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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
n
point 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, p̂ .
x 3
pˆ    0.375
n 8
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
Chapter 7: Estimation: Single Population
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
Variance
0.0512
Median
2.5500
Q3
2.7250
Maximum
3.0000
Range
0.8000
CoefVar
8.81
Sum
25.7000
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
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
7-3
7-4
7.6
Statistics for Business & Economics, 8th edition
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 ) 
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
Chapter 7: Estimation: Single Population
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
c. Relative efficiency between Y and X :
  1.25
Var ( X ) 4
Var ( Z ) 10
Relative efficiency between Z and X :
  1.111
Var ( X ) 9
7.8
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
7.9
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
= 2.33 12
ME  z 2 

n
64


b. 99% confidence interval, n = 120; standard deviation = 100
 = 23.552
= 2.58 100
ME  z 2 

n
120 

7.11
Calculate the width to estimate the population mean, for
a. 90% confidence level, n = 100, variance = 169
  = 4.277
 = 2 1.645 13
width = 2ME = 2  z 2 




100  
n



b. 95% confidence interval, n = 120, standard deviation = 25
  = 8.9461
 = 2 1.96  25
width = 2ME = 2  z 2 




120  
n



Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
7-5
7-6
Statistics for Business & Economics, 8th edition
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
= 1.645(.5) = .8225
ME  z 2 
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

Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
Chapter 7: Estimation: Single Population
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
Q1
27.25
TrMean
35.39
StDev
9.87
Variance
97.36
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
Statistics for Business & Economics, 8th edition
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:
 = 45,659.35 miles to 49,340.65 miles
 = 47500  1.753  4200
x  tv, 2  s



16 
n


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
Q1
43.256
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)
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
Chapter 7: Estimation: Single Population
7.24
Results for: Sugar.xls
Descriptive Statistics: Weights
Variable
Weights
N
100
Variable
Weights
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
n
 = 3.7246
= 1.966  37.89

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
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
7-9
7-10
Statistics for Business & Economics, 8th edition
7.30 Find the margin of error to estimate the population proportion for each of the following:
a. n = 350, p̂ = .3,  = .01
pˆ (1  pˆ )
.3(.7)
= .0631
 2.576
n
350
b. n = 275, p̂ = .45,  = .05
ME  z 2
pˆ (1  pˆ )
.45(.55)
= .0588
 1.96
n
275
c. n = 500, p̂ = .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, p̂ = .10
pˆ (1  pˆ )
.10(1  .10)
= .10  2.326
= .0671 to .1329
n
450
b. 95% confidence level; n = 240, p̂ = .01
pˆ  z 2
pˆ (1  pˆ )
.01(1  .01)
= .01  1.96
= –0.0026 to .0226
n
240
c.  = .04; n = 265, p̂ = .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
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
Chapter 7: Estimation: Single Population
n = 142, 87 answered GMAT or GRE is ‘very important’.
95% confidence interval for the population proportion:
pˆ (1  pˆ )
.6127(1  .6127)
= .6127  1.96
= .5326 to .6928
pˆ  z 2
n
142
7.33
7.34
n  95, pˆ  67 / 95  .7053, z.005  2.58
99% confidence interval:
pˆ (1  pˆ )
.7053(.2947)
= .7053  (2.58)
= .5846 up to .8260
pˆ  z / 2
n
95
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
n
350
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
pˆ (1  pˆ )
.80(1  .80)
= .80  2.326
= .75348
n
400
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%
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
7-11
7-12
Statistics for Business & Economics, 8th edition
7.39
n  540, pˆ  320 / 540  .5926, z.025  1.96
95% confidence interval:
pˆ (1  pˆ )
.5926(.4074)
= .5926  (1.96)
= .5512 up to .634
pˆ  z / 2
n
540
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:
pˆ (1  pˆ )
.3261(.6739)
= .3261  (1.645)
= .29009 up to .36209
pˆ  z / 2
n
460
7.41
a. n  246, pˆ  232 / 246  .9431, z.01  2.326
98% confidence interval:
pˆ (1  pˆ )
.9431(.0569)
= .9431  (2.326)
= .9087 up to .9775
pˆ  z / 2
n
246
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
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
Chapter 7: Estimation: Single Population
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
 2 n 1,1 /2
15(8) 2
=
= 153.3546
6.26
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
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
82.64463
7-13
7-14
7.45
Statistics for Business & Economics, 8th edition
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.
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
the Appendix,
Chapter 7: Estimation: Single Population
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
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
7.51
a. ˆ x2 
7.52
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
102 1200  80

 1.1676
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
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)
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
7-16
Statistics for Business & Economics, 8th edition
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̂ 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̂ 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)
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
Chapter 7: Estimation: Single Population
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 p̂ and ˆ p̂ .
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 p̂ and ˆ p̂ .
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)
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
7-18
Statistics for Business & Economics, 8th edition
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
ˆ


b.  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
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
Chapter 7: Estimation: Single Population
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
p̂ = 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
p̂ = 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
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
7-20
7.63
Statistics for Business & Economics, 8th edition
p̂ = 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
p̂ = 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.
2
ME
.052
c. In order to increase the confidence level for a given margin of error, the sample size
must be increased.
2
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
Chapter 7: Estimation: Single Population
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.
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
7-22
7.72
Statistics for Business & Economics, 8th edition
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
0.25N
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 
0.25N
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
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
Chapter 7: Estimation: Single Population
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 p̂ decreases. Thus, the sample size increases as the desired
standard deviation,  p̂ , and the desired variance,  p̂2 , of the sample proportion
decreases.
7.74
x 
7.75
x 
7.76
 pˆ 
7.77
 pˆ 
2000
812(20000)2
 1020.4 , n 
 261.0038 . Take 262 observations.
1.96
811(1020.4)2  (20000)2
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
.04
417(.25)
 .0243 , n 
 210.33 = 211 observations.
1.645
416(.0243)2  .25
7.78 a. n  25, x  227.60, s  41.86, t24,.05  1.711
 = 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
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
7-24
7.81
Statistics for Business & Economics, 8th edition
Results from Minitab:
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
TrMean
136.75
Q1
118.50
StDev
24.44
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
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
7.84 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 

Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
Chapter 7: Estimation: Single Population
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%
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
7-26
7.91
Statistics for Business & Economics, 8th edition
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
pˆ (1  pˆ )
.7143(1  .7143)
= ME  2.58
ME  z 2
n
350
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
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
Chapter 7: Estimation: Single Population
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 
36
.6(.4)  148  60 
 .6 , ˆ 2pˆ 

  .0024
60
60  1  148  1 
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 )(72.6772 )

 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.
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.
7-28
Statistics for Business & Economics, 8th edition
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 
.06
527(.25)
 .0306 , n 
 177.43 . Take 178 observations
1.96
526(.0306)2  .25
7.102 We know that n0 
z2 / 2 2
z2 / 2 2
2
n


,
where
.
This
gives
or
ME

z

0
 /2 X
( z / 2 X ) 2  X2
ME 2
 2  n0 X2 .
Substituting this into
N (n0 X2 )
( N  1)  n0
2
X
2
X

N 2
gives
( N  1) X2   2
 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
Copyright © 2013 Pearson Education, Inc. publishing as Prentice Hall.