Newbold Stat8 Ism 07

Chapter 7: Estimation: Single Population

7.1

a. Check for nonnormality

The distribution shows no significant evidence of nonnormality. b. Assuming normality, the unbiased and most efficient point estimator for the population mean is the sample mean:  X i  1705 = 60.89 X n 28 c. Assuming normality, the sample mean is an unbiased estimator of the population mean with variance Var ( X ) 

2

. Also, assuming normality, the unbiased and most efficient point n estimator for the population variance is the sample variance, s 2 . d. Thus, the unbiased point estimate of the variance of the sample mean: s 2 (5.231)2 Var ( X )    .9771 n 28

Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.

7-1

7-2 7.2

Statistics for Business & Economics, 8th edition

a. There appears to be no evidence of nonnormality, as shown by the normal probability plot.

b. Assuming normality, the unbiased and most efficient point estimator for the population mean is the sample mean, X .  X i  2411  301.375 thousand dollars X n 8 c. Assuming normality, the sample mean is an unbiased estimator of the population mean with variance Var ( X ) 

2

. Also, assuming normality, the unbiased and most efficient point n estimator for the population variance is the sample variance, s 2 . s 2 8373.125 Var ( X )    1046.64 n 8 d. The unbiased and most efficient point estimator for a proportion is the sample proportion, pˆ . x 3 pˆ    0.375 n 8



7.3

n = 10 economists forecast for percentage growth in real GDP in the next year

Descriptive Statistics: RGDP_Ex7.3 Variable RGDP_Ex7.3

N 10

N* 0

Variable RGDP_Ex7.3

Minimum 2.2000

Mean 2.5700

Q1 2.4000

SE Mean 0.0716

TrMean 2.5625

StDev 0.2263

Median 2.5500

Q3 2.7250

Maximum 3.0000

Variance 0.0512 Range 0.8000

CoefVar 8.81 IQR 0.3250

a. Unbiased point estimator of the population mean is the sample mean:  X i  25.7  2.57 X n 10 b. The unbiased point estimate of the population variance: s 2  .0512 c. Unbiased point estimate of the variance of the sample mean s 2 .0512 Var ( X )    .00512 n 10 x 7 d. Unbiased estimate of the population proportion: pˆ    .70 n 10 7.4

n = 12 employees. Number of hours of overtime worked in the last month: a. Unbiased point estimator of the population mean is the sample mean:  X i  293  24.42 X n 12 b. The unbiased point estimate of the population variance: s 2  85.72 c. Unbiased point estimate of the variance of the sample mean s 2 85.72 Var ( X )    7.1433 n 12 x 3 d. Unbiased estimate of the population proportion: pˆ    .25 n 12

7.5

n = 350 accounts of the company’s total portfolio a. Unbiased point estimator of the population mean is the sample mean  X i  4, 428, 043  $12,651.55 X n 350 b. Unbiased point estimate of the population variance s 2  $28,192, 687.10 c. Unbiased point estimate of the variance of the sample mean s 2 28,192, 687.10 Var ( X )    $80,550.53 n 350 d. Unbiased estimate of the population proportion x 133 pˆ    .38 n 350


Sum 25.7000

7-3

7-4 7.6


a.

No evidence of the data distribution coming from a nonnormal population. b. The minimum variance unbiased point estimate of the population mean is the sample  X i  285.59  3.8079 mean: X  n 75 Descriptive Statistics: Volumes Variable Volumes

Variable Volumes

N 75

Minimum 3.5700

Mean 3.8079 Maximum 4.1100

Median 3.7900 Q1 3.7400

TrMean 3.8054

StDev 0.1024

SE Mean 0.0118

Q3 3.8700

c. The minimum variance unbiased point estimate of the population variance is the sample variance s2 = 0.10242 = .0105. 7.7

1 1   E( X1 )  E( X 2 )     2 2 2 2 1 3  3 E (Y )  E ( X 1 )  E ( X 2 )    4 4 4 4 1 2  2 E (Z )  E ( X1 )  E ( X 2 )    3 3 3 3

a. E ( X ) 



1 1 2 2 2 b. Var ( X )  Var ( X 1 )  Var ( X 2 )    4 4 4 4 2 2 1 9 5 Var (Y )  Var ( X 1 )  Var ( X 2 )  16 16 8 2 1 4 5 Var ( Z )  Var ( X 1 )  Var ( X 2 )  9 9 9 X is most efficient since Var ( X )  Var ( Z )  Var (Y ) Var (Y ) 5   1.25 c. Relative efficiency between Y and X : Var ( X ) 4 Var ( Z ) 10   1.111 Relative efficiency between Z and X : Var ( X ) 9 7.8

7.9

Reliability factor for each of the following: a. 93% confidence level: z 2 = +/– 1.81 b. 96% confidence level: z

2

= +/– 2.05

c. 80% confidence level: z

2

= +/– 1.28

Reliability factor for each of the following: a.   .08 ; z 2 = +/– 1.75 b.  2 = .02 ; z 2 = +/– 2.05

7.10

Calculate the margin of error to estimate the population mean a. 98% confidence level, n = 64; variance = 144  = 3.495 ME  z 2  = 2.33 12  n 64   b. 99% confidence interval, n = 120; standard deviation = 100  = 23.552 ME  z 2  = 2.58 100  n 120  

