Statistics for business and economics 8th edition newbold solutions manual

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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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