ‫‪7‬‬

‫‪1‬‬ ‫‪5‬‬ ‫‪1‬‬ ‫‪6‬‬

‫‪12‬‬

‫‪11‬‬

‫‪10‬‬

‫‪9‬‬

‫‪8‬‬

‫‪7‬‬

‫‪6‬‬

‫‪5‬‬

‫‪4‬‬

‫‪13‬‬

‫‪11‬‬ ‫‪12‬‬

‫‪9‬‬ ‫‪10‬‬ ‫‪11‬‬

‫‪9‬‬ ‫‪10‬‬ ‫‪10‬‬

‫‪9‬‬ ‫‪9‬‬

‫‪1‬‬ ‫‪4‬‬ ‫‪1‬‬ ‫‪6‬‬

‫‪1‬‬ ‫‪4‬‬ ‫‪1‬‬ ‫‪5‬‬

‫‪1‬‬ ‫‪4‬‬ ‫‪1‬‬ ‫‪4‬‬

‫‪1‬‬ ‫‪3‬‬ ‫‪1‬‬ ‫‪4‬‬

‫‪1‬‬ ‫‪3‬‬ ‫‪1‬‬ ‫‪3‬‬

‫‪1‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪2‬‬

‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬

‫‪3‬‬

‫‪2‬‬

‫‪n2‬‬ ‫‪n1‬‬

‫‪٦٣٣‬‬

‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬

‫‪7‬‬ ‫‪8‬‬

‫‪18‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪23‬‬ ‫‪24‬‬ ‫‪25‬‬ ‫‪25‬‬ ‫‪26‬‬ ‫‪27‬‬ ‫‪27‬‬ ‫‪28‬‬

‫‪18‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪23‬‬ ‫‪23‬‬ ‫‪24‬‬ ‫‪25‬‬ ‫‪26‬‬ ‫‪26‬‬ ‫‪27‬‬ ‫‪27‬‬

‫‪18‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪23‬‬ ‫‪24‬‬ ‫‪25‬‬ ‫‪25‬‬ ‫‪26‬‬ ‫‪26‬‬ ‫‪27‬‬

‫‪18‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪23‬‬ ‫‪23‬‬ ‫‪24‬‬ ‫‪25‬‬ ‫‪25‬‬ ‫‪26‬‬ ‫‪26‬‬

‫‪18‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪23‬‬ ‫‪23‬‬ ‫‪24‬‬ ‫‪25‬‬ ‫‪25‬‬ ‫‪25‬‬

‫‪18‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪22‬‬ ‫‪23‬‬ ‫‪23‬‬ ‫‪24‬‬ ‫‪24‬‬ ‫‪25‬‬

‫‪17‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪22‬‬ ‫‪23‬‬ ‫‪23‬‬ ‫‪23‬‬ ‫‪24‬‬

‫‪17‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪22‬‬ ‫‪23‬‬ ‫‪23‬‬

‫‪16‬‬ ‫‪17‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪21‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪22‬‬

‫‪16‬‬ ‫‪17‬‬ ‫‪17‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪19‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪20‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪21‬‬

‫اﻟﻤﺼﺪر ‪ :‬ﻋﻦ ])‪[Daniel (1978‬‬

‫‪٦٣٤‬‬

‫‪16‬‬ ‫‪16‬‬ ‫‪17‬‬ ‫‪17‬‬ ‫‪18‬‬ ‫‪18‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪19‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪20‬‬

‫‪15‬‬ ‫‪16‬‬ ‫‪16‬‬ ‫‪16‬‬ ‫‪17‬‬ ‫‪17‬‬ ‫‪18‬‬ ‫‪18‬‬ ‫‪18‬‬ ‫‪18‬‬ ‫‪18‬‬ ‫‪18‬‬

‫‪14‬‬ ‫‪15‬‬ ‫‪15‬‬ ‫‪16‬‬ ‫‪16‬‬ ‫‪16‬‬ ‫‪16‬‬ ‫‪17‬‬ ‫‪17‬‬ ‫‪17‬‬ ‫‪17‬‬ ‫‪17‬‬

‫‪14‬‬ ‫‪14‬‬ ‫‪14‬‬ ‫‪14‬‬ ‫‪15‬‬ ‫‪15‬‬ ‫‪15‬‬

‫‪13‬‬ ‫‪13‬‬ ‫‪13‬‬ ‫‪13‬‬

‫‪9‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪13‬‬ ‫‪14‬‬ ‫‪15‬‬ ‫‪16‬‬ ‫‪17‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪20‬‬

‫ﻣﻠﺤﻖ )‪(١١‬‬ ‫ﺟدول اﻟﻘﯾم اﻟﺣرﺟﺔ ﻻﺧﺗﺑﺎر ‪Mann-Whitney-Wilcoxon‬‬

‫‪20‬‬

‫‪19‬‬

‫‪18‬‬

‫‪17‬‬

‫‪16‬‬

‫‪15‬‬

‫‪14‬‬

‫‪13‬‬

‫‪12‬‬

‫‪11‬‬

‫‪10‬‬

‫‪9‬‬

‫‪8‬‬

‫‪7‬‬

‫‪6‬‬

‫‪5‬‬

‫‪4‬‬

‫‪3‬‬

‫‪n2=2‬‬

‫‪‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪8‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪7‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪7‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪6‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪5‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪1‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪16‬‬

‫‪1‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪8‬‬ ‫‪11‬‬ ‫‪15‬‬

‫‪1‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪14‬‬

‫‪1‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪10‬‬ ‫‪13‬‬

‫‪0‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬

‫‪0‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪11‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪11‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪10‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪9‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪7‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪6‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪6‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬

‫‪001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪4‬‬ ‫‪9‬‬ ‫‪11‬‬ ‫‪15‬‬ ‫‪19‬‬ ‫‪23‬‬

‫‪4‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪14‬‬ ‫‪18‬‬ ‫‪22‬‬

‫‪4‬‬ ‫‪7‬‬ ‫‪10‬‬ ‫‪13‬‬ ‫‪17‬‬ ‫‪21‬‬

‫‪3‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪16‬‬ ‫‪19‬‬

‫‪3‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪18‬‬

‫‪2‬‬ ‫‪6‬‬ ‫‪9‬‬ ‫‪11‬‬ ‫‪13‬‬ ‫‪17‬‬

‫‪2‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪10‬‬ ‫‪12‬‬ ‫‪16‬‬

‫‪2‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪9‬‬ ‫‪11‬‬ ‫‪14‬‬

‫‪1‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪13‬‬

‫‪1‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬

‫‪1‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪11‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪10‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪8‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪7‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪6‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬

‫‪001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪8‬‬ ‫‪14‬‬ ‫‪17‬‬ ‫‪21‬‬ ‫‪26‬‬ ‫‪31‬‬