7.11

Calculate the width to estimate the population mean, for a. 90% confidence level, n = 100, variance = 169   = 2 1.645 13 width = 2ME = 2  z 2    = 4.277   100   n    b. 95% confidence interval, n = 120, standard deviation = 25   = 2 1.96  25 width = 2ME = 2  z 2    = 8.9461   120   n   


7-5

7-6 7.12


Calculate the LCL and UCL:

 = 40.2 to 59.8 = 50  1.96  40  n 64    = 81.56 to 88.44 b. x  z 2  = 85  2.58  20  n 225    = 506.27 to 513.73 c. x  z 2  = 510  1.645  50  n 485   a. x  z 2 

7.13

a. n  9, x  187.9,   32.4, z.10  1.28 80% confidence interval: 187.9  1.28(32.4/3) = 174.076 up to 201.724 b. 210.0 – 187.9 = 22.1 = z / 2 (32.4 / 3), z / 2  2.05

  2[1  Fz (2.05)]  .0404 Confidence level: 100(1–.0404) = 95.96% 7.14

a. Calculate the standard error of the mean   5 = .5 n 100 b. Calculate the margin of error of the 90% confidence interval for the population mean ME  z 2  = 1.645(.5) = .8225 n c. Calculate the width of the 98% confidence interval for the population mean

 = 2  2.33 .5  = 2.33 width = 2ME = 2  z 2     n   7.15

a. n  25, x  2.90,   .45, z.025  1.96 95% confidence interval:  = 2.90  1.96(.45/5) = 2.7236 up to 3.0764 x  z    n  b. 2.99 – 2.90 = .09 = z / 2 (.45 / 5), z / 2  1

  2[1  Fz (1)]  .3174 Confidence level: 100(1–.3174) = 68.26% 7.16

n  25, x  19.8,   1.2, z.005  2.58 99% confidence interval:  = 19.8  2.58(1.2/5) = 19.1808 up to 20.4192 x  z    n 



7.17

Find the standard error a. n = 17, 95% confidence level, s = 16; s = 3.8806  16 n 17 b. n = 25, 90% confidence level, s = 6.56; s = 1.3115  6.56 n 25

7.18 Find the ME a. n = 4, 99% confidence level, x1 = 25, x2 = 30, x3 = 33, x4 = 21 s = 5.3151 ME  tv, 2 s  5.841 5.3151  = 15.5227 n 4  b. n = 5, 90% confidence level, x1 = 15, x2 = 17, x3 = 13, x4 = 11, x5 = 14 s = 2.2361 ME  tv, 2 s  2.132  2.2361  = 2.1320 n 5  7.19 Time spent driving to work for n = 20 people Descriptive Statistics: Driving_Ex7.19 Variable Driving_Ex7.19

N 20

N* 0

Variable Driving_Ex7.19

Sum 702.00

Mean 35.10

Minimum 15.00

SE Mean 2.21

TrMean 35.39

StDev 9.87

Variance 97.36

Q1 27.25

Median 33.50

Q3 45.00

Maximum 50.00

CoefVar 28.11 Range 35.00

a. Calculate the standard error s  9.867 = 2.2063 n 20 b. Find the value of t for the 95% confidence interval tv , 2  t19,.025 = 2.093 c. Calculate the width for a 95% confidence interval for the population mean width = 2ME = 2 t 2 s   2 2.093  2.2063  = 9.2356 n  7.20

Find the LCL and UCL for each of the following: a.  = .05, n = 25, sample mean = 560, s = 45  = 560  2.064  45  = 541.424 to 578.576 x  tv, 2  s    n 25    b.  / 2 = .05, n = 9, sample mean = 160, sample variance = 36  = 160  1.860  6  = 156.28 to 163.72 x  tv, 2  s    n 9   c. 1   = .98, n = 22, sample mean = 58, s = 15  = 58  2.518 15  = 49.9474 to 66.0526 x  tv, 2  s    n 22     Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.

IQR 17.75

7-7

7-8


7.21

n = 16, sample mean = 47,500 miles, sample standard deviation = 4,200 miles a. Calculate the margin of error for 95% confidence level to estimate the population mean;  = 2237.55 = 2.131 4200 ME  tv , 2 s  n 16   b. 90% confidence interval:  = 47500  1.753  4200  = 45,659.35 miles to 49,340.65 miles x  tv, 2  s    n 16   

7.22

Calculate the width for each of the following: a.  = 0.05, n = 6, s = 40   w  2ME  2 tv , 2 s   2  2.571 40    2(41.98425)  83.9685 n 6     b.  = 0.01, n = 22, sample variance = 400     2(12.07142)  24.1428 w  2ME  2 tv , 2 s   2  2.831 20  n 22      c.  = 0.10, n = 25, s = 50     2(17.11)  34.22 w  2ME  2 tv , 2 s   2 1.711 50  n 25     

7.23

95% confidence interval: Results for: HEI Cost Data Variable Subset.xls Descriptive Statistics: HEI-2005_day2 Variable HEI-2005_day2

N 4130

Variable HEI-2005_day2

Median 54.550

Mean 54.371

SE Mean 0.227

Q3 64.929

TrMean 54.317

StDev 14.559

Minimum 12.462

Maximum 93.709

One-Sample T: HEI-2005_day2 Variable HEI-2005_day2

N 4130

Mean 54.371

StDev 14.559

SE Mean 0.227

95.0% CI (53.927, 54.815)


