Chapter 8
Interval Estimation
Learning Objectives
1.
Know how to construct and interpret an interval estimate of a population mean, a
population proportion, or both.
2.
Understand and be able to compute the margin of error.
3.
Learn about the t distribution and its use in constructing an interval estimate when  is
unknown for a population mean.
4.
Be able to determine the size of a simple random sample necessary to estimate a
population mean and/or a population proportion with a specified margin of error.
5.
Know the definition of the following terms:
confidence interval
margin of error
confidence coefficient
degrees of freedom
confidence level
Solutions
1.
a.
 x   / n  5 / 40  .79
b. At 95%, z  /
2.
n  1.96 ( 5 /
a. 32 + 1.645 ( 6 /
40 )  1.55
50 )
32 + 1.4 or 30.6 to 33.4
b. 32 + 1.96 ( 6 /
50 )
32 + 1.66 or 30.34 to 33.66
c. 32 + 2.576 ( 6 /
50 )
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in part.
32 + 2.19 or 29.81 to 34.19
3.
a. 80 + 1.96 (15 /
80 + 3.8
60 )
or
6.2 to 83.8
b. 80 + 1.96 (15 /
80 + 2.68
120 )
or
77.32 to 82.68
c. Larger sample provides a smaller margin of error.
Sample mean
x
160  152
 156
2
Margin of error = 160 – 156 = 4
1.96( /
n)  4
n  1 .9 6  / 4  1 .9 6 (1 5) / 4  7 .3 5
n = (7.35)2 = 54
5.
a. With 99% confidence z / 2  z.005  2.576
Margin of error = 2 .5 7 6 /
n  2 .5 7 6 (6 /
6 4 )  1 .9 3
b. Confidence Interval: 21.52 + 1.93 or 19.59 to 23.45
6.
A 95% confidence interval is of the form x  z.025 ( /
n) .
Using Excel and the web file TravelTax, the sample mean is x  40.31 and the sample size
is n = 400. The sample standard deviation  = 85 is known. The confidence interval is
40.31 + 1.96(8.5/
200 )
40.311.18 40.31 + 1.18 or 39.13 to 41.49
7.
a. Margin of error = z.025 ( /
n)
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in part.
z.025  1.96 ,   $255 , n  50
Margin of error = 1.96(255/
50 ) = 70.68
b. From data set, sample mean x = 1,836.84.
95% confidence interval: 1836.84 + 70.68 (1766.16 to 1907.52)
8.
a. Because n is small, an assumption that the population is at least approximately normal
is required so that the sampling distribution of x can be approximated by a normal
distribution.
9.
b. Margin of error: z .0 2 5 ( /
n )  1 .9 6 (5 .5 /
1 0 )  3 .4 1
c. Margin of error: z .0 0 5 ( /
n )  2 .5 7 6 (5 .5 /
1 0 )  4 .4 8
x + z .0 2 5 ( /
n)
19,100 + 1.96 (3027 / 55)
19,100 + 800 ($18,300 to $19,900)
The average cost to repair the smoke and fire damage that result from home fires of all
causes is $11,389. The 95% confidence interval shows the mean cost of fires resulting
from tobacco use is higher and more expensive.
10.
a.
x  z / 2

n
3,486 + 1.645 (650 / 120 )
3,486 + 98 or $3388 to $3584
b. 3,486 + 1.96 (650 / 120 )
3,486 + 116 or $3370 to $3602
c. 3,486 + 2.576 (650 / 120 )
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in part.
3,486 + 153 or $3,333 to $3,639
The confidence interval gets wider as we increase our confidence level. We need a
wider interval to be more confident that the interval will contain the population mean.
11.
a. .025
b. 1 – .10 = .90
c. .05
d. .01
e. 1 – 2(.025) = .95
f. 1 – 2(.05) = .90
12.
a. 2.179
b. –1.676
c. 2.457
d. Use .05 column, –1.708 and 1.708
e. Use .025 column, –2.014 and 2.014
13.
a.
x
 xi 80

