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About Help with Statistics Assignment: We have excellent statistics assignment tutors to help you in the field of econometrics, applied statistics, quantitative methods, mathematical statistics, business statistics and operations research. We follow a strategic approach in solving and presenting the assignment. With the comprehensive use of tools, graphs, histograms, pie charts, diagrams, tables you can be rest assured of expecting a higher grade in your assignments. As far as software based assignments are concerned we provide analysis and explanation in a word file along with necessary code file and output file. We also assist students in their online statistics quizzes, exams and homework. Sample Statistics Assignment Questions and Answer: Depreciation Sample Question Question-1. Taking the deviations of the time variable, compute the trend values for the following data by the method of the least square : Days : Sales (in $) :
1 20
2 30
3 40
4 20
5 50
6 60
7 80
Solution. Computation of the Trend Values Taking the deviations of the time variable by the method of the least Days
Sales Y
Time dvn. From mid value 4
XY
X2
1 2 2 4 5 6 7
20 30 40 20 50 60 80
-3 -2 -1 0 1 2 3
-60 -60 -40 0 50 120 240
9 4 1 0 1 4 9
Trend values =42.86 + 8.93X 16.7 25.00 33.93 42.86 51.79 60.72 69.65
Total
300
0
140
N=7
0.00
The trend values of Y are given by Yc = a + bX Where, a = =
đ?’€ đ?‘ľ đ?&#x;‘đ?&#x;Žđ?&#x;Ž đ?&#x;•
[âˆľ
đ?‘‹đ?‘Œ = a đ?‘‹ + b đ?‘‹ 2 and
đ?‘‹ = 0]
= 42.86 approx. Copyright Š 2010-2015 Tutorhelpdesk.com
And
b= =
đ?‘‹đ?‘Œ đ?‘‹2 250 8
[âˆľ
đ?‘‹đ?‘Œ = a đ?‘‹ + b đ?‘‹ 2 and
đ?‘‹ = 0]
= 8.93 approx.
Putting the values of a and b in the above, we get the required trend line equation as: Yc = 42.86 + 8.93X Where, Yc represents the computed trend value of Y, and X the deviation of the time variable. Using the above trend equation, the various trend values will be computed as under: Computation of the Trend Values When X = -3, Yc = 42.86 + 8.93 (-3) = 16.07 When X = -2, Yc = 42.86 + 8.93 (-2) = 25.00 When X = -1, Yc = 42.86 + 8.93 (-1) = 33.93 When X = 0, Yc = 42.86 + 8.93 (0) = 42.86 When X = 1, Yc = 42.86 + 8.93 (1) = 51.79 When X = 2, Yc = 42.86 + 8.93 (2) = 60.72 When X = 3, Yc = 42.86 + 8.93 (3) = 69.65 Aliter The above trend values could be obtained by simply adding 8.93 (the value of b i.e. rate of change of the slope) successively to 42.86 (the value of the trend origin at t = 4) for each time period succeeding the time of the origin, and by deducting 8.93 successively from 42.86 for each item period preceding the time of the origin as follow: When X is at the origin 4, Yc = 42.86 When X is at 5, Yc = 42.86 + 8.93 = 51.79 When X is at 6, Yc = 51.79 + 8.93 = 60.72 When X is at 7, Yc = 60.72 + 8.93 = 69.65 When X is at 3, Yc = 42.86 – 8.93 = 33.93 When X is at 2, Yc = 33.93 – 8.93 = 25.00 When X is at 1, Yc = 25 – 8.93 = 16.07 From the above it must be seen that the trend values, thus obtained on the basis of the time deviations, are the same as they were obtained on the basis of the original data in the illustration 8 before, where the values of the constants a and b were determined through the lengthy procedure of simultaneous equations. Copyright Š 2010-2015 Tutorhelpdesk.com
Question-2. Find the trend line equation and obtain the trend values for the following data using the method of the least square. Also, forecast the earning for 2006. Year : Earning in ’000 $ :
1997 38
1998 40
1999 65
2000 72
2001 69
2002 60
2003 87
2004 95
Solution. Here, the number of items being 8 (i.e. even), the time deviation X will be taken as đ?’•âˆ’đ?’Žđ?’Šđ?’… đ?’‘đ?’?đ?’Šđ?’?đ?’• đ?’?đ?’‡ đ?’•đ?’Šđ?’Žđ?’† đ?&#x;? đ?’?đ?’‡ đ?’•đ?’Šđ?’Žđ?’† đ?’Šđ?’?đ?’•đ?’†đ?’“đ?’—đ?’‚đ?’? đ?&#x;?
to avoide the decimal numbers Thus, the working will run as under:
(a) Determination of the Trend Line Equation and the Trend Values
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Year t
Earnings Y
Time dvn. i.e.
XY
X2
Trend values Yc=65.75 + 3.67X
266 200 195 -72 0 69 180 435 665 616
49 25 9 1 0 1 9 25 49
40.06 47.40 54.74 62.08 A = 65.75 69.42 76.76 84.10 91.44
168
N=8
đ?’•âˆ’đ?&#x;?đ?&#x;Žđ?&#x;Žđ?&#x;Ž.đ?&#x;“ đ?&#x;?/đ?&#x;? Ă—đ?&#x;?