Q1 43.256


7.24

Results for: Sugar.xls Descriptive Statistics: Weights Variable Weights

Variable Weights

N 100

Mean 520.95

Minimum 504.70

Maximum 544.80

Median 518.75

TrMean 520.52

Q1 513.80

StDev 9.45

SE Mean 0.95

Q3 527.28

90% confidence interval: Results for: Sugar.xls One-Sample T: Weights Variable Weights

N 100

Mean 520.948

StDev 9.451

SE Mean 0.945

90.0% CI ( 519.379, 522.517)

b. Narrower since a smaller value of t will be used in generating the 80% confidence interval. 7.25 n  400, x  357.75, s  37.89, t399,.025  1.966

ME  tv , 2 s

 = 3.7246 = 1.966  37.89  n 400  

7.26 n  28, x  60.893, s  5.2305, t27,.025  2.052

ME  tv , 2 s 7.27

 = 2.0283 = 2.052  5.2305  n 28  

n  24, x  24.375, s  8.9434, t23,.005  2.807

a.

99% confidence interval: 24.375  2.807(8.9434/ 24 ) = 19.2506 pounds up to 29.4994 pounds

b. Narrower since the t-score will be smaller for a 90% confidence interval than for a 99% confidence interval. 7.28

n  25, x  42, 740, s  4, 780, t24,.05  1.711

90% confidence interval: 42,740  1.711(4780/5) = $41,104.28 up to $44,375.72 7.29

n  9, x  16.222, s  4.790, t8,.05  1.86

We must assume a normally distributed population 90% confidence interval: 16.222  1.86(4.790/3) = 13.252 up to 19.192


7-9

7-10


7.30 Find the margin of error to estimate the population proportion for each of the following: a. n = 350, pˆ = .3,  = .01

pˆ (1  pˆ ) .3(.7) = .0631  2.576 n 350 b. n = 275, pˆ = .45,  = .05 ME  z 2

pˆ (1  pˆ ) .45(.55) = .0588  1.96 n 275 c. n = 500, pˆ = .05,  = .10 ME  z 2

ME  z 2

pˆ (1  pˆ ) .05(.95) = .01603  1.645 n 500

7.31 Find the confidence level for estimating the population proportion for each of the following: a. 98% confidence level; n = 450, pˆ = .10

.10(1  .10) pˆ (1  pˆ ) = .10  2.326 = .0671 to .1329 450 n b. 95% confidence level; n = 240, pˆ = .01 pˆ  z 2

.01(1  .01) pˆ (1  pˆ ) = .01  1.96 = –0.0026 to .0226 240 n c.  = .04; n = 265, pˆ = .50 pˆ  z 2

pˆ  z 2 7.32

pˆ (1  pˆ ) .5(.5) = .50  2.054 = .4369 to .5631 n 265

n = 250 + 75 + 25 = 350; a. Estimate the percent of alumni in favor of the program: pˆ  250 / 350  .7143  = .05 pˆ (1  pˆ ) .7143(.2857) = .667 up to .7616 pˆ  z /2  .7143  1.96 n 350 b. Estimate the percent of alumni in opposed to the program: pˆ  75 / 350  .2143 90% confidence interval pˆ (1  pˆ ) .2143(.7857) = .1782 up to .2504 pˆ  z /2  .2143  1.645 n 350



7.33

7.34

n = 142, 87 answered GMAT or GRE is ‘very important’. 95% confidence interval for the population proportion: .6127(1  .6127) pˆ (1  pˆ ) = .6127  1.96 = .5326 to .6928 pˆ  z 2 142 n

n  95, pˆ  67 / 95  .7053, z.005  2.58 99% confidence interval: .7053(.2947) pˆ (1  pˆ ) = .7053  (2.58) = .5846 up to .8260 pˆ  z / 2 95 n x 133   .38 n 350 n  350, pˆ  133 / 350  .38,   .02

7.35 pˆ 

pˆ  z / 2

pˆ (1  pˆ ) .38(.62) = .38  2.33 = .3195 up to .4405 350 n

7.36

n  350, pˆ  73 / 350  .2086, z.025  1.96 95% confidence interval: pˆ (1  pˆ ) = .2086  (1.96) .2086(.7914) / 350 = .166 to .2512 pˆ  z / 2 n

7.37

n  400,

pˆ  320 / 400  .80, z.01  2.326

.80(1  .80) pˆ (1  pˆ ) = .80  2.326 = .75348 400 n b. width of a 90% confidence interval pˆ (1  pˆ ) .8(1  .8) w  2ME  2 z 2  2 1.645  2(.0329)  .0658 n 400 a. LCL = pˆ  z / 2

7.38

width = .76 – .68 = .08; ME = .04, pˆ  180 / 250  0.72

pˆ (1  pˆ ) .72(.28)   .0284 n 250 ME = .04 = z / 2 (.0284), z / 2  1.41   2[1  Fz (1.41)]  2[.0793]  .1586 Confidence level: 100(1 – .1586) = 84.14%


7-11

7-12


7.39

n  540, pˆ  320 / 540  .5926, z.025  1.96 95% confidence interval: .5926(.4074) pˆ (1  pˆ ) = .5926  (1.96) = .5512 up to .634 pˆ  z / 2 540 n

7.40

a. n  460, pˆ  (50  120  80) / 460  .5435, z.025  1.96 95% confidence interval: pˆ (1  pˆ ) .5435(.4565) = .5435  (1.96) = .49798 up to .58898 pˆ  z / 2 n 460 b. n  460, pˆ  (60  50  40) / 460  .3261, z.05  1.645 90% confidence interval: .3261(.6739) pˆ (1  pˆ ) = .3261  (1.645) = .29009 up to .36209 pˆ  z / 2 460 n