 10
8
n
b.
xi
( xi  x )
( xi  x ) 2
10
0
0
8
–2
4
12
2
4
15
5
25
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in part.
13
3
9
11
1
1
6
–4
16
5
–5
25
84
s
n )  2 .3 6 5(3 .4 6 4 /
c.
t .025 ( s /
d.
x  t .0 2 5 ( s /
 ( xi  x ) 2

n 1
84
 3.464
7
8 )  2 .9
n)
10 ± 2.9 or 7.1 to 12.9
14.
x  t / 2 ( s /
n)
, df = 53
a. 22.5 ± 1.674 (4.4 / 54 )
22.5 ± 1 or 21.5 to 23.5
b. 22.5 ± 2.006 (4.4 / 54 )
22.5 ± 1.2 or 21.3 to 23.7
c. 22.5 ± 2.672 (4.4 / 54 )
22.5 ± 1.6 or 20.9 to 24.1
d. As the confidence level increases, there is a larger margin of error and a wider
confidence interval.
15.
x  t / 2 ( s /
n)
90% confidence
df = 64 t.05 = 1.669
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in part.
19.5 ± 1.669 (5.2 /
65 )
19.5 ± 1.08 or 18.42 to 20.58
95% confidence: df = 64
19.5 ± 1.998 (5.2 /
t.025 = 1.998
65 )
19.5 ± 1.29 or 18.21 to 20.79
16.
a. Using JMP or Excel, x = 9.7063 and s = 7.9805
The sample mean years to maturity is 9.7063 years with a standard deviation of
7.9805.
b.
x  t .0 2 5 ( s /
n ) , df = 39, t.025 = 2.023
9.7063 ± 2.023 (7 .9 8 0 5 /
40 )
9.7063 ± 2.5527 or 7.1536 to 12.2590
The 95% confidence interval for the population mean years to maturity is 7.1536 to
12.2590 years.
c. Using JMP or Excel, x = 3.8854 and s = 1.6194
The sample mean yield on corporate bonds is 3.8854% with a standard deviation of
1.6194.
d.
x  t .0 2 5 ( s /
n)
df = 39
3.8854 ± 2.023 (1 .6 1 9 4 /
t.025 = 2.023
40 )
3.8854 ± .5180 or 3.3674 to 4.4034
The 95% confidence interval for the population mean yield is 3.3674 to 4.4034%.
17.
Using JMP or Excel, x = 6.34 and s = 2.163
x  t .0 2 5 ( s /
n)
df = 49 t.025 = 2.010
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in part.
6.34 ± 2.010 ( 2 .1 6 3 /
50 )
6.34 ± .61 or 5.73 to 6.95
18.
For the JobSearch data set, x  22 and s = 11.8862
a.
x = 22 weeks
b. margin of error = t.0 25 s /
n  2 .0 2 3(1 1 .8 8 6 2 ) /
4 0  3.8020
c. The 95% confidence interval is x  margin of error
22  3.8020 or 18.20 to 25.80
d. Skewness = 1.0062, data are skewed to the right. Use a larger sample next time.
19.
a.
t .0 2 5 ( s /
n ) , df = 41
t.025 = 2.02, s = 6.827
2.02 (6.827 / 42) = 2.13
b.
x  t .0 2 5 ( s /
n)
32.66 ± 2.13 or $30.53 to $34.79
c. The mean cost for a mid-range meal for two in Tokyo ($40) is not in the 95%
confidence interval for comparable meals in Hong Kong. The mean cost of
comparable meals in Hong Kong is less than in Tokyo.
20.
a.
x   xi / n  51, 020 / 20  2551
The point estimate of the mean annual auto insurance premium in Michigan is
$2,551.
s
b.
(xi  x )2
(xi  2551)2

 301.3077
n 1
20 1
95% confidence interval: x + t.025 ( s /
n ) , df = 19
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in part.
2,551 + 2.093 (301.3077 /
20 )
$2,551 + 141.01 or $2,409.99 to $2,692.01
c. The 95% confidence interval for Michigan does not include the national average of
$1,503 for the United States. We would be 95% confident that auto insurance
premiums in Michigan are above the national average.
21.
x
xi 1420

