Chapter Eight
8.83
n  100 ,
x  $273 , , and s  $60;
s x  s / n  60 / 100  6
therefore
a. Point estimate of   x  $273
Margin of error  1.96 s x  1.96 (6)  $11.76
b. The 95 % confidence interval for μ is:
8.84
n  100 ,
x  $2640 , and s  $578;
x  zs x  273  1.96(6)  273  11.76  $261 .24 to $284.76
s x  s / n  578 / 100  $57 .80
then
a. Point estimate of   x  $2640
Margin of error  1.96 s x  1.96(57.80)  $113 .29
b. The 97% confidence interval for μ is:
x  zs x  2640  2.17 (57.80)  2640  125 .43  $2514 .57 to $2765.43
8.85
n  36, x  24.015 inches, and   .06 inches; so  x   / n  .06 / 36  .01
The 99% confidence interval for μ is:
x  z x  24.015  2.58(.01)  24.015  .026  23.989 to 24.041 inches.
Since the upper limit of the confidence interval is 24.041, which is greater than 24.025, the machine needs
an adjustment.
8.86
n  50,
x  3.99 inches, and   .04 inches; so
 x   / n  .04 / 50  .00565685
The 98% confidence interval for μ is:
x  z x  3.99  2.33(.00565685 )  3.99  .013  3.977 to 4.003 inches.
Since the lower limit of the confidence interval is less than 3.98 inches, the machine needs an adjustment.
203
204
8.87
Chapter Eight
n  32,
 x  1779,  x 2  142,545
x   x / n  1779 / 32  55.59 minutes
s
2
 x 2   x  / n  142 ,545  (1779 ) 2 / 32  37 .52148578 minutes
n 1
32  1
s x  s / n  37 .52148578 / 32  6.63292426 minutes
a. Point estimate of   x  55.59 minutes
Margin of error  1.96 s x  1.96(6.63292426 )  13.00 minutes
b. The 98% confidence interval for  is:
x  zs x  55.59  2.33(6.632924 )  55.59  15.45  40.14 to 71.04 minutes
8.88
n  36,
 x  547,  x 2  9265
x   x / n  547 / 36  15.19 hours
s
2
 x 2   x  / n  9265  (547 ) 2 / 36  5.21984917
n 1
36  1
s x  s / n  5.21984917 / 36  .86997486 hour
a. Point estimate of   x  15.19 hours
Margin of error  1.96 s x  1.96(.86997486 )  1.71 hours
b. The 99% confidence interval for μ is:
x  zs x  15.19  2.58(.86997486 )  15.19  2.24  12.95 to 17.43 hours
8.89
n  25,
x  $685 , and s  $74; therefore s x  s / n  74 / 25  $14 .80
df  n  1  25  1  24,  / 2  .5  (.99 / 2)  .005 and t = 2.797
The 99% confidence interval for μ is: x  ts x  685  2.797 (14.80)  685  41.40  $643 .60 to $726.40.
8.90
n  18,
x  24 minutes, and s  4.5 minutes; so
s x  s / n  4.5 / 18  1.06066017 minutes
df  n  1  18  1  17,  / 2  .5  (.95 / 2)  .025 and t = 2.110
The 95% confidence interval for  is:
x  ts x  24  2.110 (1.06066017 )  24  2.24  21.76 to 26.24 minutes
Mann – Introductory Statistics, Fifth Edition, Solutions Manual
8.91
n  20,
x  9.75 hours, and s  2.2 hours; so
205
s x  s / n  2.2 / 20  .49193496 hour
df  n  1  20  1  19,  / 2  .5  (.90 / 2)  .05 and t = 1.729
The 90% confidence interval for  is:
x  ts x  9.75  1.729 (.49193496 )  9.75  .85  8.90 to 10.60 hours
8.92
n  20, x  4.5 hours, and s  .75 hours; so
s x  s / n  .75 / 20  .16770510 hour
df  n  1  20  1  19,  / 2  .5  (.98 / 2)  .01 and t = 2.539
The 98% confidence interval for  is:
x  ts x  4.5  2.539 (.16770510 )  4.5  .43  4.07 to 4.93 hours.