7.41

a. n  246, pˆ  232 / 246  .9431, z.01  2.326 98% confidence interval: .9431(.0569) pˆ (1  pˆ ) = .9431  (2.326) = .9087 up to .9775 pˆ  z / 2 246 n b. n  246, pˆ  10 / 246  .0407, z.01  2.326 98% confidence interval: pˆ (1  pˆ ) .0407(.9593) = .0407  (2.326) = .0114 up to .0699 pˆ  z / 2 n 246

7.42

a. n  21, s 2  16, taking   .05,  220,.025  34.17 LCL 

(n  1) s 2



2

=

n 1, /2

20(16) = 9.3649 34.17

b. n  16, s  8, taking   .05,  215,.025  27.49 LCL 

(n  1) s 2

 2 n 1, /2

=

15(8) 2 = 34.9218 27.49

c. n  28, s  15, taking   .01,  227,.005  49.64 LCL 

(n  1) s 2

 2 n 1, /2

=

27(15) 2 = 122.3811 49.64



7.43

a. n  21, s 2  16, taking   .05,  220,.975  9.59 UCL 

(n  1) s 2



2

=

n 1,1 /2

20(16) = 33.3681 9.59

b. n  16, s  8, taking   .05,  215,.975  6.26 UCL 

(n  1) s 2

15(8) 2 = = 153.3546 6.26

 2 n 1,1 /2

c. n  28, s  15, taking   .01,  227,.995  11.81 UCL 

(n  1) s 2

=

 2 n 1,1 /2

27(15) 2 = 514.3946 11.81

7.44 Random sample from a normal population a. Find the 90% confidence interval for the population variance Descriptive Statistics: Ex7.44 Variable Ex7.44

Mean 11.00

SE Mean 1.41

StDev 3.16

Variance 10.00

Minimum 8.00

Maximum 16.00

n  5, s 2  10,  24,.95  .711,  24,.05  9.49 (n  1) s 2



2

n 1, / 2

2 

(n  1)s 2



2

=

n 1,1 / 2

4(10) 4(10) 2  = 4.21496 <  2 < 9.49 .711

56.25879

b. Find the 95% confidence interval for the population variance n  5, s 2  10,  24,.975  .484,  24,.025  11.14

(n  1) s 2



2

n 1, / 2

2 

(n  1)s 2



2

n 1,1 / 2

=

4(10) 4(10) 2  = 3.59066 <  2 < 11.14 .484


82.64463

7-13

7-14 7.45


No evidence of nonnormality

n = 50, s 2  0.000028 . Since df = 49 is not in the Chi-Square Table in we will approximate the interval using df = 50. 49(0.000028) 49(0.000028) (n  1) s 2 (n  1)s 2 2  2     2 2 71.42 32.36  n1, / 2  n1,1 / 2 = 4.27E-5 <  2 < 1.94E-5 7.46

n  10, s 2  28.3023,  29,.05  16.92,  29,.95  3.33

(n  1) s 2

 2 n1, / 2

2 

(n  1)s 2

 2 n1,1 / 2

=

9(28.3023) 9(28.3023) 2  = 15.0544 up to 76.4927 16.92 3.33

7.47

Assume that the population is normally distributed. n  20, s 2  6.62,  219,.025  32.85,  219,.975  8.91

(n  1) s 2



2

n 1, / 2

  2

(n  1)s 2



2

n 1,1 / 2

=

19(6.62) 19(6.62) 2  32.85 8.91

= 3.8289 up to 14.1167.


the Appendix,


7.48

n  18, s 2  (10.4)2  108.16,  217,.05  27.59,  217,.95  8.67

(n  1) s 2

(n  1)s 2

17(108.16) 17(108.16) 2  27.59 8.67  n1, / 2  n1,1 / 2 = 66.6444 up to 212.0784. Assume that the population is normally 2

7.49

7-15

2 

2

=

distributed.

a. n  15, s 2  (2.36)2  5.5696,  214,.025  26.12,  214,.975  5.63 14(5.5696) 14(5.5696) 2  = 2.9852 up to 13.8498 26.12 5.63 b. wider since the chi-square statistic for a 99% confidence interval is larger than for a 95% confidence interval.

7.50

7.51

7.52

n  9, s 2  .7875,  28,.05  15.51,  28,.95  2.73 8(.7875) 8(.7875) 2  = .4062 up to 2.3077 15.51 2.73

102 1200  80   1.1676 a. ˆ  80 1200  1 64 1425  90  0.6667 b. ˆ x2   90 1425  1 129 3200  200   0.6049 c. ˆ x2  200 3200  1 2 x

a. The confidence interval is x  t n 1, / 2ˆ x    x  t n 1, / 2ˆ x . Use t79,0.025  1.990, and

ˆ x2  1.1676 from exercise 7.51 part a. x  t n 1, / 2ˆ x    x  t n 1, / 2ˆ x

142  (1.990)(1.0806)    142  (1.990)(1.0806) or (139.85, 144.15)

b. The confidence interval is x  t n 1, / 2ˆ x    x  t n 1, / 2ˆ x . Use t89,0.025  1.987, and ˆ x2  0.6667 from exercise 7.51 part b. x  t n 1, / 2ˆ x    x  t n 1, / 2ˆ x

