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Chapter 2: Describing Data: Numerical
2.1 Cruise agency – number of weekly specials to the Caribbean: 20, 73, 75, 80, 82 a. Compute the mean, median and mode x 330 x i 66 n 5 median = middlemost observation = 75 mode = no unique mode exists b. The median best describes the data due to the presence of the outlier of 20. This skews the distribution to the left. The agency should first check to see if the value ‘20’ is correct. 2.2 Number of complaints: 8, 8, 13, 15, 16 a. Compute the mean number of weekly complaints: x 60 x i 12 n 5 b. Calculate the median = middlemost observation = 13 c. Find the mode = most frequently occurring value = 8 2.3 CPI percentage growth forecasts: 3.0, 3.1, 3.4, 3.4, 3.5, 3.6, 3.7, 3.7, 3.7, 3.9 x 35 a. Compute the sample mean: x i 3.5 n
10
b. Compute the sample median = middlemost observation: c. Mode = most frequently occurring observation = 3.7 2.4
3.5 3.6 3.55 2
Department store % increase in dollar sales: 2.9, 3.1, 3.7, 4.3, 5.9, 6.8, 7.0, 7.3, 8.2, 10.2 x 59.4 a. Calculate the mean number of weekly complaints: x i 5.94 n 10 5.9 6.8 b. Calculate the median = middlemost observation: 6.35 2
Copyright Š 2013 Pearson Education, Inc. publishing as Prentice Hall.
2-1
Chapter 2: Describing Data: Numerical
2.5
2-2
Percentage of total compensation derived from bonus payments: 10.2, 13.1, 15, 15.8, 16.9, 17.3, 18.2, 24.7, 25.3, 28.4, 29.3, 34.7 a. Median % of total compensation from bonus payments = 17.3 18.2 17.75 2 x 248.9 20.7417 b. Mean % x i n
12
2.6 Daily sales (in hundreds of dollars): 6, 7, 8, 9, 10, 11, 11, 12, 13, 14 a. Find the mean, median, and mode for this store x 101 Mean = x i 10.1 n
10
10 + 11 10.5 2 Mode = most frequently occurring observation = 11 b. Find the five-number summary Q1 = the value located in the 0.25(n + 1)th ordered position = the value located in the 2.75th ordered position = 7 + 0.25(8 â&#x20AC;&#x201C;7) = 7.25 Q3 = the value located in the 0.75(n + 1)th ordered position = the value located in the 8.25th ordered position = 12 + 0.75(13 â&#x20AC;&#x201C;12) = 12.75 Minimum = 6 Maximum = 14 Five - number summary: minimum < Q1 < median < Q3 < maximum 6 < 7.25 < 10.5 < 12.75 < 14 Median = middlemost observation =
2.7 Find the measures of central tendency for the number of imperfections in a sample of 50 bolts 0(35) 1(10) 2(3) 3(2) Mean number of imperfections = = 0.44 imperfections per bolt 50 Median = 0 (middlemost observation in the ordered array) Mode = 0 (most frequently occurring observation) 2.8 Ages of 12 students: 18, 19, 21, 22, 22, 22, 23, 27, 28, 33, 36, 36 x 307 a. x i 25.58 n 12 b. Median = 22.50 c. Mode = 22
Chapter 2: Describing Data: Numerical
2-3
2.9 a. First quartile, Q1 = the value located in the 0.25(n + 1)th ordered position = the value located in the 39.25th ordered position = 2.98 + 0.25(2.98 –2.99) = 2.9825 Third quartile, Q3 = the value located in the 0.75(n + 1)th ordered position = the value located in the 117.75th ordered position = 3.37 + 0.75(3.37 –3.37) = 3.37 b. 30th percentile = the value located in the 0.30(n + 1)th ordered position = the value located in the 47.1th ordered position = 3.10 + 0.1(3.10 –3.10) = 3.10 th 80 percentile = the value located in the 0.80(n + 1)th ordered position = the value located in the 125.6th ordered position = 3.39 + 0.6(3.39 –3.39) = 3.39 2.10 xi 282 8.545 n 33 b. Median = 9.0 c. The distribution is slightly skewed to the left since the mean is less than the median. d. The five-number summary Q1 = the value located in the 0.25(n + 1)th ordered position = the value located in the 8.5th ordered position = 6 + 0.5(6 – 6) = 6 Q3 = the value located in the 0.75(n + 1)th ordered position = the value located in the 25.5th ordered position = 10 + 0.5(11 –10) = 10.5 Minimum = 2 Maximum = 21 Five - number summary: minimum < Q1 < median < Q3 < maximum 2 < 6 < 9 < 10.5 < 21 a. x