8.93
n  12,  x  26 .80, and
s
 x 2  62.395 . This means x   x / n  26.80 / 12  2.23 hours,
2
 x 2   x  / n  62 .395  (26 .80 ) 2 / 12  .48068764 hours, and
n 1
12  1
s x  s / n  .48068764 / 12  .13876257 hours
df  n  1  12  1  11,  / 2  .5  (.95 / 2)  .025 and t = 2.201
The 95% confidence interval for  is:
x  ts x  2.23  2.201(.13876257 )  2.23  .31  1.92 to 2.54 hours.
8.94
n  10,  x  1514 , and
s
 x 2  229,646 ; Then
x   x / n  1514 / 10  151 .40 calories
2
 x 2   x  / n  229 ,646  (1514 ) 2 / 10  6.88315173 calories, and
n 1
10  1
s x  s / n  6.88315173 / 10  2.17664369 calories
df  n  1  10  1  9,  / 2  .5  (.99 / 2)  .005 and t = 3.250
The 99% confidence interval for  is:
x  ts x  151 .40  3.250 (2.17664369 )  151 .40  7.07  144 .33 to 158.47 calories
8.95
n  50,
pˆ  .12 , and qˆ  1  .12  .88; therefore
s pˆ 
pˆ qˆ / n  (.12)(.88) / 50  .04595650
a. Point estimate for p  pˆ  .12 or 12%
Margin of error  1.96s pˆ  1.96(.04595650)  .090 or 9.0%
206
Chapter Eight
b. The 99% confidence interval for p is:
pˆ  zs pˆ  .12  2.58(.04595650)  .12  .119  .001 to .239 or .1% to 23.9%
8.96
n  500 ,
pˆ  .44 , and qˆ  1  .44  .56 ; therefore
s pˆ 
pˆ qˆ / n  (.44)(.56) / 500  .02219910
a. Point estimate of p  pˆ  .44 or 44%
Margin of error  1.96s pˆ  1.96(.02219910)  .044 or 4.4%
b. The 90% confidence interval for p is:
pˆ  zs pˆ  .44  1.65(.02219910)  .44  .037  .403 to .477 or 40.3% to 47.7%
8.97
n  20,
pˆ  8 / 20  .40, and qˆ  1  .40  .60 ; hence, s pˆ 
pˆ qˆ / n  (.40)(.60) / 20  .10954451
The 99% confidence interval for p is:
pˆ  zs pˆ  .40  2.58(.10954451)  .40  .283  .117 to .683 or 11.7% to 68.3%
8.98
n  16,
pˆ  5 / 16  .313 , , and qˆ  1  .313  .687 ; hence s pˆ 
pˆ qˆ / n  (.313)(.687) / 16  .11592859
The 97% confidence interval for p is:
pˆ  zs pˆ  .313  2.17(.11592859)  .313  .252  .061 to .565 or 6.1% to 56.5%
8.99
n  30; 95% confidence interval: $8.46 to $9.86
95% confidence interval for a large sample: x  1.96 s x
a. x 
$9.86  $8.46
 $9.16
2
b. x  1.96 , s x  9.86 , hence s x 
9.86  x 9.86  9.16

 .35714286
1.96
1.96
Confidence level = 99%, i.e. z = 2.58 so the 99% confidence interval is x  2.58 s x  $8.24 to $10.08
8.100
a. Let x = time devoted to commercials per hour on Channel 66.
A sample of 30 twenty – minute time intervals yielded a mean of 4.68 minutes of commercials with a
standard deviation of 1.30 minutes. This is equivalent to:
x  3(4.68)  14 .04 minutes and s  3(1.30)  3.90 minutes for a 60–minute time interval.
s x  s / n  3.90 / 30  .71203932 minute.
Mann – Introductory Statistics, Fifth Edition, Solutions Manual
207
The 90% confidence interval for  is:
x  zs x  14.04  1.65(.71203932 )  14.04  1.17  12.87 to 15.21 minutes per hour
b. n  30,
ˆ  7 / 30
p
The point estimate of  is (7/30)  60 minutes per hour = 14 minutes per hour
a. Both estimates yield approximately 14 minutes per hour.