‫‪8‬‬ ‫‪13‬‬ ‫‪16‬‬ ‫‪20‬‬ ‫‪24‬‬ ‫‪29‬‬

‫‪7‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪19‬‬ ‫‪23‬‬ ‫‪28‬‬

‫‪6‬‬ ‫‪11‬‬ ‫‪14‬‬ ‫‪18‬‬ ‫‪21‬‬ ‫‪26‬‬

‫‪6‬‬ ‫‪10‬‬ ‫‪13‬‬ ‫‪16‬‬ ‫‪20‬‬ ‫‪24‬‬

‫‪5‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪19‬‬ ‫‪23‬‬

‫‪4‬‬ ‫‪8‬‬ ‫‪11‬‬ ‫‪14‬‬ ‫‪17‬‬ ‫‪21‬‬

‫‪4‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪13‬‬ ‫‪16‬‬ ‫‪19‬‬

‫‪3‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪14‬‬ ‫‪18‬‬

‫‪3‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪13‬‬ ‫‪16‬‬

‫‪2‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪14‬‬

‫‪2‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪13‬‬

‫‪1‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪11‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪9‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪6‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬

‫‪001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪13‬‬ ‫‪19‬‬ ‫‪23‬‬ ‫‪28‬‬ ‫‪33‬‬ ‫‪39‬‬

‫‪12‬‬ ‫‪18‬‬ ‫‪21‬‬ ‫‪26‬‬ ‫‪31‬‬ ‫‪37‬‬

‫‪11‬‬ ‫‪17‬‬ ‫‪20‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪35‬‬

‫‪10‬‬ ‫‪16‬‬ ‫‪19‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪32‬‬

‫‪9‬‬ ‫‪14‬‬ ‫‪17‬‬ ‫‪22‬‬ ‫‪26‬‬ ‫‪30‬‬

‫‪8‬‬ ‫‪13‬‬ ‫‪16‬‬ ‫‪20‬‬ ‫‪24‬‬ ‫‪28‬‬

‫‪7‬‬ ‫‪12‬‬ ‫‪14‬‬ ‫‪18‬‬ ‫‪22‬‬ ‫‪26‬‬

‫‪6‬‬ ‫‪11‬‬ ‫‪13‬‬ ‫‪17‬‬ ‫‪20‬‬ ‫‪24‬‬

‫‪5‬‬ ‫‪10‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪18‬‬ ‫‪22‬‬

‫‪5‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪14‬‬ ‫‪17‬‬ ‫‪20‬‬

‫‪4‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪18‬‬

‫‪3‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪11‬‬ ‫‪13‬‬ ‫‪16‬‬

‫‪2‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪11‬‬ ‫‪14‬‬

‫‪0‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬

‫‪0‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪10‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪6‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪4‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬

‫‪001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪17‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪35‬‬ ‫‪40‬‬ ‫‪47‬‬

‫‪16‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪33‬‬ ‫‪38‬‬ ‫‪44‬‬

‫‪15‬‬ ‫‪22‬‬ ‫‪25‬‬ ‫‪31‬‬ ‫‪36‬‬ ‫‪42‬‬

‫‪14‬‬ ‫‪20‬‬ ‫‪24‬‬ ‫‪29‬‬ ‫‪34‬‬ ‫‪39‬‬

‫‪12‬‬ ‫‪19‬‬ ‫‪22‬‬ ‫‪27‬‬ ‫‪31‬‬ ‫‪37‬‬

‫‪11‬‬ ‫‪17‬‬ ‫‪20‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪34‬‬

‫‪10‬‬ ‫‪16‬‬ ‫‪18‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪32‬‬

‫‪9‬‬ ‫‪14‬‬ ‫‪17‬‬ ‫‪21‬‬ ‫‪25‬‬ ‫‪29‬‬

‫‪8‬‬ ‫‪13‬‬ ‫‪15‬‬ ‫‪19‬‬ ‫‪22‬‬ ‫‪27‬‬

‫‪7‬‬ ‫‪11‬‬ ‫‪13‬‬ ‫‪17‬‬ ‫‪20‬‬ ‫‪24‬‬

‫‪6‬‬ ‫‪10‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪18‬‬ ‫‪22‬‬

‫‪4‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪13‬‬ ‫‪16‬‬ ‫‪19‬‬

‫‪3‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪11‬‬ ‫‪14‬‬ ‫‪17‬‬

‫‪2‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪14‬‬

‫‪1‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪9‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪7‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬

‫‪001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪22‬‬ ‫‪31‬‬ ‫‪35‬‬ ‫‪42‬‬ ‫‪48‬‬ ‫‪55‬‬

‫‪21‬‬ ‫‪29‬‬ ‫‪33‬‬ ‫‪39‬‬ ‫‪45‬‬ ‫‪52‬‬

‫‪19‬‬ ‫‪27‬‬ ‫‪31‬‬ ‫‪37‬‬ ‫‪42‬‬ ‫‪49‬‬

‫‪18‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪35‬‬ ‫‪40‬‬ ‫‪46‬‬

‫‪16‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪32‬‬ ‫‪37‬‬ ‫‪43‬‬

‫‪15‬‬ ‫‪21‬‬ ‫‪25‬‬ ‫‪30‬‬ ‫‪34‬‬ ‫‪40‬‬

‫‪13‬‬ ‫‪19‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪32‬‬ ‫‪37‬‬

‫‪12‬‬ ‫‪18‬‬ ‫‪21‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪34‬‬

‫‪10‬‬ ‫‪16‬‬ ‫‪18‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪31‬‬

‫‪9‬‬ ‫‪14‬‬ ‫‪16‬‬ ‫‪20‬‬ ‫‪24‬‬ ‫‪28‬‬

‫‪7‬‬ ‫‪12‬‬ ‫‪14‬‬ ‫‪18‬‬ ‫‪21‬‬ ‫‪25‬‬

‫‪6‬‬ ‫‪10‬‬ ‫‪12‬‬ ‫‪16‬‬ ‫‪19‬‬ ‫‪23‬‬

‫‪5‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪14‬‬ ‫‪16‬‬ ‫‪20‬‬

‫‪3‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪11‬‬ ‫‪14‬‬ ‫‪17‬‬

‫‪2‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪11‬‬ ‫‪14‬‬

‫‪1‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪11‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪8‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪6‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬

‫‪001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫اﻟﻤﺼﺪر ‪ :‬ﻋﻦ ])‪[Daniel (1978‬‬ ‫‪٦٣٥‬‬