 71
20
n
s
(xi  x )2
9492

 22.3513
n 1
20 1
t.025 = 2.093
df = 19
95% confidence interval: x + t.025 ( s /
71 + 2.093 (22.3513 /
n)
20 )
71 + 10.46 or $60.54 to $81.46 per visit
22.
a.
x
xi 693,000

 $23,100
30
n
The point estimate of the population mean ticket sales revenue per theater is $23,100.
(xi  x )2
s
 3981.89
n 1
95% confidence interval: x + t.025 ( s /
n ) with df = 29
t.025 = 2.045
23,100 + 2.045 3981.89
30
23,100 + 1,487
The 95% confidence interval for the population mean is $21,613 to $24,587.
b. Mean number of customers per theater = 23,100/8.11 = 2,848 customers per theater
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in part.
c. Total number of customers = 4,080(2,848) = 11,619,840
Total box office ticket sales for the weekend = 4,080(23,100) = $94,248,000
23.
24.
n
n
z.2025 2 (196
. ) 2 ( 9) 2

 34.57
32
E2
n
(1.96) 2 ( 9) 2
 77.79
22
Use n  78
n
(1.96) 2 (6.84) 2
 79.88
(1.5) 2
Use n  80
n
(1.645) 2 (6.84) 2
 31.65
22
Use n  32
a.
n
c.
26.
Use n  62
a. Planning value of  = Range/4 = 36/4 = 9
b.
25.
2
z.025
 2 (1.96) 2 (40) 2

 61.47
E2
10 2
Use n  35
2
z.025
 2 (1.96) 2 (.07) 2

 20.92
E2
(.03) 2
Use 21.
If the normality assumption for the population appears questionable, this should be
adjusted upward to at least 30.
b.
n
(1.96) 2 (.07) 2
 47.06
(.02) 2
Use 48 to guarantee a margin of error no greater than .02. However, the U.S. EIA
may choose to increase the sample size to a round number of 50.
c.
n
(1.96) 2 (.07) 2
 188.24
(.01) 2
Use 189
For reporting purposes, the U.S. EIA might decide to round up to a sample size of
200.
27.
Planning value  
60,000  45,000
 3750
4
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in part.
a.
b.
c.
n
2
 2 (1.96) 2 (3750) 2
z.025

 216.09
E2
(500) 2
n
(1.96) 2 (3750) 2
 1350.56
(200) 2
Use n  1351
n
(1.96) 2 (3750) 2
 5402.25
(100) 2
Use n  5403
Use n  217
d. Sampling 5,403 college graduates to obtain the $100 margin of error would be viewed
as too expensive and too much effort by most researchers.
28.
a.
n
z2 /2 2
(1.645) 2 (11.35) 2

 191.27
2
E
(1.35) 2
b.
n
(1.96) 2 (11.35) 2
 271.54
(1.35) 2
c.
n
(2.576) 2 (11.35) 2
 469.05
(1.35) 2
Use n  192
Use n  272
Use n  470
d. The sample size gets larger as the confidence level is increased. We would not
recommend 99% confidence. The sample size must be increased by 80 respondents
(272 – 192) to go from 90% to 95%. This may be reasonable. However, increasing
the sample size by 198 respondents (470 – 272) to go from 95% to 99% would
probably be viewed as too expensive and time consuming for the 4% gain in
confidence level.
29.
a. 75 seconds is 1.25 minutes, so
b.
30.
n
(1.96) 2 (4) 2
 39.3380
(1.25) 2
n
(1.96) 2 (4) 2
 61.4656
12
Use n = 40
Use n = 62
Planning value from previous study:   3200
n
2
z .025
 2 (1.96) 2 (3200) 2

