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Chapter 7:
Estimation: Single Population
7.1
a. Check for nonnormality
The distribution shows no significant evidence of nonnormality.
b. Point estimate of the population mean that is unbiased, efficient and
consistent.
 X i  560 = 7.0
Unbiased point estimator is the sample mean: X 
n
8
c. The unbiased point estimate of the variance of the sample mean:
s 2 (2.268) 2
Var ( X )  
 .64298
n
8
Chapter 7: Estimation: Single Population
7.2
173
a. There appears to be no evidence of non-normality, as shown by the normal
probability plot shown next.
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
n
efficient point 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
174
7.3
Statistics for Business & Economics, 7th edition
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
Sum
25.7000
IQR
0.3250
a. Unbiased point estimator of the population mean is the sample mean:
 X i  2.57
X
n
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
e. Unbiased estimate of the variance of the sample proportion:
pˆ (1  pˆ ) .7(1  .7)
Var ( pˆ ) 

 .021
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  24.42
X
n
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
e. Unbiased estimate of the variance of the sample proportion:
pˆ (1  pˆ ) .25(1  .25)
Var ( pˆ ) 

 .015625
n
12
Chapter 7: Estimation: Single Population
a. Check each variable for normal distribution:
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
45
50
55
Meals
Average: 50.1
StDev: 2.46842
N: 30
Anderson-Darling Normality Test
A-Squared: 0.413
P-Value: 0.318
Normal Probability Plot
.999
.99
.95
Probability
7.5
.80
.50
.20
.05
.01
.001
15
20
25
Attendance
Average: 21.24
StDev: 2.50466
N: 25
Anderson-Darling Normality Test
A-Squared: 0.377
P-Value: 0.383
175
176
Statistics for Business & Economics, 7th edition
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
9
10
11
12
13
14
15
16
Ages
Average: 12.24
StDev: 1.96384
N: 25
Anderson-Darling Normality Test
A-Squared: 0.569
P-Value: 0.126
No evidence of non-normality in Meals, Attendance or Ages
b. Unbiased estimates of population mean and variance:
Descriptive Statistics: Meals, Attendance, Ages
Variable
Meals
Attendan
Ages
Variable
Meals
Attendan
Ages
N
30
25
25
Minimum
45.000
15.000
9.000
Mean
50.100
21.240
12.240
Maximum
55.000
25.000
16.000
Median
50.000
21.000
12.000
Q1
48.000
19.500
10.000
TrMean
50.192
21.348
12.217
StDev
2.468
2.505
1.964
Q3
52.000
23.500
14.000
Variable
Unbiased estimate of mean Unbiased estimate of variance (s2)
Meals
50.100
(2.468)2 = 6.0910
Attendance
21.240
(2.505)2 = 6.2750
Ages
12.240
(1.964)2 = 3.8573
7.6
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 ) 
SE Mean
0.451
0.501
0.393
Chapter 7: Estimation: Single Population
1
1
12 2
 Var ( X 1 )  Var ( X 2 ) 

n 4
4
2 8
4
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
is
most
efficient
since
Var ( X )  Var (Y )  Var ( Z )
X
Var (Y ) 5
c. Relative efficiency between Y and X :
  2.5
Var ( X ) 2
Var ( Z ) 20
Relative efficiency between Z and X :

 2.222
Var ( X ) 9
b. Var ( X ) 
a. Evidence of non-normality?
Normal Probability Plot for Leak Rates (
ML Estimates
99
ML Estimates
95
Mean
0.0515
StDev
0.0216428
90
Goodness of Fit
80
Percent
7.7
2
AD*
70
60
50
40
30
20
10
5
1
0.00
0.05
0.10
Data
No evidence of nonnormality exists.
b. The minimum variance unbiased point estimator of the population
 X i  .0515
mean is the sample mean: X 
n
c. The unbiased point estimate of the variance of the sample mean:
s 2 x  (.0216428)2  .0004684
Var ( X ) 
2
n
ˆ (X ) 
; Var
s 2 .0004684