‫‪n1‬‬

‫‪2‬‬

‫‪3‬‬

‫‪4‬‬

‫‪5‬‬

‫‪6‬‬

‫‪7‬‬

‫‪8‬‬

‫ﺗﺎﺑﻊ ﻣﻠﺣق )‪(١١‬‬ ‫ﺟدول اﻟﻘﯾم اﻟﺣرﺟﺔ ﻻﺧﺗﺑﺎر ‪Mann-Whitney-Wilcoxon‬‬ ‫‪20‬‬

‫‪19‬‬

‫‪18‬‬

‫‪17‬‬

‫‪16‬‬

‫‪15‬‬

‫‪14‬‬

‫‪13‬‬

‫‪12‬‬

‫‪11‬‬

‫‪10‬‬

‫‪9‬‬

‫‪8‬‬

‫‪7‬‬

‫‪6‬‬

‫‪5‬‬

‫‪4‬‬

‫‪3‬‬

‫‪n2=2‬‬

‫‪‬‬

‫‪27‬‬ ‫‪37‬‬ ‫‪41‬‬ ‫‪49‬‬ ‫‪55‬‬ ‫‪63‬‬

‫‪26‬‬ ‫‪34‬‬ ‫‪39‬‬ ‫‪46‬‬ ‫‪52‬‬ ‫‪59‬‬

‫‪24‬‬ ‫‪32‬‬ ‫‪37‬‬ ‫‪43‬‬ ‫‪49‬‬ ‫‪56‬‬

‫‪22‬‬ ‫‪30‬‬ ‫‪34‬‬ ‫‪40‬‬ ‫‪46‬‬ ‫‪53‬‬

‫‪20‬‬ ‫‪28‬‬ ‫‪32‬‬ ‫‪38‬‬ ‫‪43‬‬ ‫‪49‬‬

‫‪18‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪35‬‬ ‫‪40‬‬ ‫‪46‬‬

‫‪16‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪32‬‬ ‫‪37‬‬ ‫‪42‬‬

‫‪15‬‬ ‫‪21‬‬ ‫‪24‬‬ ‫‪29‬‬ ‫‪34‬‬ ‫‪39‬‬

‫‪13‬‬ ‫‪19‬‬ ‫‪22‬‬ ‫‪27‬‬ ‫‪31‬‬ ‫‪36‬‬

‫‪11‬‬ ‫‪17‬‬ ‫‪19‬‬ ‫‪24‬‬ ‫‪28‬‬ ‫‪32‬‬

‫‪9‬‬ ‫‪14‬‬ ‫‪17‬‬ ‫‪21‬‬ ‫‪25‬‬ ‫‪29‬‬

‫‪8‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪18‬‬ ‫‪22‬‬ ‫‪26‬‬

‫‪6‬‬ ‫‪10‬‬ ‫‪12‬‬ ‫‪16‬‬ ‫‪19‬‬ ‫‪23‬‬

‫‪4‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪13‬‬ ‫‪16‬‬ ‫‪19‬‬

‫‪3‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪11‬‬ ‫‪13‬‬ ‫‪16‬‬

‫‪2‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪13‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪10‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪6‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪33‬‬ ‫‪43‬‬ ‫‪48‬‬ ‫‪56‬‬ ‫‪63‬‬ ‫‪71‬‬

‫‪30‬‬ ‫‪40‬‬ ‫‪45‬‬ ‫‪53‬‬ ‫‪59‬‬ ‫‪67‬‬

‫‪28‬‬ ‫‪38‬‬ ‫‪42‬‬ ‫‪49‬‬ ‫‪56‬‬ ‫‪63‬‬

‫‪26‬‬ ‫‪35‬‬ ‫‪39‬‬ ‫‪46‬‬ ‫‪52‬‬ ‫‪59‬‬

‫‪24‬‬ ‫‪32‬‬ ‫‪37‬‬ ‫‪43‬‬ ‫‪49‬‬ ‫‪55‬‬

‫‪22‬‬ ‫‪30‬‬ ‫‪34‬‬ ‫‪40‬‬ ‫‪45‬‬ ‫‪52‬‬

‫‪20‬‬ ‫‪27‬‬ ‫‪31‬‬ ‫‪37‬‬ ‫‪42‬‬ ‫‪48‬‬

‫‪18‬‬ ‫‪25‬‬ ‫‪28‬‬ ‫‪34‬‬ ‫‪38‬‬ ‫‪44‬‬

‫‪15‬‬ ‫‪22‬‬ ‫‪25‬‬ ‫‪30‬‬ ‫‪35‬‬ ‫‪40‬‬

‫‪13‬‬ ‫‪19‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪32‬‬ ‫‪37‬‬

‫‪11‬‬ ‫‪17‬‬ ‫‪20‬‬ ‫‪24‬‬ ‫‪28‬‬ ‫‪33‬‬

‫‪9‬‬ ‫‪14‬‬ ‫‪17‬‬ ‫‪21‬‬ ‫‪25‬‬ ‫‪29‬‬

‫‪7‬‬ ‫‪12‬‬ ‫‪14‬‬ ‫‪18‬‬ ‫‪21‬‬ ‫‪25‬‬

‫‪6‬‬ ‫‪10‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪18‬‬ ‫‪22‬‬

‫‪4‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪18‬‬

‫‪2‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪14‬‬

‫‪1‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪11‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪7‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪38‬‬ ‫‪49‬‬ ‫‪54‬‬ ‫‪63‬‬ ‫‪70‬‬ ‫‪79‬‬

‫‪35‬‬ ‫‪46‬‬ ‫‪51‬‬ ‫‪59‬‬ ‫‪66‬‬ ‫‪74‬‬

‫‪33‬‬ ‫‪43‬‬ ‫‪48‬‬ ‫‪56‬‬ ‫‪62‬‬ ‫‪70‬‬

‫‪30‬‬ ‫‪40‬‬ ‫‪45‬‬ ‫‪52‬‬ ‫‪58‬‬ ‫‪66‬‬

‫‪28‬‬ ‫‪37‬‬ ‫‪42‬‬ ‫‪48‬‬ ‫‪55‬‬ ‫‪62‬‬

‫‪25‬‬ ‫‪34‬‬ ‫‪38‬‬ ‫‪45‬‬ ‫‪51‬‬ ‫‪58‬‬

‫‪23‬‬ ‫‪31‬‬ ‫‪35‬‬ ‫‪41‬‬ ‫‪47‬‬ ‫‪53‬‬

‫‪21‬‬ ‫‪28‬‬ ‫‪32‬‬ ‫‪38‬‬ ‫‪43‬‬ ‫‪49‬‬

‫‪18‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪34‬‬ ‫‪39‬‬ ‫‪45‬‬

‫‪16‬‬ ‫‪22‬‬ ‫‪26‬‬ ‫‪31‬‬ ‫‪35‬‬ ‫‪41‬‬

‫‪13‬‬ ‫‪19‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪32‬‬ ‫‪37‬‬

‫‪11‬‬ ‫‪17‬‬ ‫‪19‬‬ ‫‪24‬‬ ‫‪28‬‬ ‫‪32‬‬

‫‪9‬‬ ‫‪14‬‬ ‫‪16‬‬ ‫‪20‬‬ ‫‪24‬‬ ‫‪28‬‬

‫‪7‬‬ ‫‪11‬‬ ‫‪13‬‬ ‫‪17‬‬ ‫‪20‬‬ ‫‪24‬‬

‫‪5‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪14‬‬ ‫‪17‬‬ ‫‪20‬‬

‫‪3‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪13‬‬ ‫‪16‬‬

‫‪1‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪43‬‬ ‫‪55‬‬ ‫‪61‬‬ ‫‪70‬‬ ‫‪78‬‬ ‫‪87‬‬