2.11 xi 23,699 236.99 . The mean volume of the random sample of 100 bottles n 100 (237 mL) of a new suntan lotion was 236.99 mL. b. Median = 237.00 c. The distribution is symmetric. The mean and median are nearly the same. d. The five-number summary Q1 = the value located in the 0.25(n + 1)th ordered position = the value located in the 25.25th ordered position = 233 + 0.25(234 – 233) = 233.25 Q3 = the value located in the 0.75(n + 1)th ordered position = the value located in the 75.75th ordered position = 241 + 0.75(241 –241) = 241 a. x
Chapter 2: Describing Data: Numerical
Minimum = 224 Maximum = 249 Five - number summary: minimum < Q1 < median < Q3 < maximum 224 < 233.25 < 237 < 241 < 249 2.12 The variance and standard deviation are
DEVIATION ABOUT THE MEAN, xi x
SQUARED DEVIATION ABOUT THE
–1 1 0 3 –4 –2 2 1
1 1 0 9 16 4 4 1
xi 6 8 7 10 3 5 9 8 8
xi x
MEAN,
8
8
xi 56
xi x 0
xi x 36
i1
i1
i1
2
2
8
xi
56
i1
Sample mean = x
n
7 8
8
xi x s2
Sample variance =
i1
n1
Sample standard deviation = s =
2
36 5.143 81 s2 =
5.143 = 2.268
2.13
The variance and standardABOUT deviation DEVIATION THEare SQUARED DEVIATION ABOUT THE 2 MEAN, xi x MEAN, xi x xi 3 0 –2 –1 5 10 6
0.5 –2.5 –4.5 –3.5 2.5 7.5 6
xi 15
xi x 0
i1
i1
0.25 6.25 20.25 12.25 6.25 56.25 6
xi x i1
2
101.5
2-4
Chapter 2: Describing Data: Numerical
6
xi Sample mean = x
15
i =1
n
2.5 6
101.5
6
2
xi x Sample variance = s 2
i1
n1
5 s2 =
Sample standard deviation = s =
20.3 20.3 = 4.5056
2.14
DEVIATION ABOUT THE SQUARED DEVIATION ABOUT 2 MEAN, xi x THE MEAN, xi x
xi 10 8 11 7 9
1 –1 2 –2 0
5
1 1 4 4 0
5
5
xi 45
xi x 0
xi x 10
i1
i1
i1
2
5
xi Sample mean = x
45
i1
n
9 5
10
5
2
xi x Sample variance =
s2
i1
n1
Sample standard deviation = s s
Coefficient of variation = CV
4 s2
2.5 2.5 1.581
x × 100%
1.581 × 100% 17.57% 9
2.15 Minitab Output: Descriptive Statistics: Ex2.15 Variable Ex2.15
Mean 28.77
SE Mean 2.15
Variable Ex2.15
Q3 38.00
Maximum 65.00
StDev 12.70
Variance 161.36
a. Mean = 2.15 b. Standard deviation = 12.70 c. CV = 44.15
CoefVar 44.15
Minimum 12.00
Q1 18.00
Median 27.00
2-5
Chapter 2: Describing Data: Numerical
2-6
2.16 Minitab Output Stem-and-Leaf Display: Ex2.16 Stem-and-leaf of Ex2.16 Leaf Unit = 1.0 3 10 17 (4) 14 13 7 4 2 1 1 1
1 1 2 2 3 3 4 4 5 5 6 6
N
= 35
234 5577889 0012333 7799 1 557788 002 59 3 5
IQR = Q3 – Q1 Q1 = the value located in the 0.25(35 + 1)th ordered position = the value located in the 9th ordered position = 18 Q3 = the value located in the 0.75(35 + 1)th ordered position = the value located in the 27th ordered position = 38 IQR = Q3 – Q1 = 38 – 18 = 20 years 2.17 Mean = 75, variance = 25,
2
25 5
Using the mean of 75 and the standard deviation of 5, we find the following interval: µ ± 2σ = 75 ± (2*5) = (65, 85), hence we have k = 2 a. According to the Chebyshev’s theorem, proportion must be at least 100[1 (1/ k 2 )]% = 100[1 (1/ 22 )]% = 75%. Therefore, approximately 75% of the observations are between 65 and 85 b. According to the empirical rule, approximately 95% of the observations are between 65 and 85
Chapter 2: Describing Data: Numerical
2-7