 .00000937
n
50
0.596
177
178
7.8
Statistics for Business & Economics, 7th edition
a. Evidence of non-normality?
No evidence of the data distribution coming from a non-normal
population
b. The minimum variance unbiased point estimator of the population
 X i  3.8079
mean is the sample mean: X 
n
Descriptive Statistics: Volumes
Variable
Volumes
Variable
Volumes
N
75
Minimum
3.5700
Mean
3.8079
Median
3.7900
Maximum
4.1100
Q1
3.7400
TrMean
3.8054
StDev
0.1024
Q3
3.8700
c. Minimum variance unbiased point estimate of the population variance
is the sample variance s2 = .0105
7.9
Reliability factor for each of the following:
a. 96% confidence level: z 2 = +/- 2.05
b. 88% confidence level: z 2 = +/- 1.56
c. 85% confidence level: z 2 = +/- 1.44
d.   .07 z 2 = +/- 1.81
e.  2 = .07 z 2 = +/- 1.48
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
= 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



SE Mean
0.0118
Chapter 7: Estimation: Single Population
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 

 = 506.2652 to 513.73478
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
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
100(1-.0404)% = 95.96%
7.14
a. Find the reliability factor for 92% confidence level: z 2 = +/- 1.75
b. Calculate the standard error of the mean

= .63246
 6
n
90
c. Calculate the width of the 92% confidence interval for the population mean
  = 2.2136
 = 2 1.75  6
width = 2ME = 2  z 2 



90  
n 



7.15
a. n  25, x  2.90,   .45, z.025  1.96
 = 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
100(1-.3174)% = 68.26%
7.16 a. n  16, x  4.07,   .12, z.005  2.58
4.07  2.58(.12/4) = 3.9926 up to 4.1474
b. narrower since the z score for a 95% confidence interval is smaller
than the z score for the 99% confidence interval
c. narrower due to the smaller standard error
d. wider due to the larger standard error
7.17
Find the reliability factor tv ,
a.
b.
c.
d.
2
n = 20, 90% confidence level; t = 1.729
n = 7, 98% confidence level; t = 3.143
n = 16, 95% confidence level; t = 2.131
n = 23, 99% confidence level; t = 2.819
179
180
Statistics for Business & Economics, 7th edition
7.18 Find the ME
a. n = 20, 90% confidence level, s = 36
 = 13.9182
= ME  1.729  36
ME  t 2 s

n
120 

b. n = 7, 98% confidence level, s = 16
= ME  3.143 16  = 19.007
ME  t 2 s
n
7

c. n = 16, 99% confidence level, x1 = 15, x2 = 17, x3 = 13, x4 = 11, x5 = 14
 = 7.5407
x  14, s  2.58199 ME  5.841 2.58199

4

7.19 Time spent driving to work for n = 5 people
Descriptive Statistics: Driving_Ex7.19
Variable
Driving_Ex7.19
Variable
Driving_Ex7.19
N
5
N*
0
Mean
38.40
Minimum
30.00
SE Mean
2.66
Q1
32.50
TrMean
*
Median
40.00
StDev
5.94
Q3
43.50
Variance
35.30
Maximum
45.00
CoefVar
15.47
Range
15.00
Sum
192.00
IQR
11.00
a. Calculate the standard error
s
2.6571
 5.94138
n
5
b. Find the value of t for the 95% confidence interval
tv , 2 = 2.776
c. Calculate the width for a 95% confidence interval for the population mean
ME  t 2 s
= ME  2.776  5.94138  = 14.752
n
5

7.20
Find the LCL and UCL for each of the following:
a. alpha = .05, n = 25, sample mean = 560, s = 45
 = 560  2.064  45
 = 541.424 to 578.576
x  t 2  s



n
25




b. alpha/2 = .05, n = 9, sample mean = 160, sample variance = 36
 = 160  1.860  6  = 156.28 to 163.72
x  t 2  s



n
9


c. 1   = .98, n = 22, sample mean = 58, s = 15
 = 58  2.518 15
 = 49.9474 to 66.0526
x  t 2  s



n
22




Chapter 7: Estimation: Single Population
7.21
Calculate the margin of error to estimate the population mean for:
a. 98% confidence level; n = 64, sample variance = 144
 = 3.585
= ME  2.390 12
ME  t 2 s

