Solutions - Confidence Intervals
8.6 Starting salaries:
a) The point estimate is the mean of the salaries of the sample of three students: $45,000.
b) The interval would be: 20,200 to 69,800. It includes all salaries within one margin of error on either
side of the mean.
c) A point estimate alone may be highly inaccurate, especially with a small sample. An interval estimate
gives us a sense of the accuracy of the point estimate.
8.7 Believe in heaven?:
a)
The margin of error for a 95% confidence interval would be 1.96 times the standard error of 0.01 =
0.0196 (rounds to 0.02).
b) The 95% confidence interval includes all points within one margin of error of the mean.
0.86 – 0.02 = 0.84
0.86 + 0.02 = 0.88
The confidence interval goes from 0.84 to 0.88. This is the interval containing the most believable
values for the parameter.
8.12 Crime victims:
a)
The point estimate of the proportion of the population who were victims would be 17/987 = 0.017.
b) The standard error would be
p(1  p)

n
0.017(0.983)
 0.00411.
987
c) The margin of error would be (1.96)(0.00411) = 0.00806.
d) The 95% confidence interval would include all proportions within one margin of error of the mean
proportion of 0.017, which is: 0.00894 to 0.02506. We can conclude that fewer than 10% of all adults
in the U.S. were victims.
8.15 Favor death penalty:
We can obtain the value reported under “Sample p” by dividing the number of those in favor by the
total number of respondents, 1885/2815.
b) b) We can be 95% confident that the proportion of the population who are in favor of the death penalty
is between 0.652 and 0.687, or rounding, (0.65, 0.69).
c) c) 95% confidence refers to a probability that applies to the confidence interval method. If we use this
method over and over for numerous samples, in the long run we make correct inferences (that is, the
confidence interval contains the parameter) 95% of the time.
d) We can conclude that more than half of all American adults were in favor because all the values in the
confidence interval are above 0.50.
a)
8.22 Wife doesn’t want kids:
a)
The sample mean proportion is 576/708 = 0.814. The standard error is
p(1  p)

n
0.814(1 - 0.814)
 0.015. The margin of error would be (2.58)(0.015) = 0.0387. The confidence
708
interval would range from 0.775 to 0.853. We can conclude that the population proportion exceeds 75%
because 75% is below the lowest believable value of the confidence interval.
b) The 99% confidence interval would be wider than a 95% confidence interval.
8.27 Females’ ideal number of children:
a) The point estimate of the population mean is 3.16.
b) The standard error of the sample mean: se =
s
n

1.91
1097
 .0577
c) We’re 95% confident that the population mean falls between 3.05 and 3.28.
d) It is not plausible that the population mean is 2 because it falls outside the confidence interval
8.29 Using t table:
a) 2.776
b) 2.145
c) 2.977
8.48 Abstainers:
pˆ 1  pˆ z 2
0.51  0.51.96 2

 384.16 (rounds to 384)
m2
0.05 2
pˆ 1  pˆ z 2
0.191  0.191.96 2
b) n 

 236.49 (rounds to 236)
m2
0.05 2
a)
c)
n
Strategy (a) is inappropriate if we are quite sure we’ll get a sample proportion that is far from 0.50
because it overestimates the sample size by quite a bit. The first sample size would be more costly than
needed.
8.51 Farm size:
a)
n


4s 2 4 200 2
= 256

m2
25 2
b) We can use the same formula as in part (a).


4 300 2
m2
4(300 2 )
m2 
256
2
m  1406.25
m  37.5
256 