232.4  (1.987)(0.8165)    232.4  (1.987)(0.8165) or (230.78, 234.02)


7-16


c. The confidence interval is x  t n 1, / 2ˆ x    x  t n 1, / 2ˆ x . Use t199,0.025  1.972, and

ˆ x2  0.6049 from exercise 7.51 part c. x  t n 1, / 2ˆ x    x  t n 1, / 2ˆ x

59.3  (1.972)(0.7777)    59.3  (1.972)(0.7777) or (57.77, 60.83)

7.53

a. A 100(1   )% confidence interval for the population total is obtained from the following formula: Nx  tn 1, /2 Nˆ x  N   Nx  tn 1, /2 Nˆ x As stated, N  1325, n  121, s  20, and x  182. For a 95% confidence level, note that tn 1, / 2  t120,0.025  1.98. Next calculate Nˆ x .

Nˆ x 

Ns n

N  n (1325)( 20) 1325  121   2297.325 N 1 1325  1 121

So the 95% confidence interval is Nx  tn 1, /2 Nˆ x  N   Nx  tn 1, /2 Nˆ x

or

(1325)(182)  (1.98)( 2297.325)  N  (1325)(182)  (1.98)( 2297.325)

or (236601, 245699) b. As stated, N  2100, n  144, s  50, and x  1325. For a 98% confidence level, note that tn 1, / 2  t143,0.01  2.35. Next calculate Nˆ x .

Nˆ x 

Ns n

N  n (2100)(50) 2100  144   8446.684 N 1 2100  1 144

So the 98% confidence interval is Nx  tn 1, /2 Nˆ x  N   Nx  tn 1, /2 Nˆ x

or

(2100)(1325)  (2.35)(8446.684)  N  (2100)(1325)  (2.35)(8446.684)

or (2762650, 2802350)



7.54

a

A 100(1   )% confidence interval for the population proportion is obtained from the following formula: pˆ  z / 2ˆ pˆ  P  pˆ  z / 2ˆ pˆ

As stated, N  1058, n  160, and x  40. For a 95% confidence level, note that z / 2  z0.025  1.96. Next calculate pˆ and ˆ pˆ . x 40   0.25 n 160 pˆ (1  pˆ ) ( N  n) 0.25(1  0.25) 1058  160 ˆ p2ˆ      0.0010019 n 1 ( N  1) 160  1 1058  1 pˆ 

ˆ pˆ  0.0010019  0.03165 So the 95% confidence interval is pˆ  z / 2ˆ pˆ  P  pˆ  z / 2ˆ pˆ

or

0.25  (1.96)(0.03165)  P  0.25  (1.96)(0.03165)

or (0.188, 0.312) b. As stated, N  854, n  81, and x  50. For a 99% confidence level, note that z / 2  z0.005  2.576. Next calculate pˆ and ˆ pˆ . x 50   0.6173 n 81 pˆ (1  pˆ ) ( N  n) 0.6173(1  0.6173) 854  81 ˆ p2ˆ      0.002676 n 1 ( N  1) 81  1 854  1 pˆ 

ˆ pˆ  0.002676  0.05173 So the 99% confidence interval is pˆ  z / 2ˆ pˆ  P  pˆ  z / 2ˆ pˆ

or

7-17

0.6173  (2.576)(0.05173)  P  0.6173  (2.576)(0.05173)

or (0.484, 0.751)


7-18


7.55

Answer varies Within Minitab, go to Calc  Make Patterned Data… in order to generate a simple set of numbers of size ‘n’ or ‘N’. Enter first value as 1, last value as 854which is the total number of pages in the text. Go to Calc  Random Data  Sample from Columns… in order to generate a simple random sample of size ‘n’. “Sample ____ rows from column(s):” Enter 50 as the number of rows to sample from. The results will be the observation numbers in the list to include in the sample. Then, count the number of pages which contains figure. Divide this number by 50 and the result will be the required proportion.

7.56

a. x  9.7, s  6.2 , ˆ x 

( s)2 N  n (6.2) 2 139 = .7539    n N 1 50 188

95% confidence interval: 9.7  2.01 (.7539) or (8.1847, 11.2153) b. 99% confidence interval: Nx  t /2 Nˆ x  N   Nx  t /2 Nˆ x where, Nx  (189)(9.7)  1833.30 Nˆ x  189  .7539 = 142.4871 1833.30 2.68(142.4871) 1451.4346 < N  < 2215.1654 7.57

a. x  127.43 s 2 N  n (43.27)2 760 2   b. ˆ x   = 28.9569 n N 1 60 819 c. 90% confidence interval: 127.43  1.671 ( 28.9569 ) = (118.438, 136.422) d. [137.43 – 117.43]/2 = 10 = t /2 28.9569 , solving for t: 1.858 yields a confidence level of 93.19% or an  of .0681 e. 95% confidence interval: using ˆ x 2  28.9569 Nx  t /2 Nˆ x  N   Nx  t /2 Nˆ x , where Nx  (820)(127.43)  104, 492.6

Nˆ x  820 28.9569  4412.584 104,492.6 2.001(4412.584) = 95,663.0194 < N  < 113,322.1806



7.58

7-19

a. 99% confidence interval:

(5.32) 2 85 = .6964  40 124 7.28  2.708 (.6964) = (5.3941, 9.1659)