n
64 

b. 99% confidence level; n = 120, sample standard deviation = 100
 = 24.2824
= ME  2.660 100
ME  t 2 s

n
120 

c. 95% confidence level; n = 200, sample standard deviation = 40
 = 5.65685
= ME  2.000  40
ME  t 2 s

n
200 

7.22
Calculate the width for each of the following:
a. alpha = 0.05, n = 6, s = 40
w  2ME  2t  2 s
 2  2.571 40   2(41.98425)  83.9685
n
6

b. alpha = 0.01, n = 22, sample variance = 400
  2(12.07142)  24.1428
w  2ME  2t  2 s
 2  2.831 20

n
22 

c. alpha = 0.10, n = 25, s = 50
  2(17.11)  34.22
w  2ME  2t  2 s
 2  1.711 50

n
25


7.23
181
a. 95% confidence interval:
Results for: TOC.xls
One-Sample T: Leak Rates (cc/sec.)
Variable
Leak Rates (
N
50
Mean
0.05150
StDev
0.02186
SE Mean
0.00309
95.0% CI
( 0.04529, 0.05771)
b. 98% confidence interval:
One-Sample T: Leak Rates (cc/sec.)
Variable
Leak Rates (
7.24
N
50
Mean
0.05150
StDev
0.02186
SE Mean
0.00309
98.0% CI
( 0.04406, 0.05894)
a.
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)
182
Statistics for Business & Economics, 7th edition
b. narrower since a smaller value of z will be used in generating the 80%
confidence interval.
7.25
n  9, x  157.82, s  38.89, t8,.025  2.306
margin of error:  2.306(38.89/3) =  29.8934
7.26
n  7, x  74.7143, s  6.3957, t6,.025  2.447
margin of error:  2.447(6.3957/ 7 ) =  5.9152
7.27
a. n  10, x  15.90, s  5.30, t9,.005  3.25
15.90  3.25(5.30/ 10 ) = 10.453 up to 21.347
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
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,.10  1.86
We must assume a normally distributed population
16.222  1.86(4.790/3) = 13.252 up to 19.192
7.30 Find the standard error of the proportion for
pˆ (1  pˆ )
.3(.7)
a. n = 250, p̂ = 0.3
= .02898

n
250
pˆ (1  pˆ )
.45(.55)
b. n = 175, p̂ = 0.45
= .03761

n
175
pˆ (1  pˆ )
.05(.95)
c. n = 400, p̂ = 0.05
= .010897

n
400
7.31 Find the margin of error for
a. n = 250, p̂ = 0.3, α = .05 z 2
b. n = 175, p̂ = 0.45, α = .08 z 2
c. n = 400, p̂ = 0.05, α = .04 z 2
pˆ (1  pˆ )
.3(.7)
= .056806
 1.96
n
250
pˆ (1  pˆ )
.45(.55)
= .05810
 1.75
n
175
pˆ (1  pˆ )
.05(.95)
= .02234
 2.05
n
400
Chapter 7: Estimation: Single Population
7.32
Find the confidence level for estimating the population proportion for
a. 92.5% confidence level; n = 650, p̂ = .10
pˆ (1  pˆ )
.10(1  .10)
= .10  1.78
= .079055 to .120945
n
650
b. 99% confidence level; n = 140, p̂ = .01
pˆ  z 2
pˆ (1  pˆ )
.01(1  .01)
= .01  2.58
= 0.0 to .031696
n
140
c. alpha = .09; n = 365, p̂ = .50
pˆ  z 2
pˆ  z 2
7.33
7.34
.5(.5)
pˆ (1  pˆ )
= .50  1.70
= .4555 to .5445
650
n
n = 142, 87 answered GMAT or GRE is ‘very important’. 95% confidence
interval for the population proportion:
pˆ (1  pˆ )
.61268(1  .61268)
= .61268  1.96
= .53255 to .6928
pˆ  z 2
n
142
n  95,
pˆ  67 / 95  .7053, z.005  2.58
pˆ (1  pˆ )
.7053(.2947)
= .7053  (2.58)
=
n
95
99% confidence interval: .5846 up to .8260
pˆ  z / 2
7.35 a. unbiased point estimate of proportion:
Tally for Discrete Variables: Adequate Variety
Adequate Variety
1
2
N=
Count CumCnt
135
135
221
356
356
Percent CumPct
37.92 37.92
62.08 100.00
x 135