 1536.64
E2
(160) 2
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in part.
Use n = 1537 to guarantee the margin of error will not exceed 160.
31.
a. p = 100/400 = .25
b.
c.
p (1  p )

n
p  z .025
.25(.75)
 .0217
400
p (1  p )
n
.25 + 1.96 (.0217)
.25 + .0424 or .2076 to .2924
32.
.70(.30)
800
a. .70 + 1.645
.70 + .0267 or .6733 to .7267
b. .70 + 1.96
.70(.30)
800
.70 + .0318 or .6682 to .7318
33.
34.
n
Use n  350
Use planning value p* = .50
n
35.
2
z.025
p  (1  p  ) (1.96) 2 (.35)(.65)

 349.59
E2
(.05) 2
(1.96) 2 (.50)(.50)
 1067.11
(.03) 2
Use n  1068
a. p = 574/1007 = .57
b. Margin of Error
z.05
p (1  p )
n
 1.645
0.57(0.43)
1007
 0.0257
c. Confidence interval:
.57 ± .0257 or .5443 to .5927
d. Margin of Error
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in part.
z.025
p (1  p )
n
 1.96
0.57(0.43)
1007
 0.0306
95% confidence interval
.57 + .0306 or .5394 to .6006
36.
a.
b.
p =
46/200 = .23
p (1  p )

n
p  z .025
.23(1  .23)
 .0298
200
p (1  p )
n
.23 + 1.96(.0298)
.23 + .0584 or .1716 to .2884
37.
a.
p  384 /1007  .3813
b. Margin of error:
z.025
c.
p (1  p )
.3813(1  .3813)
 1.96
 .0300
n
1007
p  .0300
confidence interval: .3813 + .0300 or .3513 to .4113
d.
p  553 /1007  .5492
z.025
p (1  p )
.5492(1  .5492)
 1.96
 .0307
n
1007
confidence interval: .5492 .0307 or .5185 to .5799
e. Part c margin of error = .0300
Part d margin of error = .0307
The confidence interval in part c has the smaller margin of error. This is because p is
farther away from 0.5 in part c (the point at which the standard error is the largest)
than it is in part d.
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in part.
38.
a.
p  81 / 142  .5704
b. Margin of error = 1.96
(.5704)(.4296)
p (1  p )
 1.96
 .0814
n
142
Confidence interval: .5704 + .0814 or .4890 to .6518
c.
39.
a.
b.
40.
n
1.96 2 (.5704)(.4296)
 1046
(.03) 2
n
2
z .025
p  (1  p  ) (1.96) 2 (.16)(1  .16)

 573.68
E2
(.03) 2
n
2
z.005
p  (1  p  ) (2.576) 2 (.16)(1  .16)

 990.94
E2
(.03) 2
Margin of error: z.025
Use 574
Use 991
p (1  p )
.52(1  .52)
 1.96
 .0346
n
800
95% confidence interval: p ± .0346
.52 + .0346 or .4854 to .5546
41.
a. Margin of error = z.025
.639(1  .639)
p (1  p )
 1.96
 .0228
1200
n
Interval estimate: .639  .0228 or .6162 to .6618
b. Margin of error = z.025
.417(1  .417)
p (1  p )
 1.96
 .0234
1200
n
Interval estimate: .417 ± .0234 or .3936 to .4404
c. The margin of error is larger in part b. This is because the sample proportion is closer
to .50 in part b than in part a. This also leads to a larger interval estimate in part b.
42.
a.
p * (1  p *)