2.18 Mean = 250, σ = 20 a. To determine k, use the lower or upper limit of the interval: Range of observation is 190 to 310. k 310 or k 190 250 20k 310 or 250 20k 190 Solving both the equations we arrive at k = 3. According to the Chebyshev’s theorem, proportion must be at least 100[1 (1/ k 2 )]% = 100[1 (1 / 32 )]% = 75%. Therefore, approximately 88.89% of the observations are between 190 and 310. b. To determine k, use the lower or upper limit of the intervals: Range of observation is 190 to 310. k 290 or k 210 250 20k 290 or 250 20k 210 Solving both the equations we arrive at k = 2. According to the Chebyshev’s theorem, proportion must be at least 100[1 (1/ k 2 )]% = 100[1 (1 / 22 )]% = 75%. Therefore, approximately 75%
of the
observations are between 210 and 290. 2.19 Since the data is Mound shaped with mean of 450 and variance of 625, use the empirical rule . a. Greater than 425: Since approximately 68% of the observations are within 1 standard deviation from the mean that is 68% of the observations are between (425, 475). Therefore, approximately 84% of the observations will be greater than 425. b. Less than 500: Approximately 97.5% of the observations will be less than 500. c. Greater than 525: Since all or almost all of the distribution is within 3 standard deviations from the mean, approximately 0% of the observations will be greater than 525. 2.20 Compare the annual % returns on common stocks vs. U.S. Treasury bills Minitab Output: Descriptive Statistics: Stocks_Ex2.20, TBills_Ex2.20
Variable Stocks_Ex2.20 TBills_Ex2.20
N 7 7
N* 0 0
Variable Stocks_Ex2.20 TBills_Ex2.20
Q1 -14.70 4.400
Mean 8.16 5.786
SE Mean 8.43 0.556
Median 14.30 5.800
Q3 23.80 6.900
TrMean * *
Maximum 37.20 8.000
StDev 22.30 1.471
Range 63.70 4.200
Variance 497.39 2.165 IQR 38.50 2.500
CoefVar 273.41 25.43
Minimum -26.50 3.800
2-8
Chapter 2: Describing Data: Numerical
a. Compare the means of the populations Using the Minitab output µstocks = 8.16, µTbills = 5.786 Therefore, the mean annual % return on stocks is higher than the return for U.S. Treasury bills b. Compare the standard deviations of the populations Using the Minitab output, σstocks = 22.302, σTbills = 1.471 Standard deviations are not sufficient for comparision. We need to compare the coefficient of variation rather than the standard deviations. 8.16 s × 100 70.93% CVStocks × 100 22.302 x CVTbills
5.79 s × 100 6.60% × 100 1.471 x
Therefore, the variability of the U.S. Treasury bills is much smaller than the return on stocks. 2.21 xi
xi x
20 35 28 22 10 40 23 32 28 30 10
SQUARED DEVIATION ABOUT THE MEAN, DEVIATION ABOUT
xi2
–6.8 8.2 1.2 –4.8 –16.8 13.2 –3.8 5.2 1.2 3.2
400 1225 784 484 100 1600 529 1024 784 900 10
xi 268 1
10
2 i
x 7830
i1
268
i1
n
46.24 67.24 1.44 23.04 282.24 174.24 14.44 27.04 1.44 10.24
15
xi x
10
xi a. Sample mean = x
2
26.8 10
xi x
THE MEAN,
7.110
10
0
i i1
x x
2
647.6
Chapter 2: Describing Data: Numerical
2-9
b. Using equation 2.13:
10
xi x
s2
Sample standard deviation = s
i1
2
647.6
n1
8.483
9
c. Using equation 2.14: x
10
x
2
2
7830
i
i
Sample standard deviation = s
n
i1
71824
10
n1
8.483
9
d. Using equation 2.15: 10
2
xi n x2
Sample standard deviation = s
7830 10 26.8
i1
n1 e. Coefficient of variation = CV
s
x × 100
2
8.483
9 8.483 × 100 31.65% 26.8
2.22 Minitab Output: Descriptive Statistics: Weights Variable Weights
N 75
N* 0
Mean 3.8079
SE Mean 0.0118
StDev 0.1024
Variable Weights
Median 3.7900
Q3 3.8700
Maximum 4.1100
Range 0.5400
Variance 0.0105
CoefVar 2.69
Minimum 3.5700
Q1 3.7400
a. Using the Minitab output, range = 4.11 – 3.57 = 0.54, standard deviation = variance = 0.010486 b. IQR = Q3 – Q1 = 3.87 – 3.74 = .13. This tells that the range of the 50% of the distribution is 0.13 0.1024 s × 100 2.689% c. Coefficient of variation = CV x × 100 3.8079
middle
2.23 Minitab Output: Descriptive Statistics: Time (in seconds) Variable Time(in seconds)
Mean 261.05