‫‪41‬‬ ‫‪52‬‬ ‫‪57‬‬ ‫‪66‬‬ ‫‪73‬‬ ‫‪82‬‬

‫‪38‬‬ ‫‪48‬‬ ‫‪54‬‬ ‫‪62‬‬ ‫‪69‬‬ ‫‪78‬‬

‫‪35‬‬ ‫‪45‬‬ ‫‪50‬‬ ‫‪58‬‬ ‫‪65‬‬ ‫‪73‬‬

‫‪32‬‬ ‫‪42‬‬ ‫‪47‬‬ ‫‪54‬‬ ‫‪61‬‬ ‫‪68‬‬

‫‪29‬‬ ‫‪38‬‬ ‫‪43‬‬ ‫‪50‬‬ ‫‪56‬‬ ‫‪64‬‬

‫‪26‬‬ ‫‪35‬‬ ‫‪39‬‬ ‫‪46‬‬ ‫‪52‬‬ ‫‪59‬‬

‫‪24‬‬ ‫‪32‬‬ ‫‪36‬‬ ‫‪42‬‬ ‫‪48‬‬ ‫‪54‬‬

‫‪21‬‬ ‫‪28‬‬ ‫‪32‬‬ ‫‪38‬‬ ‫‪43‬‬ ‫‪50‬‬

‫‪18‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪34‬‬ ‫‪39‬‬ ‫‪45‬‬

‫‪15‬‬ ‫‪22‬‬ ‫‪25‬‬ ‫‪30‬‬ ‫‪35‬‬ ‫‪40‬‬

‫‪13‬‬ ‫‪19‬‬ ‫‪22‬‬ ‫‪27‬‬ ‫‪31‬‬ ‫‪36‬‬

‫‪10‬‬ ‫‪16‬‬ ‫‪18‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪31‬‬

‫‪8‬‬ ‫‪13‬‬ ‫‪15‬‬ ‫‪19‬‬ ‫‪22‬‬ ‫‪27‬‬

‫‪5‬‬ ‫‪10‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪18‬‬ ‫‪22‬‬

‫‪3‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪14‬‬ ‫‪18‬‬

‫‪1‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪13‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪9‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪49‬‬ ‫‪61‬‬ ‫‪68‬‬ ‫‪77‬‬ ‫‪85‬‬ ‫‪95‬‬

‫‪46‬‬ ‫‪58‬‬ ‫‪64‬‬ ‫‪73‬‬ ‫‪81‬‬ ‫‪90‬‬

‫‪43‬‬ ‫‪54‬‬ ‫‪60‬‬ ‫‪68‬‬ ‫‪76‬‬ ‫‪85‬‬

‫‪39‬‬ ‫‪50‬‬ ‫‪56‬‬ ‫‪64‬‬ ‫‪71‬‬ ‫‪80‬‬

‫‪36‬‬ ‫‪46‬‬ ‫‪52‬‬ ‫‪60‬‬ ‫‪66‬‬ ‫‪75‬‬

‫‪33‬‬ ‫‪43‬‬ ‫‪48‬‬ ‫‪55‬‬ ‫‪62‬‬ ‫‪69‬‬

‫‪30‬‬ ‫‪39‬‬ ‫‪44‬‬ ‫‪51‬‬ ‫‪57‬‬ ‫‪64‬‬

‫‪27‬‬ ‫‪35‬‬ ‫‪40‬‬ ‫‪46‬‬ ‫‪52‬‬ ‫‪59‬‬

‫‪24‬‬ ‫‪32‬‬ ‫‪36‬‬ ‫‪42‬‬ ‫‪48‬‬ ‫‪54‬‬

‫‪21‬‬ ‫‪28‬‬ ‫‪32‬‬ ‫‪38‬‬ ‫‪43‬‬ ‫‪49‬‬

‫‪18‬‬ ‫‪25‬‬ ‫‪28‬‬ ‫‪34‬‬ ‫‪38‬‬ ‫‪44‬‬

‫‪15‬‬ ‫‪21‬‬ ‫‪24‬‬ ‫‪29‬‬ ‫‪34‬‬ ‫‪39‬‬

‫‪12‬‬ ‫‪18‬‬ ‫‪21‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪34‬‬

‫‪9‬‬ ‫‪14‬‬ ‫‪17‬‬ ‫‪21‬‬ ‫‪25‬‬ ‫‪29‬‬

‫‪6‬‬ ‫‪11‬‬ ‫‪13‬‬ ‫‪17‬‬ ‫‪20‬‬ ‫‪24‬‬

‫‪4‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪13‬‬ ‫‪16‬‬ ‫‪19‬‬

‫‪2‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪9‬‬ ‫‪11‬‬ ‫‪14‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪10‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪55‬‬ ‫‪68‬‬ ‫‪74‬‬ ‫‪84‬‬ ‫‪93‬‬ ‫‪103‬‬

‫‪51‬‬ ‫‪64‬‬ ‫‪70‬‬ ‫‪79‬‬ ‫‪88‬‬ ‫‪98‬‬

‫‪47‬‬ ‫‪59‬‬ ‫‪66‬‬ ‫‪75‬‬ ‫‪83‬‬ ‫‪92‬‬

‫‪44‬‬ ‫‪55‬‬ ‫‪61‬‬ ‫‪70‬‬ ‫‪78‬‬ ‫‪86‬‬

‫‪40‬‬ ‫‪51‬‬ ‫‪57‬‬ ‫‪65‬‬ ‫‪72‬‬ ‫‪81‬‬

‫‪37‬‬ ‫‪47‬‬ ‫‪52‬‬ ‫‪60‬‬ ‫‪67‬‬ ‫‪75‬‬

‫‪33‬‬ ‫‪43‬‬ ‫‪48‬‬ ‫‪56‬‬ ‫‪62‬‬ ‫‪70‬‬

‫‪30‬‬ ‫‪39‬‬ ‫‪44‬‬ ‫‪51‬‬ ‫‪57‬‬ ‫‪64‬‬

‫‪26‬‬ ‫‪35‬‬ ‫‪39‬‬ ‫‪46‬‬ ‫‪52‬‬ ‫‪59‬‬

‫‪23‬‬ ‫‪31‬‬ ‫‪35‬‬ ‫‪41‬‬ ‫‪47‬‬ ‫‪53‬‬

‫‪20‬‬ ‫‪27‬‬ ‫‪31‬‬ ‫‪37‬‬ ‫‪42‬‬ ‫‪48‬‬

‫‪16‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪32‬‬ ‫‪37‬‬ ‫‪42‬‬

‫‪13‬‬ ‫‪19‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪32‬‬ ‫‪37‬‬

‫‪10‬‬ ‫‪16‬‬ ‫‪18‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪32‬‬

‫‪7‬‬ ‫‪12‬‬ ‫‪14‬‬ ‫‪18‬‬ ‫‪22‬‬ ‫‪26‬‬

‫‪4‬‬ ‫‪8‬‬ ‫‪11‬‬ ‫‪14‬‬ ‫‪17‬‬ ‫‪21‬‬

‫‪2‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪10‬‬ ‫‪12‬‬ ‫‪16‬‬

‫‪0‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪11‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪5‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪60‬‬ ‫‪74‬‬ ‫‪81‬‬ ‫‪91‬‬ ‫‪101‬‬ ‫‪111‬‬