 .3792
n 356
b. 90% confidence interval:
n  356, pˆ  135 / 356  .3792, z.05  1.645
pˆ 
pˆ (1  pˆ )
= .3792  (1.645) .3792(.6208) / 356 =
n
.3369 up to .4215
pˆ  z / 2
7.36
n  320,
pˆ  80 / 320  .25, z.025  1.96
pˆ (1  pˆ )
= .25  (1.96) .25(.75) / 320 =
n
95% confidence interval: .2026 up to .2974
pˆ  z / 2
183
184
7.37
Statistics for Business & Economics, 7th edition
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 = .545-.445 = .100; ME = 0.05 pˆ  0.495
pˆ (1  pˆ )
.495(.505)

 .0355
n
198
.05 = z / 2 (.0355), z / 2  1.41
  2[1  Fz (1.41)]  .0793
100(1-.1586)% = 84.14%
7.39
n  420,
pˆ  223/ 420  .5310, z.025  1.96
pˆ (1  pˆ )
= .5310  (1.96) .5310(.4690) / 420 =
n
95% confidence interval: .4833 up to .5787
The margin of error is .0477
pˆ  z / 2
7.40
n  246,
pˆ  40 / 246  .1626, z.01  2.326
pˆ (1  pˆ )
= .1626  (2.326) .1626(.8374) / 246 =
n
98% confidence interval: .1079 up to .2173
pˆ  z / 2
7.41
a. n  246,
pˆ  232 / 246  .9431, z.01  2.326
pˆ (1  pˆ )
= .9431  (2.326) .9431(.0569) / 246 =
n
98% confidence interval: .9087 up to .9775
b. n  246, pˆ  10 / 246  .0407, z.01  2.326
pˆ  z / 2
pˆ (1  pˆ )
= .0407  (2.326) .0407(.9593) / 246 =
n
98% confidence interval: .0114 up to .0699
pˆ  z / 2
Chapter 7: Estimation: Single Population
7.42
a. n  21, s 2  16, taking   .05,  220,.975  9.59,  220,.025  34.17
(n  1) s 2

2
n 1, / 2
2 
(n  1)s 2

2
=
n 1,1 / 2
21(16)
21(16)
2 
= 9.8332 <  2
34.17
9.59
<
35.036
b. n  16, s  8,  215,.975  6.26,  215,.025  27.49
(n  1) s 2
 2 n1, / 2
 
2
(n  1)s 2
 2 n1,1 / 2
15(8)2
15(8)2
2
=
= 34.9218 <  2
 
27.49
6.26
<
153.3546
c. n  28, s  15,  227,.995  11.81,  227,.005  49.64
(n  1) s 2
 2 n1, / 2
 
2
(n  1)s 2
 2 n1,1 / 2
=
28(15)2
28(15)2
= 126.9138 <  2 <
2 
49.64
11.81
533.446
7.43
a. n  21, s 2  16, taking   .05,  220,.975  9.59,  220,.025  34.17
(n  1) s 2

2
n 1, / 2
2 
(n  1)s 2

2
=
n 1,1 / 2
21(16)
21(16)
2 
= 9.8332 <  2
34.17
9.59
<
35.036
b. n  16, s  8,  215,.975  6.26,  215,.025  27.49
(n  1) s 2
 2 n1, / 2
2 
(n  1)s 2
 2 n1,1 / 2
=
15(8)2
15(8)2
= 34.9218 <  2
2 
27.49
6.26
<
153.3546
c. n  28, s  15,  227,.995  11.81,  227,.005  49.64
(n  1) s 2
 2 n1, / 2
2 
(n  1)s 2
 2 n1,1 / 2
=
28(15)2
28(15)2
= 126.9138 <  2 <
2 
49.64
11.81
533.446
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
56.25879
 