Using the Minitab output
StDev 17.51
Variance 306.44
CoefVar Q1 6.71 251.75
Median 263.00
Q3 271.25
0.1024,
Chapter 2: Describing Data: Numerical
a. Samp le mean = x = 261.0 5
2-10
Chapter 2: Describing Data: Numerical
b. Sample variance = s2 = 306.44; s =
2-11
306.44 = 17.51
c. Coefficient of variation =
2.24 a. Standard deviation (s) of the assessment rates: n
( xi x ) 2 s
s2
583.75 39
i1
n1
14.974 3.8696
b. The distribution is approximately mounded. Therefore, the empirical rule applies. Approximately 95% of the distribution is expected to be within +/- 2 standard deviations of the mean. 2.25 Mean dollar amount and standard deviation of the amounts charged to a Visa account at Florin’s Flower Shop. Descriptive Statistics: Cost of Flowers Variable Cost of Flowers
Method of Payment American Express Cash Master Card Other Visa
N 23 16 24 23 39
N* 0 0 0 0 0
Mean 52.99 51.34 54.58 53.42 52.65
StDev 10.68 16.19 15.25 14.33 12.71
Median 50.55 50.55 55.49 54.85 50.65
Mean dollar amount = $52.65, standard deviation = $12.71 2.26 xi 21 4.2 n 5
a. mean without the weights x b. weighted mean wi
xi
wi xi
8 3 6 2 5 24
4.6 3.2 5.4 2.6 5.2
36.8 9.6 32.4 5.2 26.0 110.0
x
wi xi 110 4.583 wi 24
Chapter 2: Describing Data: Numerical
2-12
2.27 a. Calculate the sample mean of the frequency distribution for n = 40 mi fi f i mi Class 0-4 5-9 10-14 15-19 20-24
2 7 12 17 22
5 8 11 9 7 40
observations
10 56 132 153 154 505
f i mi 505 12.625 n 40
x
b. Calculate the sample variance and sample standard deviation Class 0-4 5-9 10-14 15-19 20-24 K
f mi x s2
i1
s
s2
mi
fi
f i mi
2 7 12 17 22
5 8 11 9 7 40
10 56 132 153 154 505
(mi x) -10.625 -5.625 -0.625 4.375 9.375
(mi x) 2 f i (m i x) 2 112.8906 31.64063 0.390625 19.14063 87.89063
564.4531 253.125 4.296875 172.2656 615.2344 1609.375
2 i
i
1609.375 41.266 n1 39 41.266 6.424
2.28 Class
mi
fi
mifi
(mi x)
4 < 10 10 < 16 16 < 22 22 < 28
7 13 19 25
8 15 10 7
56 195 190 175
–8.4 –2.4 3.6 9.6
f i (mi x )2
(mi x )2 70.56 5.76 12.96 92.16
564.48 86.4 129.6 645.12 2 i
a. Sample mean = x
mi f i 616 n
15.4 40
K
f m x b. Sample variance =
s2
i1
i
i
n1
i
2
1425.6 36.554 39
Chapter 2: Describing Data: Numerical
Sample standard deviation = s ď&#x20AC;
s2 ď&#x20AC;
36.554 6.046
2-13
Chapter 2: Describing Data: Numerical
2.29 Calculate the for the number of defects per n = 50 mi standard deviation fi f i mi (mi x) ) (mi x) f2i (mi x # of Defects # of Radios 0 1 2 3
s
2
f i (mi x )
12 15 17 6 50 2
47.22
0 15 34 18 67
-1.34 -0.34 0.66 1.66
.96367 ; s ď&#x20AC;
n1
1.7956 0.1156 0.4356 2.7556
radios
21.5472) 1.734 7.4052 16.5336 47.22
2
s 2 .9817
49
2.30 Based on a sample of n=50: mi fi f i mi 0 1 2 3 4 5 6 Sum
21 13 5 4 2 3 2 50
(mi x) 2
(mi x)
0 13 10 12 8 15 12 70
-1.4 -0.4 0.6 1.6 2.6 3.6 4.6
1.96 0.16 0.36 2.56 6.76 12.96 21.16
a. Sample mean number of claims per day = X
f i mi
x
Sample standard deviation = s =
= 1.40 50
3.0612 )2
( n1
41.16 2.08 1.8 10.24 13.52 38.88 42.32 150
f i mi 70 n
b. Sample variance = s 2
f i (m i x) 2
150
49 s = 1.7496 2
2.31 Estimate the sample mean and standard deviation (mi x) mi fi f i mi Class 0<4 4 <8 8 < 12 12 < 16 16 < 20 Sum
2 6 10 14 18
3 7 8 5 2 25
6 42 80 70 36 234
-7.36 -3.36 0.64 4.64 8.64
(mi x) 2 f i (m i x) 2 54.1696 11.2896 0.4096 21.5296 74.6496
162.5088 79.0272 3.2768 107.648 149.2992 501.76
a. Sample mean =
2-14
Chapter 2: Describing Data: Numerical
X
f i mi 234
= 9.36 n
25
2-15
Chapter 2: Describing Data: Numerical
)2
( f i mi
b. Sample variance = s ď&#x20AC; 2
501.76
x
20.9067
n1
24
Sample standard deviation = s =
s 2 = 4.572
2.32 Estimate the sample mean and sample standard deviation Class (mi x) mi fi f i mi (mi x) 2 9.95-10.45 10.45-10.95 10.95-11.45 11.45-11.95 11.95-12.45 Sum