 7  23 
pˆ qˆ / n     / 30  .07722022 and pˆ  7 / 30  .233
 30  30 
d. s pˆ 
So the 90% confidence interval for p̂ is pˆ  1.65s pˆ  .233  1.65(.07722022)  .106 to .360. Hence
the 90% confidence interval for   p × 60minutes per hour = 6.36 to 21.60 minutes per hour.
e. Maximum error of Waldo’s estimate:
z  s p  60 minutes per hour = 1.65  (.07722022 )  (60 ) minutes per hour = 7.64 minutes per hour
Maximum error of class estimate: z  s x  1.65  (.71203932 ) = 1.17 minutes per hour
f. We need a new s pˆ such that s x  s pˆ  60 minutes per hour; hence,
.71203932 
(7 / 30 )  (23 / 30 )
 60 , so n 
n
(7 / 30 )  (23 / 30 )
(.71203932 / 60 ) 2
 1270 .2  1271 .
It does not seem to be feasible to achieve the same accuracy as that of the class using Waldo’s method.
8.101
Let: p1= proportion of 12–18 year old females who expect a female president within 10 years
p2= proportion of 12–18 year old females who expect a female president within 15 years
p3= proportion of 12–18 year old females who expect a female president within 20 years
p4 = proportion of 12–18 year old females who do not expect a female president not within their
lifetime
n  1250 ,
pˆ 1  .40,
pˆ 2  .25,
pˆ 3  .21,
pˆ 4  .14
ˆ and nqˆ exceed 5 for all these proportions, so the sample is considered large.
np
s pˆ1  pˆ 1qˆ1 / n  .40(.60) / 1250  .01385641
The 95% confidence interval for p1 is:
pˆ 1  zs pˆ1  .40  1.96(..01385641)  .40  .027  .373 to .427 or 37.3% to 42.7%
s pˆ 2 
pˆ 2 qˆ 2 / n  .25(.75) / 1250  .01224745
208
Chapter Eight
The 95% confidence interval for p2 is:
pˆ 2  zs pˆ 2  .25  1.96(.01224745)  .25  .024  .226 to .274 or 22.6% to 27.4%
s pˆ 3 
pˆ 3 qˆ 3 / n  .21(.79) / 1250  .01152042
The 95% confidence interval for p3 is:
pˆ 3  zs pˆ 3  .21 1.96(.01152042)  .21 .023  .187 to .233 or 18.7% to 23.3%
s pˆ 4 
pˆ 4 qˆ 4 / n  .14(.86) / 1250  .00981428
The 95% confidence interval for p4 is:
pˆ 4  zs pˆ 4  .14  1.96(.00981428)  .14  .019  .121 to .159 or 12.1% to 15.9%
A confidence interval is a range of numbers (in this particular case proportions or percentages) which give
an estimate for the true value (i.e. proportion of 12–18 year old females who feel this way). The 95%
means that we are 95% confident that this interval actually contains the true value. A single percentage that
we assign as an estimate would almost always differ from the true value, hence a range with the associated
confidence level is more informative. We assume that the 1200 people are a random sample of 12–18 year
old females.
8.102
n  100 , and pˆ  10 / 100  .10
a. s pˆ 
.10 (.90 )
 .030
100
The 95% confidence interval for p is:
pˆ  1.96s pˆ  .10  1.96(.030)  .10  .059  .041 to .159 or 4.1% to 15.9%
b. No, 18% does not lie within the confidence interval in part a. This suggests that the vaccine is effective
to some degree.
c. There are several things that may have distorted the outcome of this experiment: we only point out two.
It is not mentioned, although it is reasonable to assume that all the dogs that were vaccinated did not
have Lyme disease to begin with. Assuming this, the vaccinated dogs were exposed only 1 year to
possible tick bites, whereas the other dogs in the area probably were exposed much longer. Also, the
owners who agreed to participate in this experiment may be more concerned about their dogs’ health
and restrict the area that the dogs can access, thereby decreasing the exposure to the ticks as well.
8.103
s  4.1 miles, E  1.0 and z  1.96 .
 may be estimated by s.
Hence, n  z 2 s 2 / E 2  (1.96) 2 (4.1) 2 /(1.0) 2  64.58  65
Thus, an additional 65–20=45 observations must be taken.
Mann – Introductory Statistics, Fifth Edition, Solutions Manual
8.104
209
The major problem with this procedure is that the sample is not drawn from the whole target population.
This introduces nonsampling error.
i. Households with no cars would be excluded from this sample. Furthermore, households with more than
2 cars could be counted more than once in the sample. Both of these problems would result in an
upwardly biased estimate of p.
ii. Because of the characteristics of this particular gas station, the clientele may not be typical of all gas
station clientele (with respect to multiple car ownership). Other problems may be present, such as the
next 200 gasoline customers may not be typical of all customers, or there may be dependency between
gasoline customers from the same household. If the attendant took a random sample of 200 households,
and determined how many of these households owned more than 2 cars, the sampling error would still
be present, but could be reduced by taking a larger sample.