2
(n  1)s 2

2
n 1,1 / 2
=
4(10)
4(10)
2 
=
9.49
.711
4.21496 <  2
<
185
186
Statistics for Business & Economics, 7th edition
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
<
82.6446
7.45
No evidence of non-nomality
Probability Plot of Leak Rates (cc/sec.)
Normal
99
Mean
StDev
N
AD
P-Value
95
90
0.0515
0.02186
50
0.359
0.437
Percent
80
70
60
50
40
30
20
10
5
1
0.00
0.02
0.04
0.06
0.08
Leak Rates (cc/sec.)
0.10
0.12
n = 50, s 2  0.000478 Since df = 49 is not in the Chi-Square Table in
the Appendix, we will approximate the interval using df = 50.
49(0.000478)
49(0.000478)
(n  1) s 2
(n  1)s 2
2

 2 



2
2
71.42
32.36
 n1, / 2
 n1,1 / 2
= 3.279E-4 <  2 < 7.238E-4
7.46
n  10, s 2  28.898,  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(28.898)
9(28.898)
2 
= 15.3713 up to 78.1027
16.92
3.33
7.47
n  20, s 2  6.62,  219,.025  32.85,  219,.975  8.91
(n  1) s 2
(n  1)s 2
19(6.62)
19(6.62)
2 
32.85
8.91
 n1, / 2
 n1,1 / 2
= 3.8289 up to 14.1167. Assume that the population is normally
distributed.
2
2 
2
=
Chapter 7: Estimation: Single Population
7.48
n  18, s 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
distributed.
2
2 
2
=
7.49
a. n  15, s 2  (2.36)2  5.5696
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
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
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)(1.1676)    142  (1.990)(1.1676)
or (139.68, 144.32)
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)(0.6667)    232.4  (1.987)(0.6667)
or (231.08, 233.72)
187
188
Statistics for Business & Economics, 7th 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)(0.6049)    59.3  (1.972)(0.6049)
or (58.11, 60.49)
7.53
a. A 100(1   )% confidence interval for the population total is obtained from the
following formula.
Nx  t n 1, / 2 Nˆ x  N  Nx  t n 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  t n 1, / 2 Nˆ x  N  Nx  t n 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  t n 1, / 2 Nˆ x  N  Nx  t n 1, / 2 Nˆ x
or
(2100)(1325)  (2.35)(8446.684)  N  (2100)(1325)  (2.35)(8446.684)
or
(2762650, 2802350)
Chapter 7: Estimation: Single Population
7.54
a
189
A 100(1   )% confidence interval for the population total 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 



 0.0009956
n 1
( N  1)
160  1
1058  1
pˆ 
ˆ pˆ  0.0009956  0.03155
So the 95% confidence interval is
pˆ  z / 2ˆ pˆ  P  pˆ  z / 2ˆ pˆ
or
0.25  (1.96)(0.03155)  P  0.25  (1.96)(0.03155)
or
(0.1882, 0.3118)
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 80
pˆ (1  pˆ ) ( N  n) 0.6173(1  0.6173) 854  81
ˆ p2ˆ 



 0.002643
n 1
( N  1)
81  1
854  1
pˆ 
ˆ pˆ  0.002643  0.05141
So the 99% confidence interval is
pˆ  z / 2ˆ pˆ  P  pˆ  z / 2ˆ pˆ
or
0.6173  (2.576)(0.05141)  P  0.6173  (2.576)(0.05141)
or
(0.4849, 0.7497)
190
Statistics for Business & Economics, 7th edition
7.55
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 984 which 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.
7.56
a. x  9.7, s  6.2 , ˆ x 
( s)2 N  n
(6.2) 2 139
= .7519

n
N
50 189
9.7  1.96 (.7519) (8.2262, 11.1738)
b. 99% confidence interval:
Nx  Z / 2 Nˆ x  N   Nx  Z / 2 Nˆ x
where, Nx  (189)(9.7)  1833.30
s2
(6.2)2
N ( N  n) 
189(189  50) = 142.1167
n
50
1833.30 2.58(142.1167)
1466.6390 < N  < 2199.9610
Nˆ x 
7.57
a. x  127.43
s 2 N  n (43.27)2 760
2
b. ˆ x 
= 28.9216