10.2 10.7 11.2 11.7 12.2
a. sample mean = X
2 8 6 3 1 20
20.4 85.6 67.2 35.1 12.2 220.5
-0.825 -0.325 0.175 0.675 1.175
f i mi 220.5 n
b. sample variance = s 2 ď&#x20AC;
1.361 0.845 0.184 1.367 1.381 5.138
= 11.025
20 ( )2 5.138 f i mi x ď&#x20AC; n1
sample standard deviation = s =
0.681 0.106 0.031 0.456 1.381
f i (m i x) 2
= 0.2704
19 2
s = 0.520
2.33 Find the mean and standard deviation of the number of errors per page (mi x) mi fi f i mi (mi x) 2 f i (m i x) 2 0 1 2 3 4 5 Sum
f i mi 827
2
n f i (mi x )
102 138 140 79 33 8 500
0 138 280 237 132 40 827
-1.654 -0.654 0.346 1.346 2.346 3.346
= 1.654 500 2 769.142
= 1.5383
2.735716 0.427716 0.119716 1.811716 5.503716 11.19572
279.043 59.02481 16.76024 143.1256 181.6226 89.56573 769.142
2-16
Chapter 2: Describing Data: Numerical
n
500
Sample standard deviation = Ď&#x192; =
2
= 1.240
2-17
Chapter 2: Describing Data: Numerical
2-18
2.34 Using Table 1.7Minutes_
mi
220<230 230<240 240<250 250<260 260<270 270<280 280<290 290<300
225 235 245 255 265 275 285 295
fi
f i mi
(mi x)
5 1125 8 1880 13 3185 22 5610 32 8480 13 3575 10 2850 7 2065 110 28770
a. Using Equation 2.21, Sample mean, x
-36.545 -26.545 -16.545 -6.5455 3.45455 13.4545 23.4545 33.4545
(mi x) 2 f i (m i x) 2 1335.57 704.6612 273.7521 42.84298 11.93388 181.0248 550.1157 1119.207
f i mi 28770 n
6677.851 5637.289 3558.777 942.5455 381.8843 2353.322 5501.157 7834.446 32887.27
261.54545 110
b. Using Equation 2.22, sample variance s2
2
f i (mi x )
32887.27
n1
301.718 ; s ď&#x20AC; 109
s 2 17.370
c. From Exercise 2.23, x = 261.05 and s2 = 306.44. Therefore, the mean value obtained in both the Exercises are almost same, however variance is slightly by 4.7219 compared to Exercise 2.23. 2.35 a. Compute the sample covariance ( xi x ) xi yi (xi x) 2 ( y i y) ( yi y) 2 ( xi x ) ( yi y) 1 3 4 5 7 3 5 28
5 7 6 8 9 6 7 48
x = 4.0 y = 6.8571
-3 -1 0 1 3 -1 1 0
9 1 0 1 9 1 1 22
-1.85714 0.14286 -0.85714 1.14286 2.14286 -0.85714 0.14286 2.7E-15
s x2 = 3.667 s x =1.9149
Cov( x, y)
( xi x )( yi y ) 14 2.3333 n1 6
b. Compute the sample correlation coefficient
3.4489796 0.0204082 0.7346939 1.3061224 4.5918367 0.7346939 0.0204082 10.857143
5.571428571 -0.142857143 0 1.142857143 6.428571429 0.857142857 0.142857143 14
s y2 = 1.8095 Cov(x,y) = 2.333 s y =1.3452
lower
Chapter 2: Describing Data: Numerical
rxy
Cov( x, y) sx s y
2.3333 .9059 (1.9149)(1.3452)
2-19
Chapter 2: Describing Data: Numerical
2.36 a. Compute the sample covariance xi
yi
12 30 15 24 14 95
200 600 270 500 210 1780
x 19.00
y 356.00
Cov( x, y)
( xi x )
(xi x) 2
-7 11 -4 5 -5 0
( yi y) 2
( y i y)
49 121 16 25 25 236 x
-156 244 -86 144 -146 0
24336 59536 7396 20736 21316 s y133320 = 33330
( xi x ) ( yi y) 1092 2684 344 720 730 5570
2
s 2 = 59
Cov(x,y) = 1392.5
s y =182.5650569
( xi x )( yi y ) 5570s x =7.681146 1392.5 n1 4
b. Compute the sample correlation coefficient Cov( x, y) 1392.5 r 0.9930 (7.6811)(182.565) sx s y
2.37 a. Compute the sample covariance xi y i ( xi x ) (xi x) 2 6 7 8 9 10 40
80 60 70 40 0 250
x = 8.00 y = 50.00
-2 -1 0 1 2 0
( y i y) 4 1 0 1 4 10
s x2 = 2.5 s x = 1.5811
( xi x )( yi y ) Cov( x, y)
n1
180 45 4
b. Compute the sample correlation coefficient Cov( x, y) 45 rxy .90 sx s y (1.58114)(31.6228)
30 10 20 -10 -50 0
( yi y) 2 ( xi x ) ( yi y) 900 100 400 100 2500 4000
-60 -10 0 -10 -100 -180
s y2 =1000 Cov(x,y) = -45 s y =31.623
2-20
Chapter 2: Describing Data: Numerical
2.38
2-21
Minitab output Covariances: x_Ex2.38, y_Ex2.38 x_Ex2.38 y_Ex2.38
x_Ex2.38 31.8987 4.2680
y_Ex2.38
34.6536
Correlations: x_Ex2.38, y_Ex2.38