1997 1998 1999 2000 2000.5 (mid time)
38 40 65 72 69 60 87 95
X -7 -5 -3 -1 0 1 3 5 7
Total
526
0
Working The trend line equation is given by Y = a + bX Where, a = = And
đ?&#x;“đ?&#x;?đ?&#x;” đ?&#x;–
b= =
đ?&#x;”đ?&#x;?đ?&#x;” đ?&#x;?đ?&#x;”đ?&#x;–
đ?’€ đ?‘ľ
[âˆľ
đ?‘‹đ?‘Œ = Na + b đ?‘‹, and
đ?‘‹=0]
= 65.75 đ?‘żđ?’€ đ?‘żđ?&#x;?
[âˆľ
đ?‘‹đ?‘Œ = a đ?‘‹ + b đ?‘‹ 2 , and
đ?‘‹=0]
= 3.67 approx.
Putting the above values of a and b in the equation we get the required trend line equation as Yc = 65.75 + 3.67 X Where, trend origin is 2000.5, Y unit = annual earning, and X unit = time deviation Putting the respective values of X in the above equation, we get the different trend values as under: Copyright Š 2010-2015 Tutorhelpdesk.com
Trend Values For 1997 When X = -7, Yc = 65.75 + 3.67 (-7) = 40.06 1998 When X = -5, Yc = 65.75 + 3.67 (-5) = 47.40 1999 When X =-3, Yc = 65.75 + 3.67 (-3) = 54.74 2000 When X = -1, Yc = 65.75 + 3.67 (-1) = 62.08 2001 When X =1, Yc = 65.75 + 3.67 (1) = 69.42 2002 When X = 3, Yc = 65.75 + 3.67 (3) = 76.76 2003 When X = 5, Yc = 65.75 + 3.67 (5) = 84.10 2004 When X = 7, Yc = 65.75 + 3.67 (7) = 91.44 (b) Forecasting of earnings for 2006 For 2006, X =
đ?’•âˆ’đ?’Žđ?’Šđ?’… đ?’‘đ?’?đ?’Šđ?’?đ?’• đ?’?đ?’‡ đ?’•đ?’Šđ?’Žđ?’† đ?&#x;? đ?’?đ?’‡ đ?’•đ?’Šđ?’Žđ?’† đ?’Šđ?’?đ?’•đ?’†đ?’“đ?’—đ?’‚đ?’? đ?&#x;?
=
đ?&#x;?đ?&#x;Žđ?&#x;Žđ?&#x;”−đ?&#x;?đ?&#x;Žđ?&#x;Žđ?&#x;Ž.đ?&#x;“ đ?&#x;? Ă—đ?&#x;? đ?&#x;?
= 11
Thus, Yc = 65.75 + 3.67 (11) = 106.12 Hence, the earnings for 2005 is expected to be = $ 106.12 Ă— 100 = $106120
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Question-3. Obtain the straight line trend equation for the following data by the method of the least square. Year : 1995 1997 Sales in ’000 140 144 $: Also, estimate the sales for 2002
1998 160
1999 152
2000 168
2001 176
2004 180
Solution (a) Determination of the straight line trend equation by the method of least square Year t 1995 1997 1998 1999 2000 2001 2004 Total 13994
Sales Y 140 144 160 152 168 176 180 1120
Time dvn. i.e. t-1999 X -4 -2 -1 0 1 2 5 1
đ?‘‹2
XY -560 -288 -1 0 1 2 5 412
-16 4 1 0 1 4 25 51
Note. *In the above case, the average of the time variable is given by đ?‘‹ =
N=7 đ?’• đ?‘ľ
=
đ?&#x;?đ?&#x;‘đ?&#x;—đ?&#x;—đ?&#x;’ đ?&#x;•
=
1999 approx. Hence, 1999 has been taken as the year of origin in the above table. Working The trend line equation is given by Yc = a + bX Here, since đ?‘‹ ≠0, the value of the two constants a, and b are to be found out by solving simultaneously the following two normal equations: đ?‘Œ = Na + b đ?‘‹ đ?‘‹đ?‘Œ = a đ?‘‹ + b đ?‘‹ 2 Substituting the respective values in the above we get 1120 = 7a + b 412 = a + 51b Multiplying the eqn (ii) by 7 under the eqn (iii) and getting the same deducted from the equation (i) we get 7a + b =1120
=
− đ?&#x;•đ?’‚+đ?&#x;‘đ?&#x;“đ?&#x;•đ?’ƒ=đ?&#x;?đ?&#x;–đ?&#x;–đ?&#x;’ −đ?&#x;‘đ?&#x;“đ?&#x;”đ?’ƒ= −đ?&#x;?đ?&#x;•đ?&#x;”đ?&#x;’
b=
1764 356
= 4.96
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putting the above value of b in the equation (i) we get, 7a + 4.96 = 1120 7a = 1120 – 4.96 = 1115.04 or a = 1115.04/7 = 159.29 Putting the above values of a and b in the format of the equation we get the straight line for trend as under : Yc = 159.29 + 4.96X Where, the year of working origin = 1999, Y unit = annual sales (in ’000 $) and X unit = time deviations. (b)Estimation of the Sale for 2002 For 2002, X = 2002 – 1999 =3 Thus when, X = 3, Yc = 159.29 + 4.96 (3) = 159.29 + 14.88 = 174.17 Hence, the sales for 2002 are expected to be 174.17 × 103 = $ 174170.
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