n N
60 820
c. 127.43  1.645 ( 28.9216 )
(118.5834, 136.2766)
d. [137.43 – 117.43]/2 = 10 = z / 2 28.9216 , solving for z: 1.86
tabled value of .9686 yields a confidence level of 93.72% or an  of .0628
s 2 N  n (43.27)2 760

e. 95% confidence interval: using ˆ x 2 
n N
60 820
Nx  Z / 2 Nˆ x  N   Nx  Z / 2 Nˆ x , where
Nx  (820)(127.43)  104, 492.6
s2
(43.27)2
N ( N  n) 
820(820  60) = 4,409.8619
n
60
104,492.6 1.96(4409.8619)
95,849.2706 < N  < 113,135.9294
Nˆ x 
Chapter 7: Estimation: Single Population
7.58
(5.32) 2 85
= .6936
40 125
7.28  2.58 (.6936) (5.4904, 9.0696)
b. 90% confidence interval:
Nx  Z / 2 Nˆ x  N   Nx  Z / 2 Nˆ x , where Nx  (125)(7.28)  910
a. ˆ x 
s2
(5.32)2
N ( N  n) 
125(125  40) = 86.7054
n
40
910 1.645(86.7054)
767.3696 < N  < 1,052.6304
Nˆ x 
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  Z / 2 Nˆ x  N   Nx  Z / 2 Nˆ x , where
Nx  (120)(4.0857)  490.2857
s2
(3.1) 2
Nˆ x 
N ( N  n) 
120(120  35) = 52.9210
n
35
490.2857 1.645(52.9210)
403.2307 < N  < 577.3407
7.61
p̂ = x/n = 39/400 = .0975
 pˆ  [ pˆ (1  pˆ ) /(n  1)][( N  n) / N ]
 [(.0975)(.9025) /(399)][(1395  400) /1395] = .0125
95% confidence interval: .0975  1.96(.0125): .073 up to .1220
7.62
p̂ = 56/100 = .56
 pˆ  [ pˆ (1  pˆ ) /(n  1)][( N  n) / N ]
 [(.56)(.44) / 99][(420  100) / 420] = .0435
90% confidence interval: .56  1.645(.0435): .4884 up to .6316
191
192
7.63
Statistics for Business & Economics, 7th edition
p̂ = 37/120 = .3083
 pˆ  [ pˆ (1  pˆ ) /(n  1)][( N  n) / N ]
 [(.3083)(.6917) /(119)][(257  120) / 257] = .0309
95% confidence interval: .3083  1.96(.0309): .2477 up to .3689
7.64
p̂ = 31/80 = .3875
 pˆ  [ pˆ (1  pˆ ) /(n  1)][( N  n) / N ]
 [(.3875)(.6125) /(79)][(420  80) / 420] = .0493
90% confidence interval: .3875  1.645(.0493): .3064 up to .4686
128.688 < Np < 196.812 or between 129 and 197 students intend to take
the final.
a. n  10, x  257, s  37.2, t9,.05  1.833
7.65
 = 257  1.833(37.2/ 10 ) = 235.4318 up to 278.5628
x  t 2  s

n

assume that the population is normally distributed
b. 95% and 98% confidence intervals:
 = 257  2.262(37.2/ 10 ) = 230.39 up to 283.61
[95%]: x  t 2  s

n

 = 257  2.821(37.2/ 10 ) = 223.815 up to
[98%]: x  t 2  s

n

290.185
n  16, x  150, s  12, t15,.025  2.131
7.66
 = 150  2.131(12/4) = 143.607 up to 156.393
x  t 2  s

n

It is recommended that he stock 157 gallons.
n  50, x  30, s  4.2, z.05  1.645
7.67
= 30  1.645(4.2/ 50 ) = 29.0229 up to 30.9771
7.68
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
SE Mean
3.46
Q3
152.00
SE Mean
3.46
95% confidence interval: 129.27 up to 143.17
StDev
24.44
(
95.0% CI
129.27, 143.17)
Chapter 7: Estimation: Single Population
7.69
a. Use a 5% risk. Incorrect labels = 8/48 = 0.1667
0.1667(0.8333)
0.1667  1.96
 0.1667  0.1054 = 0.0613 up to 0.2721
48
b. For a 90% confidence interval,
0.1667(0.8333)
0.1667  1.645
 0.1667  0.0885 = 0.0782 up to
48
0.2552
7.70
a. The minimum variance unbiased point estimator of the population
 X i  27  3.375. The unbiased point
mean is the sample mean: X 
n
8
2
2
x