Pearson correlation of x_Ex2.38 and y_Ex2.38 = 0.128
Using Minitab outputa. Cov(x,y) = 4.268 b. r = 0.128 c. Weak positive association between the number of drug units and the number of days to complete recovery. Recommend low or no dosage units. 2.39
Minitab output Covariances: x_Ex2.39, y_Ex2.39 x_Ex2.39 y_Ex2.39
x_Ex2.39 9.28571 -5.50000
y_Ex2.39 5.40952
Correlations: x_Ex2.39, y_Ex2.39
Pearson correlation of x_Ex2.39 and y_Ex2.39 = -0.776
Using Minitab output a. Cov(x,y) = - 5.5, r = -.776 b. Higher prices are associated with fewer days to deliver, i.e., faster delivery time. 2.40 a. Compute the covariance xi y i ( xi x ) 5 6 7 8 9 35
55 53 45 40 20 213
µx = 7.00 µy= 42.60
(xi x) 2 -2 -1 0 1 2 0
( y i y) 4 1 0 1 4 10
12.4 10.4 2.4 -2.6 -22.6 0
( yi y)
2
153.76 108.16 5.76 6.76 510.76 785.2
σ x2 = 2.0
σ 2y = 157.04
σ x = 1.4142
σ y = 12.532
=
b. Compute the correlation coefficient Cov( x, y) 16.6 rxy .937 σxσ y (1.4142)(12.5316)
( xi x ) ( yi y) -24.8 -10.4 0 -2.6 -45.2 -83
Cov( x, y) = -16.6
Chapter 2: Describing Data: Numerical
2.41 Minitab output Covariances: Temperature (F), Time(hours) Temperature (F) Time(hours)
Temperature (F) 145.67273 2.80136
Time(hours) 0.05718
Correlations: Temperature (F), Time(hours)
Pearson correlation of Temperature (F) and Time(hours) = 0.971
Using Minitab output a. Covariance = 2.80136 b. Correlation coefficient = 0.971 2.42 Scatter plot â&#x20AC;&#x201C; Advertising expenditures (thousands of $s) vs. Monthly Sales (thousands of units) Scatterplot of Advertising Expenditures vs. Monthly Sales 200
Sales (thousands of $s)
180 160 140 120 100 80 6
7
8
9
10
11
12
13
14
15
A dvertising Expenditures (thousands of $s)
xi 10 15 7 12 14 58
y i ( xi x ) 100 200 80 120 150 650
(xi x) 2
-1.6 3.4 -4.6 0.4 2.4
2.56 11.56 21.16 0.16 5.76 41.2 x
x = 11.60 y = 130.00
( yi y) -30 70 -50 -10 20
( yi y)
2
900 4900 2500 100 400 8800
s 2 = 10.3
s y2 =2200
s x = 3.2094
s y = 46.9042
( xi x ) ( yi y) 48 238 230 -4 48 560 Cov(x,y) = 140
2-22
Chapter 2: Describing Data: Numerical
( xi x )( yi y )
Covariance = C ov( x, y) ď&#x20AC;
Correlation =
n1
Cov( x, y) sx s y
140
2-23
= 560 / 4 = 140
.93002
(3.2094)(46.9042)
2.43 Compute covariance and correlation between retail experience (years) and weekly sales (hundreds of dollars) ( y i y) ( yi y) 2 ( xi x ) ( yi y) xi y i ( xi x ) (xi x) 2 2 4 3 6 3 5 6 2 31
5 10 8 18 6 15 20 4 86
-1.875 0.125 -0.875 2.125 -0.875 1.125 2.125 -1.875
x = 3.875 y = 10.75
3.515625 0.015625 0.765625 4.515625 0.765625 1.265625 4.515625 3.515625 18.875
33.0625 0.5625 7.5625 52.5625 22.5625 18.0625 85.5625 45.5625 265.5
10.78125 -0.09375 2.40625 15.40625 4.15625 4.78125 19.65625 12.65625 69.75
s x2 = 2.6964
s 2y = 37.9286 Cov(x,y) = 9.964286
s x = 1.64208
s y = 6.15862
( xi x )( yi y ) n1
Covariance = C ov( x, y) ď&#x20AC;
Correlation =
-5.75 -0.75 -2.75 7.25 -4.75 4.25 9.25 -6.75
Cov( x, y) sx s y
=69.75 / 7 = 9.964286
9.964286
.9853
(1.64208)(6.15862)
2.44 Air Traffic Delays (Number of Minutes Late) (mi x) mi f i f i mi 5 15 25 35 45 55
30 25 13 6 5 4 83
150 375 325 210 225 220 1505
-13.133 -3.133 6.867 16.867 26.867 36.867
(mi x) 2 172.46 9.81 47.16 284.51 721.86 1359.21
f i (m i x) 2 5173.90 245.32 613.11 1707.07 3609.30 5436.84 16785.54
Chapter 2: Describing Data: Numerical
x 18.13
variance =
a. Sample mean number of minutes late = 1505 / 83 = 18.1325 b. Sample variance = 16785.54/82 = 204.7017 Sample standard deviation = s = 14.307
204.7017
2-24
Chapter 2: Describing Data: Numerical
2.45 Minitab Output Descriptive Statistics: Cost ($) Variable Cost ($)
Total Count 50
Mean 43.10
StDev 10.16
Variance 103.32
Minimum 20.00
Q1 35.75