ˆ x 

b. 90% confidence interval: Nx  t /2 Nˆ x  N   Nx  t /2 Nˆ x , where Nx  (125)(7.28)  910 Nˆ x  125  .6964 = 87.05 910 1.685(87.05) = 763.3208 < N  < 1,052.6793

7.59

a. false: as n increases, the confidence interval becomes narrower for a given N and s2 b. true c. true: the finite population correction factor is larger to account for the fact that a smaller proportion of the population is represented as N increases relative to n. d. true

7.60

x = 143/35 = 4.0857 90% confidence interval: Nx  t /2 Nˆ x  N   Nx  t /2 Nˆ x , where Nx  (120)(4.0857)  490.2857

(3.1)2 (120  35) = 53.1429 Nˆ x  120  35 (120  1) 490.2857 1.691(53.1429) = 400.4211< N  < 580.1503

7.61

pˆ = x/n = 39/400 = .0975

ˆ pˆ  [ pˆ (1  pˆ ) / (n  1)][( N  n) / ( N  1)]  [(.0975)(.9025) / (399)][(1395 400) / (1395 1)] = .0125 95% confidence interval: .0975  1.96(.0125): .073 up to .1220 7.62

pˆ = x/n = 56/100 = .56

ˆ pˆ  [ pˆ (1  pˆ ) / (n  1)][( N  n) / ( N  1)]  [(.56)(.44) / 99][(420 100) / (420 1)] = .0436 90% confidence interval: .56  1.645(.0436): .4883 up to .6317


7-20 7.63


pˆ = x/n = 110/120 = .9167

ˆ pˆ  [ pˆ (1  pˆ ) / (n  1)][( N  n) / ( N  1)]  [(.9167)(.0833) / (119)][(1200 120) / (1200 1)] = .024 95% confidence interval: .9167  1.96(.024): .8696 up to .9638 7.64

pˆ = x/n = 31/80 = .3875

ˆ pˆ  [ pˆ (1  pˆ ) / (n  1)][( N  n) / ( N  1)]  [(.3875)(.6125) / (79)][(420  80) / (420 1)] = .0494 90% confidence interval: .3875  1.645(.0494): .3063 up to .4687 Therefore, 128.646 < NP < 196.854 or between 129 and 197 students intend to take the final. 7.65

To estimate the mean of a normally distributed population:

z   n 2

a.

 2

2

=

ME 2

z   n 2

(2.582 )(402 ) = 426.01. Take a sample of size n = 427. 52

2

(2.582 )(402 ) b. = = 106.502. Take a sample of size n = 107. 102 ME 2 c. In order to cut the ME in half, the sample size must be quadrupled.  2

7.66

To estimate the population proportion: a. n 

.25  z 2 

2

ME 2

.25  z 2 

2

.25 1.96  = = 1067.111. Take a sample of size n = 1068. .032 2

.25 1.96  b. n  = = 384.16. Take a sample of size n = 385. 2 .052 ME c. In order to reduce the ME in half, the sample size must be increased by a larger proportion. 7.67

2

To estimate the population proportion: a. n 

.25  z 2 

2

ME 2

.25  z 2 

2

.25  2.58  = = 665.64. Take a sample of size n = 666. .052 2

.25 1.645  b. n  = = 270.6. Take a sample of size n = 271. ME 2 .052 c. In order to increase the confidence level for a given margin of error, the sample size must be increased. 2



7.68

7-21

a. z.05  1.645, ME  .04

.25( z / 2 )2 (.25)(1.645) 2 =  422.8 , take n = 423 ME 2 (.04) 2 (.25)(1.96)2 b.  600.25 , take n = 601 (.04)2 (.25)(2.33)2 c.  542.89 , take n = 543 (.05)2 n

7.69

z.005  2.58, ME  .05

.25( z / 2 )2 (.25)(2.58) 2 n =  665.64 , take n = 666 ME 2 (.05) 2 7.70

z.05  1.645, ME  .03

n

7.71

.25( z / 2 )2 (.25)(1.645) 2 =  751.7 , take n = 752 ME 2 (.03) 2

N 2 Use the equation n  to find the sample size needed. ( N  1) x2   2 50  25.51. a. Since 1.96 x  50,  x  1.96 N 2 (1650)(500) 2 n   311.8, take n  312 ( N  1) x2   2 (1650  1)(25.51) 2  (500) 2 100  51.02. 1.96 N 2 (1650)(500)2 n   90.8, take n  91 ( N  1) x2   2 (1650  1)(51.02)2  (500)2

b. Since 1.96 x  100,  x 

200  102.04. 1.96 N 2 (1650)(500)2 n   23.7, take n  24 ( N  1) x2   2 (1650  1)(102.04)2  (500)2

c. Since 1.96 x  200,  x 

d. The required sample size for part (b) is larger than that for part (c), and the required sample size for part (a) is larger than that for parts (b) and (c). This shows that the required sample size increases as 1.96 X decreases. Thus, the required sample size increases as the desired standard deviation,  X , and the desired variance,  X2 , of the sample mean decreases.