nx
94.62  8(3.375) 2

i
2

 .4993
estimate of the variance: s 
n 1
7
x 3
b. pˆ    .375
n 8
7.71
3.69 – 3.59 = 0.10 = z / 2 (1.045/ 457), z / 2  2.05
  2[1  Fz (2.05)]  .0404
100(1-.0404)% = 95.96%
7.72
n  174, x  6.06, s  1.43
6.16 – 6.06 = .1 = z / 2 (1.43/ 174), z / 2  .922
  2[1  Fz (.92)]  .3576
100(1-.3576)% = 64.24%
7.73 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. Find the sampling error for a 95% confidence interval; Using degrees of
freedom = 30,
 = 1.3568
ME  2.042  3.817

33 

7.74 n = 25 patient records – the average length of stay is 6 days with a standard
deviation of 1.8 days
a. find the reliability factor for a 95% interval estimate t / 2  2.064
b. Find the LCL for a 99% confidence interval estimate of the population
mean
 . The LCL = 5.257
ME  t 2 s
= 6  2.064 1.8

n
25 

193
194
Statistics for Business & Economics, 7th edition
7.75 n = 250, x = 100
a. Find the standard error to estimate population proportion of first timers
pˆ (1  pˆ )
.4(1  .4)
= .03098

n
250
b. Find the sampling error. Since no confidence level is specified, we find
the sampling error (Margin of Error) for a 95% confidence interval.
ME = 1.96 (0.03098) = 0.0607
c. For a 92% confidence interval,
ME = 1.75 (0.03098) = 0.05422
0.40  .05422 giving 0.3457 up to 0.4542
a. 90% confidence interval reliability factor = t / 2  1.729
b. Find the LCL for a 99% confidence interval
LCL = 60.75 – 2.861 21.83159
= 46.78 or approximately 47
20
passengers.
7.76
7.77
a. Find a 95% confidence interval estimate for the population proportion
of students who would like supplements in their smoothies.
Tally for Discrete Variables: Supplements, Health Consciousness
Supplements
No
0
Yes
1
N=
n  113,
Count
42
71
113
Percent
37.17
62.83
Health
Consciousness
Very
1
Moderately 2
Slight
3
Not Very
4
N=
Count
29
55
20
9
113
Percent
25.66
48.67
17.70
7.96
pˆ  71/113  .62832, z.05  1.96
pˆ (1  pˆ )
.62832(1  .62832)
= .62832  1.96
n
113
0.0891 = 0.5392 up to 0.71742.
pˆ  z / 2
= 0.62832 ±
b. pˆ  29 / 113  0.2566 For 98% confidence level,
pˆ  2.33
(0.2566)(1  0.2566)
 0.2566  0.09573
113
or 0.1609 up to
0.3523
c. pˆ  77 / 113  0.6814
0.6814  1.645
0.7535
(0.6814)(1  0.6814)
 0.6814  0.0721 or 0.6093 up to
113
Chapter 7: Estimation: Single Population
7.78
n = 100 students at a small university.
a. Estimate the population grade point average with 95% confidence level
One-Sample T: GPA
Variable
GPA
N
100
Mean
3.12800
StDev
0.36184
SE Mean
0.03618
95% CI
(3.05620, 3.19980)
b. Estimate the population proportion of students who were very dissatisfied
(code 1) or moderately dissatisfied (code 2) with parking. Use a 90%
confidence level.
Tally for Discrete Variables: Parking
Parking
1
2
3
4
5
N=
Count
19
26
18
18
19
100
n  100,
Percent
19.00
26.00
18.00
18.00
19.00
pˆ  45 /100  .45, z.05  1.645 , pˆ  z / 2
.45  1.645
pˆ (1  pˆ )
n
.45(1  .45)
= .45 ± .08184 = .368162 up to .53184
100
c. Estimate the proportion of students who were at least moderately
satisfied
(codes 4 and 5) with on-campus food service
Tally for Discrete Variables: Dining
Dining
1
2
3
4
5
N=
Count
14
26
21
20
19
100
Percent
14.00
26.00
21.00
20.00
19.00
n  100,
pˆ  39 /100  .39, z.05  1.645
pˆ  z / 2
pˆ (1  pˆ )
.39(1  .39)
= .39  1.645
= .39 ± .08023 = .30977 up
n
100
to .47023.
7.79