‫‪56‬‬ ‫‪70‬‬ ‫‪76‬‬ ‫‪86‬‬ ‫‪95‬‬ ‫‪105‬‬

‫‪52‬‬ ‫‪65‬‬ ‫‪71‬‬ ‫‪81‬‬ ‫‪89‬‬ ‫‪99‬‬

‫‪48‬‬ ‫‪61‬‬ ‫‪67‬‬ ‫‪76‬‬ ‫‪84‬‬ ‫‪93‬‬

‫‪44‬‬ ‫‪56‬‬ ‫‪62‬‬ ‫‪71‬‬ ‫‪78‬‬ ‫‪87‬‬

‫‪41‬‬ ‫‪52‬‬ ‫‪57‬‬ ‫‪65‬‬ ‫‪73‬‬ ‫‪81‬‬

‫‪37‬‬ ‫‪47‬‬ ‫‪52‬‬ ‫‪60‬‬ ‫‪67‬‬ ‫‪75‬‬

‫‪33‬‬ ‫‪43‬‬ ‫‪48‬‬ ‫‪55‬‬ ‫‪62‬‬ ‫‪69‬‬

‫‪29‬‬ ‫‪38‬‬ ‫‪43‬‬ ‫‪50‬‬ ‫‪56‬‬ ‫‪64‬‬

‫‪25‬‬ ‫‪34‬‬ ‫‪38‬‬ ‫‪45‬‬ ‫‪51‬‬ ‫‪58‬‬

‫‪22‬‬ ‫‪30‬‬ ‫‪34‬‬ ‫‪40‬‬ ‫‪45‬‬ ‫‪52‬‬

‫‪18‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪35‬‬ ‫‪40‬‬ ‫‪46‬‬

‫‪15‬‬ ‫‪21‬‬ ‫‪25‬‬ ‫‪30‬‬ ‫‪34‬‬ ‫‪40‬‬

‫‪11‬‬ ‫‪17‬‬ ‫‪20‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪34‬‬

‫‪8‬‬ ‫‪13‬‬ ‫‪16‬‬ ‫‪20‬‬ ‫‪24‬‬ ‫‪28‬‬

‫‪5‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪19‬‬ ‫‪23‬‬

‫‪2‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪11‬‬ ‫‪13‬‬ ‫‪17‬‬

‫‪0‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪11‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪6‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪66‬‬ ‫‪80‬‬ ‫‪88‬‬ ‫‪99‬‬ ‫‪108‬‬ ‫‪120‬‬

‫‪61‬‬ ‫‪75‬‬ ‫‪83‬‬ ‫‪93‬‬ ‫‪102‬‬ ‫‪113‬‬

‫‪57‬‬ ‫‪71‬‬ ‫‪77‬‬ ‫‪87‬‬ ‫‪96‬‬ ‫‪107‬‬

‫‪53‬‬ ‫‪66‬‬ ‫‪72‬‬ ‫‪82‬‬ ‫‪90‬‬ ‫‪100‬‬

‫‪49‬‬ ‫‪61‬‬ ‫‪67‬‬ ‫‪76‬‬ ‫‪84‬‬ ‫‪94‬‬

‫‪44‬‬ ‫‪56‬‬ ‫‪62‬‬ ‫‪71‬‬ ‫‪78‬‬ ‫‪87‬‬

‫‪40‬‬ ‫‪51‬‬ ‫‪57‬‬ ‫‪65‬‬ ‫‪72‬‬ ‫‪81‬‬

‫‪36‬‬ ‫‪46‬‬ ‫‪52‬‬ ‫‪60‬‬ ‫‪66‬‬ ‫‪75‬‬

‫‪32‬‬ ‫‪42‬‬ ‫‪47‬‬ ‫‪54‬‬ ‫‪61‬‬ ‫‪68‬‬

‫‪28‬‬ ‫‪37‬‬ ‫‪42‬‬ ‫‪48‬‬ ‫‪55‬‬ ‫‪62‬‬

‫‪24‬‬ ‫‪32‬‬ ‫‪37‬‬ ‫‪43‬‬ ‫‪49‬‬ ‫‪55‬‬

‫‪20‬‬ ‫‪28‬‬ ‫‪32‬‬ ‫‪38‬‬ ‫‪43‬‬ ‫‪49‬‬

‫‪16‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪32‬‬ ‫‪37‬‬ ‫‪43‬‬

‫‪12‬‬ ‫‪19‬‬ ‫‪22‬‬ ‫‪27‬‬ ‫‪31‬‬ ‫‪37‬‬

‫‪9‬‬ ‫‪14‬‬ ‫‪17‬‬ ‫‪22‬‬ ‫‪26‬‬ ‫‪30‬‬

‫‪6‬‬ ‫‪10‬‬ ‫‪13‬‬ ‫‪16‬‬ ‫‪20‬‬ ‫‪24‬‬

‫‪3‬‬ ‫‪6‬‬ ‫‪8‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪18‬‬

‫‪0‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪6‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪٦٣٦‬‬

‫‪n1‬‬

‫‪9‬‬

‫‪10‬‬

‫‪11‬‬

‫‪12‬‬

‫‪13‬‬

‫‪14‬‬

‫‪15‬‬

‫‪16‬‬

‫ﺗﺎﺑﻊ ﻣﻠﺤﻖ )‪(١١‬‬ ‫ﺟدول اﻟﻘﯾم اﻟﺣرﺟﺔ ﻻﺧﺗﺑﺎر‬ ‫‪Mann-Whitney-Wilcoxon‬‬ ‫‪20‬‬

‫‪19‬‬

‫‪71‬‬ ‫‪87‬‬ ‫‪94‬‬ ‫‪106‬‬ ‫‪116‬‬ ‫‪128‬‬

‫‪67‬‬ ‫‪82‬‬ ‫‪89‬‬ ‫‪100‬‬ ‫‪110‬‬ ‫‪121‬‬

‫‪53 58 62‬‬ ‫‪66 71 76‬‬ ‫‪72 78 83‬‬ ‫‪82 88 94‬‬ ‫‪90 97 103‬‬ ‫‪100 107 114‬‬

‫‪77‬‬ ‫‪93‬‬ ‫‪101‬‬ ‫‪113‬‬ ‫‪124‬‬ ‫‪136‬‬

‫‪72‬‬ ‫‪88‬‬ ‫‪95‬‬ ‫‪107‬‬ ‫‪117‬‬ ‫‪129‬‬

‫‪57 62 67‬‬ ‫‪71 76 82‬‬ ‫‪77 83 89‬‬ ‫‪87 94 100‬‬ ‫‪96 103 110‬‬ ‫‪107 114 121‬‬