n
z.025
b.
n
.50(1  .50)
 .0226
491
p * (1  p *) = 1.96(.0226) = .0442
n
2
z.025
p  (1  p  )
E2
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in part.
September
n
1.96 2 (.50)(1  .50)
 600.25
.04 2
Use 601
October
n
1.96 2 (.50)(1  .50)
 1067.11
.032
Use 1068
November
n
1.96 2 (.50)(1  .50)
 2401
.02 2
 .50)
Preelection n  1.96 (.50)(1
 9604
2
2
.01
43.
a. Margin of Error = z / 2
(.90)(.10)
p (1  p )
 1.96
 .0201
857
n
95% confidence interval: .90 + .0201 or .8799 to .9201
b. Margin of error = 1.96
(.67)(.33) = .0315
857
95% confidence interval: .67 + .0315 or .6385 to .7015
c. Margin of error = 1.96
(.56)(.44) = .0332
857
95% confidence interval: .56 + .0332 or .5268 to .5932
d. The margin of error increases as p gets closer to .50.
44.
a. 𝑥̅ = 326.6674
b. degrees of freedom = 10,000 and 𝑠 ̅ = 12,318.01, so 𝑡 ⁄ 𝑠 ̅ =
1.9600(12,318.01/√10,001) = 241.4165.
c. 𝑥̅
45.
𝑡 ⁄ 𝑠 ̅ = 326.6674 ± 241.4165 = (85.25, 568.08)
a. 𝑥̅ = 60.3964
b. 𝑧 ⁄ 𝜎 ̅ = 2.576(33.4460/√3500) = 1.4999
c. 𝑥̅
𝑧 ⁄ 𝜎 ̅ = 60.40 ± 1.5023 = (58.8965, 61.8963)
d. The 99% confidence interval for the mean sick hours taken last year does not include
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in part.
the value for the mean sick hours taken two years ago. This suggests that the mean
sick hours taken changed from two years ago to last year.
46.
a. 𝑝̅ = 65120/102519 = .6352
b. 𝑧 ⁄ 𝜎 ̅ = 1.96(.0015) = .0029
c. 𝑥̅
47.
𝑧 ⁄ 𝜎 ̅ = .6352 ± .0029 = (.6323, .6381)
a. 𝑝̅ = 9252/20000 = .4626
b. 𝑧 ⁄ 𝜎 ̅ = 2.576(.0035) = .0090
c. 𝑥̅
𝑧 ⁄ 𝜎 ̅ = .4626 ± .0090 = (.4536, .4716)
d. The 99% confidence interval for the proportion of vehicles on U.S. roads does not
include the value for the proportion reported by ABC News. This suggests that the
proportion of vehicles on U.S. roads that are speeding differs from the proportion
reported by ABC News. However, the sample was taken at 6 P.M. EST on a recent
Tuesday afternoon. We must question whether the drivers on U.S. roads at this time
are representative of all U.S. drivers. If it doesn’t, then our sample suffers from
nonsampling error.
48.
a. Margin of error: z.025   1.96 15  4.00
n
54
b. Confidence interval: x + margin of error
33.77 + 4.00 or $29.77 to $37.77
49.
a. x + t.025 ( s /
df = 63 t.025 = 1.998
n)
252.45 + 1.998 ( 74 .50 /
64 )
252.45 + 18.61 or $233.84 to $271.06
b. Yes. The lower limit for the population mean at Niagara Falls is $233.84, which is
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in part.
greater than $215.60.
50.
a. Margin of error = t.025 ( s /
n)
df = 79 t.025 = 1.990 s = 550
1.990 (5 5 0 /
8 0 ) = 122
b. x ± margin of error
1873 + 122 or $1,751 to $1,995
c. Because 92 million Americans were age 50 and older, the estimate of total
expenditures = 92(1873) = 172,316.
In dollars, we estimate that $172,316 million dollars are spent annually by Americans
of age 50 and older on restaurants and carryout food.
d. We would expect the median to be less than the mean of $1,873. The few individuals
who spend much more than the average cause the mean to be larger than the median.
This is typical for data of this type.
51.
a. From the sample of 7,044 people who have never been married, the proportion who
have a healthy sleep duration is p  4437 / 7044  .63 . This is our point estimate of the
proportion for the population.
The margin of error is
Margin of error = z.025
p (1  p )
.63(1  .63)
 1.96
 .0113
n
6044
95% confidence interval: .63 + .0113 or .6186 to .6413
b. From the sample of 7044 people who have never been married, the sample mean
number of hours of sleep is x  7.2257 . Thus, our point estimate of the population
mean number of hours of sleep for people who have never been married is 7.226.
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in part.
Using Excel, we find that the sample standard deviation is s  .8211 .
Therefore,
Margin of error = t.025 s  1.96 .8211  .0192
n
7044
95% confidence interval: 7.2257 + .0192 or 7.2065 to 7.2449
c. From the sample, we see that 4,437 people had a healthy sleep duration. The mean
number of hours of sleep for those who had a healthy sleep duration is given by
x  34, 295.1/ 4437  7.7293 .
52.
a. Using Excel shows that the sample mean and sample standard deviation are x  773
and s  738.3835 .
Margin of error = t.05  s   1.645  738.3835   60.73