a. Estimate the average age of the store’s customers by the sample mean
 xi  6310  50.48
x
n
125
To find a confidence interval estimate we will assume a 95% confidence level:
13.06
50.48  1.96
 50.48  2.29 ; 48.19 up to 52.77 years
125
195
196
Statistics for Business & Economics, 7th edition
b. Estimate the population proportion of customers dissatisfied with the delivery
system
Tally for Discrete Variables: Dissatisfied with Delivery
Dissatisfied
with
Delivery
1
2
N=
Count
9
116
125
Percent
7.20
92.80
pˆ  9 /125  .072 ; Assuming a 95% confidence level, we find:
0.072  1.96
(0.072)(1  0.072)
 0.072  0.0453 or 0.0267 up to 0.1173
125
c. Estimate the population mean amount charged to a Visa credit card
Descriptive Statistics: Cost of Flowers
Variable
Cost of Flowers
Method of
Payment
American Express
Cash
Master Card
Other
Visa
Mean
52.99
51.34
54.58
53.42
52.65
SE Mean
2.23
4.05
3.11
2.99
2.04
TrMean
52.83
51.46
54.43
53.72
52.58
StDev
10.68
16.19
15.25
14.33
12.71
The population mean can be estimated by the sample mean amount
charged to a Visa credit card = $52.65.
7.80 n = 500 motor vehicle registrations, 200 were mailed, 160 paid in person,
remainder paid on-line.
a. Estimate the population proportion to pay for vehicle registration
renewals in person, use a 90% confidence level.
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 28.56856% up to 35.4314%
b. Estimate the population proportion of on-line renewals, use 95%
confidence.
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%
Median
50.55
50.55
55.49
54.85
50.65
Chapter 7: Estimation: Single Population
7.81
From the data in 7.80, 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 = 0.12 / 2 =
0.06
pˆ (1  pˆ )
or
ME  z  2
n
(0.4)(0.6)
0.06  z  2
500
Solving for z: z  2  2.74
Area from the z-table = .4969 x 2 = .9938. The confidence level is
99.38%
7.82 From the data in 7.80, 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 = .086 / 2 = .043 and pˆ  .28
pˆ (1  pˆ )
.28(1  .28)
solving for z: z 2  2.14
.043  z 2
n
500
Area from the z-table = .4838 x 2 = .9676. The confidence level is
96.76%
ME  z 2
7.83
a. What is the margin of error for a 99% confidence interval
pˆ (1  pˆ )
pˆ  x  250
 .7143 ,
ME  z 2
n
350
n
.7143(1  .7143)
ME  2.58
350
ME = .0623
=
b. Is the margin of error for a 95% confidence larger, smaller or the same as
the 99% confidence level? 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.
7.84
Compute the 98% confidence interval of the mean age of on-line 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

n
460 

43.18664
747
(11.44)2 90  10
 74.7 , s = 11.44, ˆ 2 x 
 11.633
10
10
90
90% confidence interval: 74.7  1.645 11.633 : 69.089 up to 80.311
b. The interval would be wider; the z-score would increase to 1.96
7.85 a. x 
197
198
Statistics for Business & Economics, 7th edition
Chapter 7: Estimation: Single Population
(149.92)2 272  50
 366.888
50
272
99% confidence interval: 492.36  2.58 366.888 :
442.9139 up to 541.7501
b. 95% confidence interval: 492.36  1.96 366.888 :
454.8175 up to 529.9025
c. The 90% interval is narrower; the z-score would decline to 1.645
7.86
a. ˆ 2 x 
7.87
pˆ 
36
.6(.4) 148  60
 .6 , ˆ pˆ 
 .0024
60
60
148
95% confidence interval: .6  1.96 .0024 : .504 up to .696
199
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