‫‪83‬‬ ‫‪100‬‬ ‫‪108‬‬ ‫‪120‬‬ ‫‪131‬‬ ‫‪144‬‬

‫‪67 72 78‬‬ ‫‪82 88 94‬‬ ‫‪89 95 102‬‬ ‫‪100 107 114‬‬ ‫‪110 117 124‬‬ ‫‪121 129 136‬‬

‫‪56 61‬‬ ‫‪70 75‬‬ ‫‪76 83‬‬ ‫‪86 93‬‬ ‫‪95 102‬‬ ‫‪105 113‬‬

‫‪89‬‬ ‫‪106‬‬ ‫‪115‬‬ ‫‪128‬‬ ‫‪139‬‬ ‫‪152‬‬

‫‪71‬‬ ‫‪87‬‬ ‫‪94‬‬ ‫‪106‬‬ ‫‪116‬‬ ‫‪128‬‬

‫‪60 66‬‬ ‫‪74 80‬‬ ‫‪81 88‬‬ ‫‪91 99‬‬ ‫‪101 108‬‬ ‫‪111 120‬‬

‫‪83‬‬ ‫‪100‬‬ ‫‪108‬‬ ‫‪120‬‬ ‫‪131‬‬ ‫‪144‬‬

‫‪18‬‬

‫‪77‬‬ ‫‪93‬‬ ‫‪101‬‬ ‫‪113‬‬ ‫‪124‬‬ ‫‪136‬‬

‫‪17‬‬

‫‪16‬‬

‫‪15‬‬

‫‪14‬‬

‫‪13‬‬

‫‪12‬‬

‫‪11‬‬

‫‪10‬‬

‫‪9‬‬

‫‪8‬‬

‫‪7‬‬

‫‪6‬‬

‫‪5‬‬

‫‪4‬‬

‫‪3‬‬

‫‪n2=2‬‬

‫‪P‬‬

‫‪48‬‬ ‫‪61‬‬ ‫‪67‬‬ ‫‪76‬‬ ‫‪84‬‬ ‫‪93‬‬

‫‪44‬‬ ‫‪55‬‬ ‫‪61‬‬ ‫‪70‬‬ ‫‪78‬‬ ‫‪86‬‬

‫‪39‬‬ ‫‪50‬‬ ‫‪56‬‬ ‫‪64‬‬ ‫‪71‬‬ ‫‪80‬‬

‫‪35‬‬ ‫‪45‬‬ ‫‪50‬‬ ‫‪58‬‬ ‫‪65‬‬ ‫‪73‬‬

‫‪30‬‬ ‫‪40‬‬ ‫‪45‬‬ ‫‪52‬‬ ‫‪58‬‬ ‫‪66‬‬

‫‪26‬‬ ‫‪35‬‬ ‫‪39‬‬ ‫‪46‬‬ ‫‪52‬‬ ‫‪59‬‬

‫‪22‬‬ ‫‪30‬‬ ‫‪34‬‬ ‫‪40‬‬ ‫‪46‬‬ ‫‪53‬‬

‫‪18‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪35‬‬ ‫‪40‬‬ ‫‪46‬‬

‫‪14‬‬ ‫‪20‬‬ ‫‪24‬‬ ‫‪29‬‬ ‫‪34‬‬ ‫‪39‬‬

‫‪10‬‬ ‫‪16‬‬ ‫‪19‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪32‬‬

‫‪6‬‬ ‫‪11‬‬ ‫‪14‬‬ ‫‪18‬‬ ‫‪21‬‬ ‫‪26‬‬

‫‪3‬‬ ‫‪7‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪16‬‬ ‫‪19‬‬

‫‪1‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪10‬‬ ‫‪13‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪7‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪52‬‬ ‫‪65‬‬ ‫‪71‬‬ ‫‪81‬‬ ‫‪89‬‬ ‫‪99‬‬

‫‪47‬‬ ‫‪59‬‬ ‫‪66‬‬ ‫‪75‬‬ ‫‪83‬‬ ‫‪92‬‬

‫‪43‬‬ ‫‪54‬‬ ‫‪60‬‬ ‫‪68‬‬ ‫‪76‬‬ ‫‪85‬‬

‫‪38‬‬ ‫‪48‬‬ ‫‪54‬‬ ‫‪62‬‬ ‫‪69‬‬ ‫‪78‬‬

‫‪33‬‬ ‫‪43‬‬ ‫‪48‬‬ ‫‪56‬‬ ‫‪62‬‬ ‫‪70‬‬

‫‪28‬‬ ‫‪38‬‬ ‫‪42‬‬ ‫‪49‬‬ ‫‪56‬‬ ‫‪63‬‬

‫‪24‬‬ ‫‪32‬‬ ‫‪37‬‬ ‫‪43‬‬ ‫‪49‬‬ ‫‪56‬‬

‫‪19‬‬ ‫‪27‬‬ ‫‪31‬‬ ‫‪37‬‬ ‫‪42‬‬ ‫‪49‬‬

‫‪15‬‬ ‫‪22‬‬ ‫‪25‬‬ ‫‪31‬‬ ‫‪36‬‬ ‫‪42‬‬

‫‪11‬‬ ‫‪17‬‬ ‫‪20‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪35‬‬

‫‪7‬‬ ‫‪12‬‬ ‫‪15‬‬ ‫‪19‬‬ ‫‪23‬‬ ‫‪28‬‬

‫‪4‬‬ ‫‪7‬‬ ‫‪10‬‬ ‫‪13‬‬ ‫‪17‬‬ ‫‪21‬‬

‫‪1‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪14‬‬

‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪7‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪51‬‬ ‫‪64‬‬ ‫‪70‬‬ ‫‪79‬‬ ‫‪88‬‬ ‫‪98‬‬

‫‪46‬‬ ‫‪58‬‬ ‫‪64‬‬ ‫‪73‬‬ ‫‪81‬‬ ‫‪90‬‬

‫‪41‬‬ ‫‪52‬‬ ‫‪57‬‬ ‫‪66‬‬ ‫‪73‬‬ ‫‪82‬‬

‫‪35‬‬ ‫‪46‬‬ ‫‪51‬‬ ‫‪59‬‬ ‫‪66‬‬ ‫‪74‬‬

‫‪30‬‬ ‫‪40‬‬ ‫‪45‬‬ ‫‪53‬‬ ‫‪59‬‬ ‫‪67‬‬

‫‪26‬‬ ‫‪34‬‬ ‫‪39‬‬ ‫‪46‬‬ ‫‪52‬‬ ‫‪59‬‬

‫‪21‬‬ ‫‪29‬‬ ‫‪33‬‬ ‫‪39‬‬ ‫‪45‬‬ ‫‪52‬‬

‫‪16‬‬ ‫‪23‬‬ ‫‪27‬‬ ‫‪33‬‬ ‫‪38‬‬ ‫‪44‬‬

‫‪12‬‬ ‫‪18‬‬ ‫‪21‬‬ ‫‪26‬‬ ‫‪31‬‬ ‫‪37‬‬

‫‪8‬‬ ‫‪13‬‬ ‫‪16‬‬ ‫‪20‬‬ ‫‪24‬‬ ‫‪29‬‬

‫‪4‬‬ ‫‪8‬‬ ‫‪10‬‬ ‫‪14‬‬ ‫‪18‬‬ ‫‪22‬‬

‫‪1‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪8‬‬ ‫‪11‬‬ ‫‪15‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪8‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪55‬‬ ‫‪68‬‬ ‫‪74‬‬ ‫‪84‬‬ ‫‪93‬‬ ‫‪103‬‬