7-22

7.72


N 2 to find the sample size needed. ( N  1) x2   2 50  25.51. a. Since 1.96 x  50,  x  1.96 Use the equation n 

N 2 (3300)(500) 2 n   344.2 , take n  345 ( N  1) x2   2 (3300  1)( 25.51) 2  (500) 2 b. From part (a),  x  25.51.

n

N 2 (4950)(500)2   356.6, take n  357 ( N  1) x2   2 (4950  1)(25.51)2  (500)2

c. From part (a),  x  25.51.

n

N 2 (5000000)(500) 2   384.1, take n  385 ( N  1) x2   2 (5000000  1)(25.51) 2  (500) 2

d. The required sample size for part (b) is larger than that for part (a), and the required sample size for part (c) is larger than that for parts (a) and (b). This shows that the required sample size increases as N increases. 7.73

0.25 N to find the maximum sample size needed. ( N  1) p2ˆ  0.25 0.05  0.02551. a. Since 1.96 pˆ  0.05,  pˆ  1.96 Use the equation nmax 

nmax 

0.25 N 0.25(2500)   333.1, take n  334 2 ( N  1) pˆ  0.25 (2500  1)(0.02551)2  0.25

b. Since 1.96 pˆ  0.03,  pˆ 

nmax 

0.03  0.015306. 1.96

0.25 N 0.25(2500)   748.1, take n  749 2 ( N  1) pˆ  0.25 (2500  1)(0.015306) 2  0.25



7-23

c. The sample size for part (b) is larger than that for part (a). This shows that the required sample size increases as 1.96 pˆ decreases. Thus, the sample size increases as the desired standard deviation,  pˆ , and the desired variance,  p2ˆ , of the sample proportion decreases. 7.74

2000 812(20000)2 x   1020.4 , n   261.0038 . Take 262 observations. 1.96 811(1020.4)2  (20000)2

7.75

x 

7.76

 pˆ 

7.77

 pˆ 

7.78

a. n  25, x  227.60, s  41.86, t24,.05  1.711

2000  1215.8055 1.645 N 2 (400)(10, 000) 2 n  = 57.988. Take 58 observations. 2 2 2 ( N  1) x   2 (399)(1215.8055)  (10, 000)

.05  .0194 2.575 .25 N (.25)320 n   216.18 = 217 observations. 2 ( N  1) pˆ  .25 319(.0194)2  .25

417(.25) .04  210.33 = 211 observations.  .0243 , n  1.645 416(.0243)2  .25

 = 227.60  1.711(41.86/ 25 ) = 213.2741 up 90% confidence interval: x  t 2  s  n  to 241.9259. Assume that the population is normally distributed. b. Width for 95% and 98% confidence intervals: [95%]: Width = 2ME = 2 t 2 s   2 2.064 8.3728  = 34.563 n  [98%]: Width = 2ME = 2 t 2 s   2 2.492 8.3728  = 41.73 n  7.79

n  16, x  150, s  12, t15,.025  2.131

 = 150  2.131(12/4) = 143.607 up to 95% confidence interval: x  t 2  s  n  156.393..It is recommended that he stock 157 gallons. 7.80

n  25, x  28.5, s  6.8, t24,.05  1.711

90% confidence interval = 28.5  1.711(6.8/ 25 ) = 26.173 up to 30.827 7.81

Results from Minitab:


7-24


Descriptive Statistics: Passengers7_68 Variable Passenge Variable Passenge

N 50 Minimum 86.00

Mean 136.22 Maximum 180.00

One-Sample T: Passengers7_68 Variable Passengers8_

N 50

Mean 136.22

Median 141.00 Q1 118.50

StDev 24.44

TrMean 136.75

StDev 24.44

SE Mean 3.46

Q3 152.00

SE Mean 3.46

(

95.0% CI 129.27, 143.17)

95% confidence interval: 129.27 up to 143.17 n  25, x  12.5, s  3.8, t24,.025  2.064

7.82

b. For a 95% confidence interval, 12.5  2.064(3.8/5) = 10.9314 up to 14.0686 7.83

7.84

a. The minimum variance unbiased point estimator of the population mean is the  X i  27  3.375. The unbiased point estimate of the variance: sample mean: X  n 8 2 2  xi  nx  94.62  8(3.375)2  .4993 s2  n 1 7 x 3 b. pˆ    .375 n 8 Width = 3.69 – 3.49 = .2, ME = .1 = z / 2 (1.045/ 457),

z / 2  2.05

  2[1  Fz (2.05)]  .0404 100(1–.0404) = 95.96% 7.85

n  174, x  6.06, s  1.43

Width = 6.16 – 5.96 = .2, ME = .1 = z /2 (1.43 / 174), z /2  .92

  2[1  Fz (.92)]  .3576 100(1–.3576) = 64.24% 7.86 n = 33 accounting students who recorded study time a. An unbiased, consistent, and efficient estimator of the population mean is the sample mean, x = 8.545 b. The sampling error for a 95% confidence interval using degrees of freedom = 32,  = 1.3536 ME  2.037  3.817  33  



7-25

7.87 n = 25 patient records – the average length of stay is 6 days with a standard deviation of 1.8 days a. The reliability factor for a 95% interval estimate, t24,.025  2.064 b. The LCL for a 99% confidence interval estimate of the population mean  . The LCL = 4.9931 days. = 2.797 1.8 ME  t 2 s  n 25   x 100   .4 n 250 a. The standard error to estimate population proportion of first timers pˆ (1  pˆ ) .4(1  .4) = .03098  n 250 b. Since no confidence level is specified, we find the sampling error (Margin of Error) for a 95% confidence interval. ME = 1.96 (.03098) = .0607 c. For a 92% confidence interval, ME = 1.75 (.03098) = .05422 92% confidence interval for estimating the proportion of repeat fans, .6  .05422 giving .5458 up to .6542.