Using the Minitab output a. Mean charge = $43.10 b. Standard deviation = $10.16 c. Five - number summary: minimum < Q1 < median < Q3 < maximum 20 < 35.75 < 45 < 50.25 < 60 2.46 For Location 2:
xi x
1
-9.2
84.64
19
8.8
77.44
2
-8.2
67.24
18
7.8
60.84
11
0.8
0.64
10
-0.2
0.04
3
-7.2
51.84
17
6.8
46.24
4
-6.2
38.44
17 102
6.8
46.24 473.6
xi
Mean = x
xi 102 n
Variance = s 2
xi x
10.2 10 xi x n1
Standard deviation = s
2
473.6 52.622 9 s 2 7.254
2
Median 45.00
Q3 50.25
Maximum 60.00
2-25
Chapter 2: Describing Data: Numerical
For Location 3: xi x
xi
Mean = x
xi x
2
-16.4
268.96
3
-15.4
237.16
25
6.6
43.56
20
1.6
2.56
22
3.6
12.96
19
0.6
0.36
25
6.6
43.56
20
1.6
2.56
22
3.6
12.96
26 184 18.4
7.6
57.76 682.4
xi 184
10 xi x
Variance = s 2
2
682.4 75.822 9
n1
Standard deviation = s
s 2 8.708
For Location 4:
xi x
22
9.5
90.25
20
7.5
56.25
10
-2.5
6.25
13
0.5
0.25
12
-0.5
0.25
10
-2.5
6.25
11
-1.5
2.25
9
-3.5
12.25
10
-2.5
6.25
8 125
-4.5
20.25 200.5
xi x
xi
xi 125
12.5
2
18.4
n
Mean = x
2
2-26
Chapter 2: Describing Data: Numerical
n
10
2-27
Chapter 2: Describing Data: Numerical
Variance = s
2
xi x
n1 Standard deviation = s ď&#x20AC;
2
200.5 22.278 9 s 2 4.720
2.47 Describe the data numerically
Covariances: X_Ex2.47, Y_Ex2.47 X_Ex2.47 Y_Ex2.47
X_Ex2.47 8.81046 5.18954
Y_Ex2.47
50.92810
Correlations: X_Ex2.47, Y_Ex2.47
Pearson correlation of X_Ex2.47 and Y_Ex2.47 = 0.245 P-Value = 0.327
There is a very weak positive relationship between the variables.
2-28
Chapter 2: Describing Data: Numerical
2.48 a. Describe the data graphically between graduating GPA vs. entering SAT Verbal scores Fitted Line Plot GPA = 1.638 + 0.02744 SATverb 4.0
S R-Sq R-Sq(adj)
GPA
3.5
3.0
2.5
2.0 30
40
50 SA Tverb
60
b. Correlations: GPA, SATverb
Pearson correlation of GPA and SATverb = 0.560 P-Value = 0.000
70
0.330317 31.4% 30.3%
2-29
Chapter 2: Describing Data: Numerical
2-30
2.49 Arrange the populations according to their variances and calculate the variances manually (a) has the least variability, then population (c), followed by (b) and then (d) Population standard deviation
2
a 1 2 3 4 5 6 7 8 36
b 1 1 1 1 8 8 8 8 36
c 1 1 4 4 5 5 8 8 36
d -6 -3 0 3 6 9 12 15 36
x 4.5
x 4.5
x 4.5
x 4.5
i
( x x )2 N
(a a )2 12.25 6.25 2.25 0.25 0.25 2.25 6.25 12.25 a = 5.25 42 2
(b b )2
(c c )2
12.25 12.25 12.25 12.25 12.25 12.25 12.25 12.25 b = 12.25 98
12.25 12.25 0.25 0.25 0.25 0.25 12.25 12.25 c = 6.25 50
2
2
(d d )2 110.25 56.25 20.25 2.25 2.25 20.25 56.25 110.25 = 47.25 d 378 2
2.50 Mean of $295 and standard deviation of $63. a. Find a range in which it can be guaranteed that 60% of the values lie. Use Chebyshev’s theorem: at least 60% = [1-(1/k2)]. Solving for k, k = 1.58. The interval will range from 295 +/- (1.58)(63) = 295 +/- 99.54. 195.46 up to 394.54 will contain at least 60% of the observations. b. Find the range in which it can be guaranteed that 84% of the growth figures lie Use Chebyshev’s theorem: at least 84% = [1-(1/k2)]. Solving for k, k = 2.5. The interval will range from 295 +/- (2.50)(63) = 295 +/- 157.5. 137.50 up to 452.50 will contain at least 84% of the observations. 2.51 Growth of 500 largest U.S. corporations had a mean of 9.2%, standard deviation of 3.5%. a. Find the range in which it can be guaranteed that 84% of the growth figures lie. Use Chebyshev’s theorem: at least 84% = [1-(1/k2)]. Solving for k, k = 2.5. The interval will range from 9.2 +/- (2.50)(3.5) = 9.2 +/- 8.75. 0.45% up to 17.95% will contain at least 84% of the observations. b. Using the empirical rule, approximately 68% of the earnings growth figures lie within 9.2 +/- (1)(3.5). 5.7% up to 12.7% will contain at least 68% of the observations.