‫‪49‬‬ ‫‪61‬‬ ‫‪68‬‬ ‫‪77‬‬ ‫‪85‬‬ ‫‪95‬‬

‫‪43‬‬ ‫‪55‬‬ ‫‪61‬‬ ‫‪70‬‬ ‫‪78‬‬ ‫‪87‬‬

‫‪38‬‬ ‫‪49‬‬ ‫‪54‬‬ ‫‪63‬‬ ‫‪70‬‬ ‫‪79‬‬

‫‪33‬‬ ‫‪43‬‬ ‫‪48‬‬ ‫‪56‬‬ ‫‪63‬‬ ‫‪71‬‬

‫‪27‬‬ ‫‪37‬‬ ‫‪41‬‬ ‫‪49‬‬ ‫‪55‬‬ ‫‪63‬‬

‫‪22‬‬ ‫‪31‬‬ ‫‪35‬‬ ‫‪42‬‬ ‫‪48‬‬ ‫‪55‬‬

‫‪17‬‬ ‫‪25‬‬ ‫‪29‬‬ ‫‪35‬‬ ‫‪40‬‬ ‫‪47‬‬

‫‪13‬‬ ‫‪19‬‬ ‫‪23‬‬ ‫‪28‬‬ ‫‪33‬‬ ‫‪39‬‬

‫‪8‬‬ ‫‪14‬‬ ‫‪17‬‬ ‫‪21‬‬ ‫‪26‬‬ ‫‪31‬‬

‫‪4‬‬ ‫‪9‬‬ ‫‪11‬‬ ‫‪15‬‬ ‫‪19‬‬ ‫‪23‬‬

‫‪1‬‬ ‫‪4‬‬ ‫‪6‬‬ ‫‪9‬‬ ‫‪12‬‬ ‫‪16‬‬

‫‪0‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪8‬‬

‫‪.001‬‬ ‫‪.005‬‬ ‫‪.01‬‬ ‫‪.025‬‬ ‫‪.05‬‬ ‫‪.10‬‬

‫‪٦٣٧‬‬

‫‪n1‬‬

‫‪17‬‬

‫‪18‬‬

‫‪19‬‬

‫‪20‬‬

(١٢) ‫ﻣﻠﺤﻖ‬ Kruskal – Wallis‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ﻻﺧﺘﺒﺎر‬ Sample Sizes

Sample Sizes n1

2 2 2

1 2 2

1 1 2

3 3

1 2

1 1

4 4

1 2

1 1

Critical value 2.7000 3.6000 4.5714 3.7143 3.2000 4.2857 3.8571 5.3572 4.7143 4.5000 4.4643 5.1429 4.5714 4.000 6.2500 5.3611 5.1389 4.5556 4.2500 7.2000 6.4889 5.6889 5.6000 5.0667 4.6222 3.5714 4.8214 4.5000 4.0179 6.0000 5.3333 5.1250 4.4583 4.1667 5.8333 5.2083

 0.500 0.200 0.067 0.200 0.300 0.100 0.133 0.029 0.048 0.067 0.105 0.043 0.100 0.129 0.011 0.032 0.061 0.100 0.121 0.004 0.011 0.029 0.050 0.086 0.100 0.200 0.057 0.076 0.114 0.014 0.033 0.052 0.100 0.105 0.021 0.050

5 5

1 2

1 1

٦٣٨

Critical value 4.7000 6.6667 6.1667 4.9667 4.8667 4.1667 4.0667 7.0364 6.8727 5.4545 5.2364 4.5545 4.4455 7.1439 7.1364 5.5985 5.5758 4.5455 4.4773 7.6538 7.5385 5.6923 5.6538 4.6539 4.5001 3.8571 5.2500 5.0000 4.4500 4.2000 4.0500 6.5333 6.1333 5.1600 5.0400 4.3733

 0.101 0.010 0.022 0.048 0.054 0.082 0.102 0.006 0.011 0.046 0.052 0.098 0.103 0.010 0.011 0.049 0.051 0.099 0.102 0.008 0.011 0.049 0.054 0.097 0.104 0.143 0.036 0.048 0.071 0.095 0.119 0.008 0.013 0.034 0.056 0.090

5.0000 4.0556 4.8889 6.4444 6.3000 5.4444 5.4000 4.5111 4.4444 6.7455 6.7091 5.7909 5.7273 4.7091

0.057 0.093 0.129 0.008 0.011 0.046 0.051 0.098 0.102 0.010 0.013 0.046 0.050 0.092

٦٣٩

4.2933 6.4000 4.9600 4.8711 4.0178 3.8400 6.9091 6.8218 5.2509 5.1055 4.6509 4.4945 7.0788 6.9818

0.122 0.012 0.048 0.052 0.095 0.123 0.009 0.010 0.049 0.052 0.091 0.101 0.009 0.011

(١٢) ‫ ﻣﻠﺤﻖ‬: ‫ﺗﺎﺑﻊ‬ Kruskal – Wallis‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ ﻻﺧﺘﺒﺎر‬

Sample Sizes

Sample Sizes n1

Critical value 5.6485 5.5152 4.5333 4.4121 6.9545 6.8400 4.9855 4.8600 3.9873 3.9600 7.2045 7.1182 5.2727 5.2682 4.5409 4.5182 7.4449 7.3949 5.6564 5.6308 4.5487 4.5231 7.7604 7.7440 5.6571 5.6176 4.6187 4.5527 7.3091



0.049 0.051 0.097 0.109 0.008 0.011 0.044 0.056 0.098 0.102 0.009 0.010 0.049 0.050 0.098 0.101 0.010 0.011 0.049 0.050 0.099 0.103 0.009 0.011 0.049 0.050 0.100 0.102 0.009

Critical value 6.8364 5.1273 4.9091 4.1091 4.0364 7.3385 7.2692 5. 3385 5.2462 4.6231 4.5077 7.5780 7.5429 5.7055 5.6264 4.5451 4.5363 7.8229 7.7914 5.6657 5.6429 4.5229 4.5200 8.000 7.9800 5.7800 5.6600 4.5600 4.5000