8.105
A.
Here,   170 , E  100 , and 1    .99 , and z  2.58
Thus, the required sample size is n 
z 2 2
E2

(2.58) 2 (170 ) 2
(100 ) 2
 19 .24 or 20 days
Note that since n < 30, we must assume that the number of cars passing each day is approximately
normally distributed. Or we may take a large (n  30 ) sample.
b. Since n  z 2 2 / E 2 , then,
z  E n /   100 20 / 272  1.64 which corresponds to a confidence
level of approximately 90%.
c. Since, n 
z 2 2
E
2
, then,
E
z
n

2.58(130 )
 75 .00
20
Thus, they can be 99% confident that their point estimate is within 75 cars of the true average.
8.106
No, the student’s analysis does not make sense. The relevant parameter, p, is the proportion of all U.S.
senators in favor of the bill. The value, .55, is not a sample proportion: instead it is the population
proportion, p. Since, p = .55 is known there is no need to estimate it.
Self – Review Test for Chapter Eight
1.
a. Estimation means assigning values to a population parameter based on the value of a sample statistic.
b. An estimator is the sample statistic used to estimate a population parameter.
210
Chapter Eight
c. The value of a sample statistic is called the point estimate of the corresponding population parameter.
2.
7.
b
3.
a
4.
a
5.
d
6.
b
n  36 , x  $159,000, and s  $27,000; hence s x  s / n  27 ,000 / 36  $4500
a. Point estimate of   x  $159 ,000
Margin of error  1.96 s x  1.96(4500 )  $8820
b. The 99% confidence interval for  is:
x  zs x  159 ,000  2.58(4500 )  159 ,000  11,610  $147 ,390 to $170,610
8.
n  25 , x  $410,425, and s  $74,820
df  n  1  25  1  24,  / 2  .5  (.95 / 2)  .025 and t = 2.064
s x  s / n  74,820 / 25  $14,964
The 95% confidence interval for  is:
x  ts x  410 ,425  2.064 (14,964 )  410 ,425  30,885 .70  $379 ,539 .30 to $441,310.70
9.
n  612 , pˆ  .41, and qˆ  1  .41  .59,
so
s pˆ 
pˆ qˆ / n  (.41)(.59) / 612  .01988118
a. Point estimate of p  pˆ  .41
Margin of error  1.96s pˆ  1.96(.01988118)  .039
pˆ  zs pˆ  .41 1.96(.01988118)  .41 .039  .371 to .449
b. The 95% confidence interval for p is:
10.
E  .65 houses,   2.2 houses, and z  1.96 , then
11.
E  .05, z  1.65,
12.
E  .05,
p  q  .50 , then n 
z 2 pq
E2

n
z 2 2
E2

(1.65) 2 (.50 )(. 50 )
(.05) 2
pˆ  .70, qˆ  1  pˆ  1  .70  .30, and z  1.65; so n 
z 2 pˆ qˆ
E2
(1.96 ) 2 (2.2) 2
(.65) 2
 44 .01 ≈ 45.
 272 .25  273 .

(1.65) 2 (.70 )(. 30 )
(.05) 2
 228 .69  229 .
Mann – Introductory Statistics, Fifth Edition, Solutions Manual
211
13. The width of the confidence interval can be reduced by:
1. Lowering the confidence level
2. Increasing the sample size
The second alternative is better because by lowering the confidence level we will simply lower the
probability that our confidence interval includes  .
14. To estimate the mean number of hours that all students at your college work per week:
1. Take a random sample of 12 students from your college who hold jobs
2. Record the number of hours each of these students worked last week
3. Calculate x , s, and s x from these data
4. After choosing the confidence level, find the value for the t distribution with 11df and for an area of  / 2
in the right tail.
5. Obtain the confidence interval for  by using the formula x  ts x
You are assuming that the hours worked by all students at your college have a normal distribution.
15. To estimate the proportion of people who are happy with their current jobs:
1. Take a random sample of 35 workers
2. Determine whether or not each worker is happy with his or her job
3. Calculate pˆ , qˆ, and s p
4. Choose the confidence level and find the required value of z from the normal distribution table
5. Obtain the confidence interval for p by using the formula pˆ  zs p