Chapter 2: Describing Data: Numerical
2-31
2.52 Tires have a lifetime mean of 29,000 miles and a standard deviation of 3,000 miles. a. Find a range in which it can be guaranteed that 75% of the lifetimes of tires lies Use Chebyshevâ&#x20AC;&#x2122;s theorem: at least 75% = [1-(1/k2)]. Solving for k = 2.0. The interval will range from 29,000 (2.0)(3,000) = 29,000 6,000 23,000 to 35,000 will contain at least 75% of the observations . b. 95%: solve for k = 4.47. The interval will range from 29,000 (4.47)(3000) = 29,000 13,416.41. 15,583.59 to 42,416.41 will contain at least 95% of the observations. 2.53 Minitab Output: Descriptive Statistics: Time (in seconds) Variable Time (in seconds) Variable Time (in seconds)
Total Count 110
Mean 261.05
Median 263.00
Q3 271.25
StDev 17.51
Variance 306.44
Maximum 299.00
Minimum 222.00
Q1 251.75
IQR 19.50
Using the Minitab output a. Interquartile Range = 19.50. This tells that the range of the middle 50% of the distribution is 19.50. b. Five - number summary: minimum < Q1 < median < Q3 < maximum 222 < 251.75 < 263 < 271.25 < 299 2.54 Minitab Output: Descriptive Statistics: Time Variable Time Variable Time
Total Count 104
Q3 56.50
Mean 41.68
StDev 16.86
Variance 284.35
CoefVar 40.46
Minimum 18.00
Q1 28.50
Maximum 73.00
Using the Minitab output a. Mean shopping time = 41.68 b. Variance = 284.35 Standard deviation = 16.86 c. 95th percentile = the value located in the 0.95(n + 1)th ordered position = the value located in the 99.75th ordered position = 70 + 0.75(70 â&#x20AC;&#x201C;70) = 70. d. Five - number summary: minimum < Q1 < median < Q3 < maximum 18 < 28.50 < 39 < 56.50 < 73 e. Coefficient of variation = 40.46
Median 39.00
Chapter 2: Describing Data: Numerical
2-32
f. Find the range in which ninety percent of the shoppers complete their shopping. Use Chebyshevâ&#x20AC;&#x2122;s theorem: at least 90% = [1-(1/k2)]. Solving for k, k = 3.16. The interval will range from 41.68 +/- (3.16)(16.86) = 41.88 +/- 53.28. -11.60 up to 94.96 will contain at least 90% of the observations. 2.55 xi
yi
3.5 2.4 4 5 1.1 16
88 76 92 85 60 401
x = 3.2
y = 80.2
( xi x )
( yi y)
( xi x ) ( yi y)
0.3 -0.8 0.8 1.8 -2.1
7.8 -4.2 11.8 4.8 -20.2
2.34 3.36 9.44 8.64 42.42 66.2
(xi x) 2 0.09 0.64 0.64 3.24 4.41 9.02
( y i y) 2 60.84 17.64 139.24 23.04 408.04 648.8
2 s x2 = 2.255 s y = 162.2
s x = 1.5017 s y = 12.7358 Covariance = Cov( x, y) Correlation coefficient =
( xi x )( yi y ) 66.2 16.55 n1 4 Cov( x, y) 16.55 sx s y
0.8654
(1.5017)(12.7358)
2.56 xi
yi
( xi x )
(xi x) 2
( y i y)
( yi y) 2
12 30 15 24 14 18 28 26 19 27 213
20 60 27 50 21 30 61 54 32 57 412
-9.3 8.7 -6.3 2.7 -7.3 -3.3 6.7 4.7 -2.3 5.7
86.49 75.69 39.69 7.29 53.29 10.89 44.89 22.09 5.29 32.49 378.1
-21.20 18.80 -14.20 8.80 -20.20 -11.20 19.80 12.80 -9.20 15.80
449.44 353.44 201.64 77.44 408.04 125.44 392.04 163.84 84.64 249.64 2505.6
x 21.3
y 41.2
s x2 = 42.01 s x = 6.4816
s 2y sy
= 278.4
=16.6853
( xi x ) ( yi y) 197.16 163.56 89.46 23.76 147.46 36.96 132.66 60.16 21.16 90.06 962.4
Chapter 2: Describing Data: Numerical
( xi x )( yi y ) 962.4 106.9333 n1 9 Cov(x, y) 106.9333 Correlation coefficient = =
2-33
Covariance = Cov( x, y)
= 0.9888 sx s y
Chapter 2: Describing Data: Numerical
2-34
(6.4816)(16.6853)
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