 0.011 0.046 0.053 0.086 0.105 0.010 0.010 0.047 0.051 0.097 0.100 0.010 0.010 0.046 0.051 0.100 0.102 0.010 0.010 0.049 0.050 0.099 0.101 0.009 0.010 0.049 0.051 0.100 0.102

[Daniel (1978)]‫ ﻋﻦ‬: ‫اﻟﻤﺼﺪر‬

٦٤٠

‫ﻣﻠﺤﻖ )‪(١٣‬‬ ‫ﺟﺪول اﻟﻘﯿﻢ اﻟﺤﺮﺟﺔ‬ ‫‪.100‬‬ ‫‪.8000‬‬ ‫‪.7000‬‬ ‫‪.6000‬‬ ‫‪.5357‬‬ ‫‪.5000‬‬ ‫‪.4667‬‬ ‫‪.4424‬‬ ‫‪.4182‬‬ ‫‪.3986‬‬ ‫‪. 3791‬‬ ‫‪.3626‬‬ ‫‪.3500‬‬ ‫‪.3382‬‬ ‫‪.3260‬‬ ‫‪.3148‬‬ ‫‪.3070‬‬ ‫‪.2977‬‬ ‫‪.2909‬‬ ‫‪.2829‬‬ ‫‪.2767‬‬ ‫‪.2704‬‬ ‫‪.2646‬‬ ‫‪.2588‬‬ ‫‪.2540‬‬ ‫‪.2490‬‬ ‫‪.2443‬‬ ‫‪.2400‬‬

‫‪.050‬‬ ‫‪.8000‬‬ ‫‪.8000‬‬ ‫‪.7714‬‬ ‫‪.6786‬‬ ‫‪.6190‬‬ ‫‪.5833‬‬ ‫‪.5515‬‬ ‫‪.5273‬‬ ‫‪.4965‬‬ ‫‪.4780‬‬ ‫‪.4593‬‬ ‫‪.4429‬‬ ‫‪.2465‬‬ ‫‪.4118‬‬ ‫‪.3994‬‬ ‫‪.3895‬‬ ‫‪.3789‬‬ ‫‪.3688‬‬ ‫‪.3597‬‬ ‫‪.3518‬‬ ‫‪.3435‬‬ ‫‪.3362‬‬ ‫‪.3299‬‬ ‫‪.3236‬‬ ‫‪.3175‬‬ ‫‪.3113‬‬ ‫‪.3059‬‬

‫‪.025‬‬ ‫‪-‬‬‫‪.9000‬‬ ‫‪.8286‬‬ ‫‪.7450‬‬ ‫‪.7143‬‬ ‫‪.6833‬‬ ‫‪.6364‬‬ ‫‪.6091‬‬ ‫‪.5804‬‬ ‫‪.5549‬‬ ‫‪.5341‬‬ ‫‪.5179‬‬ ‫‪.5000‬‬ ‫‪.4853‬‬ ‫‪.4716‬‬ ‫‪.4579‬‬ ‫‪.4451‬‬ ‫‪.4351‬‬ ‫‪.4241‬‬ ‫‪.4150‬‬ ‫‪.4061‬‬ ‫‪.3977‬‬ ‫‪.3894‬‬ ‫‪.3822‬‬ ‫‪.3749‬‬ ‫‪.3685‬‬ ‫‪.3620‬‬

‫*‬ ‫‪rs, ‬‬

‫ﻻﺧﺘﺒﺎر ﺳﺒﯿﺮﻣﺎن‬

‫‪.010‬‬ ‫‪-‬‬‫‪.9000‬‬ ‫‪.8857‬‬ ‫‪.8571‬‬ ‫‪.8095‬‬ ‫‪.7667‬‬ ‫‪.7333‬‬ ‫‪.7000‬‬ ‫‪.6713‬‬ ‫‪.6429‬‬ ‫‪.6220‬‬ ‫‪.6000‬‬ ‫‪.5824‬‬ ‫‪.5637‬‬ ‫‪.5480‬‬ ‫‪.5333‬‬ ‫‪.5203‬‬ ‫‪.5078‬‬ ‫‪.4963‬‬ ‫‪.4852‬‬ ‫‪.4748‬‬ ‫‪.4654‬‬ ‫‪.4564‬‬ ‫‪.4481‬‬ ‫‪.4401‬‬ ‫‪.4320‬‬ ‫‪.4251‬‬

‫اﻟﻤﺼﺪر ‪ :‬ﻋﻦ ])‪[Daniel (1978‬‬

‫‪٦٤١‬‬

‫‪.005‬‬ ‫‪-‬‬‫‪-‬‬‫‪.9429‬‬ ‫‪.8929‬‬ ‫‪.8571‬‬ ‫‪.8167‬‬ ‫‪.7818‬‬ ‫‪.7545‬‬ ‫‪.7273‬‬ ‫‪.6978‬‬ ‫‪.6747‬‬ ‫‪.6536‬‬ ‫‪.6324‬‬ ‫‪.6152‬‬ ‫‪.5975‬‬ ‫‪.5825‬‬ ‫‪.5684‬‬ ‫‪.5545‬‬ ‫‪.5426‬‬ ‫‪.5306‬‬ ‫‪.5200‬‬ ‫‪.5100‬‬ ‫‪.5002‬‬ ‫‪.4915‬‬ ‫‪.4828‬‬ ‫‪.4744‬‬ ‫‪.4665‬‬

‫‪.001‬‬ ‫‪-‬‬‫‪-‬‬‫‪-‬‬‫‪.9643‬‬ ‫‪.9286‬‬ ‫‪.9000‬‬ ‫‪.8667‬‬ ‫‪.8364‬‬ ‫‪.8182‬‬ ‫‪.7912‬‬ ‫‪.7670‬‬ ‫‪.7464‬‬ ‫‪.7265‬‬ ‫‪.7083‬‬ ‫‪.6904‬‬ ‫‪.6737‬‬ ‫‪.6586‬‬ ‫‪.6455‬‬ ‫‪.6318‬‬ ‫‪.6186‬‬ ‫‪.6070‬‬ ‫‪.5962‬‬ ‫‪.5856‬‬ ‫‪.5757‬‬ ‫‪.5660‬‬ ‫‪.5567‬‬ ‫‪.5479‬‬

‫‪n‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫‪7‬‬ ‫‪8‬‬ ‫‪9‬‬ ‫‪10‬‬ ‫‪11‬‬ ‫‪12‬‬ ‫‪13‬‬ ‫‪14‬‬ ‫‪15‬‬ ‫‪16‬‬ ‫‪17‬‬ ‫‪18‬‬ ‫‪19‬‬ ‫‪20‬‬ ‫‪21‬‬ ‫‪22‬‬ ‫‪23‬‬ ‫‪24‬‬ ‫‪25‬‬ ‫‪26‬‬ ‫‪27‬‬ ‫‪28‬‬ ‫‪29‬‬ ‫‪30‬‬

٦٤٢