7.88 n = 250, x = 100, pˆ 

7.89 n  20, x  60.75, s  21.8316 a. 90% confidence interval reliability factor = t19,.05  1.729 b. For a 99% confidence interval,  21.8316  LCL = 60.75 – 2.861   = 46.78 or approximately 47 20  

passengers.

7.90 n = 500 motor vehicle registrations, 200 were mailed, 160 paid in person, remainder paid online. a. 90% confidence interval to estimate the population proportion to pay for vehicle registration renewals in person. Test and CI for One Proportion

Test of p = 0.5 vs p not = 0.5 Sample X N Sample p 90% CI 1 160 500 0.320000 (0.285686, 0.354314)

The 90% confidence interval is from 28.5686% up to 35.4314% b. 95% confidence interval to estimate the population proportion of online renewals. Test and CI for One Proportion

Test of p = 0.5 vs p not = 0.5 Sample X N Sample p 95% CI 1 140 500 0.280000 (0.240644, 0.319356)

The 95% confidence interval is from 24.0644% up to 31.9356%


7-26 7.91


From the data in 7.90, find the confidence level if the interval extends from 0.34 up to 0.46. ME = ½ the width of the confidence interval = (0.46 – 0.34)/2 = 0.12 / 2 = 0.06 (0.4)(0.6) pˆ (1  pˆ ) or 0.06  z  2 ME  z  2 500 n z  2 . 74 Solving for z: 2 Area from the z-table = (.5 – .0031) × 2 = .4969 × 2 = .9938. The confidence level is 99.38%

7.92 From the data in 7.90, find the confidence level if the interval extends from 23.7% up to 32.3%. ME = ½ the width of the confidence interval = (.323 – .237)/2 = .086 / 2 = .043 and pˆ  .28

pˆ (1  pˆ ) n .28(1  .28) .043  z 2 500 solving for z: z 2  2.14 ME  z 2

Area from the z-table = (.5 – .0162) × 2 = .4838 × 2 = .9676. The confidence level is 96.76% 7.93

a. The margin of error for a 99% confidence interval pˆ  x  250  .7143 , n 350 .7143(1  .7143) pˆ (1  pˆ ) = ME  2.58 ME  z 2 350 n ME = .0623 b. The margin of error will be smaller (more precise) for a lower confidence level. The difference in the equation is the value for z which would drop from 2.58 down to 1.96.

7.94

The 98% confidence interval of the mean age of online renewal users. n = 460, sample mean = 42.6, s = 5.4.  = 42.6 ± .58664 = 42.0134 up to x  t 2 s = 42.6  2.33  5.4 43.18664  n 460  

7.95 a. x 

747 (11.44)2  90  10   74.7 , s = 11.44, ˆ 2x     11.766 10 10  90  1 

90% confidence interval: 74.7  1.833 11.766 : 68.412 up to 80.988 b. The interval would be wider; the t-score would increase to 2.262



7.96

7-27

(149.92)2  272  50     368.242 50  272  1  99% confidence interval for the population mean: 492.36  2.68 368.242 440.9319 up to 543.7881 a. ˆ 2x 

b. Nˆ x  (272)( 368.242)  5219.577 95% confidence interval for the population total: (272 × 492.36)  2.01(5219.577) 123430.57 up to 144413.27 c. The 90% interval is narrower; the t-score would decline to 1.677 7.97

pˆ 

.6(.4)  148  60  36  .6 , ˆ 2pˆ     .0024 60  1  148  1  60

95% confidence interval: .6  1.96 .0024 : .504 up to .696 7.98. N  20, n  10, x  257, s  37.16, t9,.05  1.833

(37.16)2  20  10     72.677 10  20  1  a. 90% confidence interval for the average number of new prescriptions: 257  1.833 72.677 = 241.3735 up to 272.6265, assuming that the population of the number of new prescriptions written for the new drug is normal.

ˆ 2x 

b. Reducing ME in part (a) by half = 15.6265/2 = 7.8133.

z2 /2 2 (1.642 )(72.677 2 )   3.202 ME 2 7.81332 n0 N 3.202  20 n   2.88  3 n0  ( N  1) 3.202  (20  1)

We know that n0 

7.99

Sample taken to estimate approval ratings for a 95% confidence level given a 45% approval rating and a margin of error of .035: ( pˆ )(1  pˆ )( z /2 ) 2 (.45)(.55)(1.96) 2 n =  776.16 . ME 2 (.035) 2 Take n = 777 observations.


7-28


7.100  x 

2000 328(12000)2  1215.8 , n   75.28 . Take 76 observations 1.645 327(1215.8)2  (12000)2

7.101  pˆ 

527(.25) .06  177.43 . Take 178 observations  .0306 , n  1.96 526(.0306)2  .25

z2 / 2 2 z2 / 2 2 2 7.102 We know that n0  , where ME  z / 2 X . This gives n0  or  ( z / 2 X ) 2  X2 ME 2

 2  n0 X2 . N 2 Substituting this into gives ( N  1) X2   2

N (n0 X2 ) ( N  1)  n0 2 X

2 X



 X2  Nn0  [( N  1)  n0 ] 2 X



n0 N . n0  ( N  1)

7.103 n  75, x  3.81, s  .10, t74,.025  1.993 For a 95% confidence interval, 3.81  1.993(.10/ 75 ) = 3.7843 up to 3.8314 gallons


Newbold Stat8 Ism 07

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