n 

400

90% Confidence Interval: 773  60.73 or $712.27 to $833.73
b. Using Excel shows that the sample mean and sample standard deviation are x  187
and s  178.6207 .
Margin of error = t.05  s   1.645  178.6207   14.69

n 

400

90% confidence interval: 187 + 14.69 or $172.31 to $201.69
c. There were 136 employees who had no prescription medication cost for the year:
p  136 / 400  .34 .
d. The margin of error in part a is 60.73; the margin of error in part c is 14.69. The
margin of error in part a is larger because the sample standard deviation in part a is
larger. The sample size and confidence level are the same in both parts.
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in part.
53.
a. Using Excel or JMP: x  27.63
The estimated mean BMI for adults in the U.S. is 27.63. This is less than the standard
established by the National Institute of Diabetes and Digestive and Kidney Diseases.
b. Using Excel or JMP: s = 10.8484
c.
x  t .0 2 5 s /
df = 1009
n
t.025
= 1.984
27.63 + 1.984 (10.8484/ 101 )
27.63 + 2.14 or 25.49 to 29.77
54.
n
55.
n
56.
57.
(2.33) 2 (1) 2
 33.93
(0.4) 2
(1.96) 2 (8) 2
 61.47
22
Use n  34
Use n  62
n
( 2.576) 2 (8) 2
 106.17
22
Use n  107
n
(196
. ) 2 (675) 2
 175.03
100 2
Use n  176
a.
p  1.96
p (1  p )
n
.47 + 1.96
(.47)(.53)
450
.47 + .0461 or .4239 to .5161
b. .47 + 2.576
(.47)(.53)
450
.47 + .0606 or .4094 to .5306
c. The margin of error becomes larger.
58.
a. p = 200/369 = .5420
b.
1.96
(.5420)(.4580)
p (1  p )
 1.96
 .0508
n
369
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in part.
c. .5420 + .0508 or .4912 to .5928
59.
a. p = .78
Margin of error = z.025
(.78)(.22)
p (1  p )
 1.96
 .0363
500
n
95% confidence interval: .78 + .0363 or .7437 to .8163
b. p = .45
Margin of error = z.005
(.45)(.55)
p (1  p )
 2.576
 .0573
500
n
95% confidence interval: .45 + .0573 or .3927 to .5073
c. The margin of error is larger in part b for two reasons: (1) with p = .45, the estimate
of the standard error is larger than with p = .78, (2) And z .0 0 5 = 2.576 for 99%
confidence is larger than z .0 2 5 = 1.96 for 95% confidence.
60.
a. With 165 out of 750 respondents rating the economy as good or excellent,
p= 165/750 = .22
b. Margin of error  1.96
.22(1  .22)
p (1  p )
 1.96
 .0296
n
750
95% Confidence interval: .22 + .0296 or .1904 to .2496
c. With 315 out of 750 respondents rating the economy as poor,
p= 315/750 = .42
Margin of error  1.96
.42(1  .42)
p (1  p )
 1.96
 .0353
n
750
95% Confidence interval: .42 + .0353 or .3847 to .4553
d. The confidence interval in part c is wider. This is because the sample proportion is
closer to .5 in part c.
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in part.
61.
a.
n
(1.96) 2 (.3)(.7)
 2016.84
(.02) 2
Use n  2017
b. p = 520/2017 = .2578
c.
p  1.96
p (1  p )
n
.2578 + 1.96
(.2578)(.7422)
2017
.2578 + .0191 or .2387 to .2769
62.
a.
b.
63.
n
(2.33) 2 (.70)(.30)
 1266.74
(.03) 2
Use n  1267
n
(2.33) 2 (.50)(.50)
 1508.03
(.03) 2
Use n  1509
a. For the 18–24 age group:
95% margin of error: 1.96 .67(1  .67)  .0432
455
95% confidence interval: .67  .0432 or .6268 to .7132
For the 25–34 age group:
95% margin of error: 1.96 .83(1  .83)  .0345
455
95% Confidence interval: .83  .0345 or .7955 to .8645
For the 35–49 age group
95% margin of error: 1.96 .76(1  .76)  .0392
455
95% confidence interval: .76 .0392 or .7208 to .7992
For the 50+ age group
95% margin of error: 1.96 .78(1  .78)  .0380
455
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in part.
95% confidence interval: .78 .0381 or .7420 to .8180
b. The 18–24 age group has the largest margin of error, so the sample size must be large
enough to reduce the margin of error for the 18–24 age group to .05 or less. Using the
proportions found in the study reported by CreditCards.com as a planning value
p *  .67.
2
n
(1.96) (0.67)(0.33)
(0.03)
2
 943.75
Use n  944
This is an increase of 489 people surveyed in each age group. The added cost of the
larger sample size would have to be taken into account before deciding whether the
smaller margin of error is worth the added cost of obtaining the data.
64.
a. p = 618/1993 = .3101
b.
p  1.96
p (1  p )
1993
.3101 + 1.96
(.3101)(.6899)
1993
.3101 + .0203 or .2898 to .3304
c.
n
z 2 p  (1  p  )
E2
z
(1.96) 2 (.3101)(.6899)
 8218.64
(.01) 2
Use n  8219
No, the sample appears unnecessarily large. The .02 margin of error reported in part b
should provide adequate precision.
65.
a. 𝑥̅ = 68.5316
b. degrees of freedom = 15,716 and s = 12.5239,
so 𝑡 ⁄ 𝑠 ̅ = 1.9600(12.5239/√15,717) = .1958
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in part.
c. 𝑥̅
66.
𝑡 ⁄ 𝑠 ̅ = 68.5316 ± .1958 = (68.3358, 68.7274)
a. 𝑥̅ = 34.9872
b. degrees of freedom = 28,584 and s = 2.9246,
so 𝑡 ⁄ 𝑠 ̅ = 2.58(2.9246/√28,584) = .0446
c. 𝑥̅
𝑡 ⁄ 𝑠 ̅ = 34.9872 ± .0446 = (34.9426, 35.0318)
d. The 99% confidence interval for the mean hours worked hours during the past week
does not include the value for the mean hours worked during the same week one year
ago. This suggests that the mean hours worked taken changed from last year to this
year.
67.
a. 𝑝̅ = 18,233/42,296 = .4311
b. 𝑧 ⁄ 𝜎 ̅ = 1.96(.0024) = .0047
c. 𝑥̅
𝑧 ⁄ 𝜎 ̅ = .4311 ± .0047 = (.4264, .4358)
The 95% confidence interval for the proportion of reports filed from Florida that dealt
with instances of fraud does not include the value for the proportion filed by the
entire country. This suggests that the proportion of instances of fraud in Florida
differs from the proportion for the United States.
68.
a. 𝑝̅ = 490/8749 = .0560
b. 𝑧 ⁄ 𝜎 ̅ = 1.64(.0025) = .0040
c. 𝑥̅
𝑧 ⁄ 𝜎 ̅ = .0560 ± .0040 = (.0520, .0600)
d. The 90% confidence interval for the proportion of California bridges that are deficient
does not include the value for the proportion of deficient bridges in the entire country.
This suggests that the proportion of California bridges that is deficient differs from
the proportion for the U.S.
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in part.
Even though the IRC report indicates that California has a large proportion of the
nation’s deficient bridges, California has a large total number of bridges, so the
proportion of bridges in California that are deficient is smaller than the proportion of
deficient bridges nationwide.
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in part.