Stat 250.3
November 5, 2003
HOMEWORK 5– SOLUTIONS
10.33
At least n =1,112. Solve for n in the equation
1
 .03 , and round the number up to get n.
n
10.38
10.39
pˆ (1  pˆ )
.59 (1  .59 )

 .016
n
1000
a. s.e.( pˆ ) 
b. The 95% confidence interval is pˆ  2 s.e.( pˆ ) which is .59  (2.016) or .558 to .622.
With 95% confidence, we can say that between .558 and .622 of American adults think the world will come to an end.
56
 .295
a. p̂ 
190
b. s.e.( pˆ )  .295 (1  .295 )  .033
190
c. For 90% confidence, the multiplier is z * = 1.645 (see Table 10.1). The interval is pˆ  z * s.e.( pˆ ) which is .295 
(1.645.033). This is .295  .054, or .241 to .349.
d. For 95% confidence, the multiplier is approximately z * = 2 (see Table 10.1). The interval is pˆ  z * s.e.( pˆ ) which
is .295  (2.033). This is .295  .066, or .229 to .361.
e. For 98% confidence, the multiplier is z * = 2.33 (see Table 10.1). The interval is pˆ  z * s.e.( pˆ ) which is .295 
(2.33.033). This is 0.295  0.077, or .218 to 0.372.
f. As the confidence level is increased, the width of the interval increases.
g. If numbers are chosen randomly, the true proportion who pick the number “7” would be 1/10=.1. None of the
intervals calculated in the parts (c-e) include .1, so with any of those confidence levels, it’s reasonable to conclude that
the student picks are not made randomly.
10.45
a.
1

1
= .036, which verifies that the margin of error is about 3.5%.
n
757
b. A 95% confidence interval is 3%  3.5% or 0.5% to 6.5%. Not all of these values are valid estimates of the
population percentage since the population percentage cannot be less than 0.
pˆ (1  pˆ )
.03(1  .03)

 .006 .
c. The standard error is s.e.( pˆ ) 
n
757
The margin of error is 2  standard error = 2 (.006) = .012 or 1.2%.
d. A 95% confidence interval based on the margin of error in part c is 3%  1.2% or 1.8% to 4.2%. This means that it
is highly likely that the percentage of all American women who would say they are “Not at all satisfied” with their
overall physical appearance is between 1.8% and 4.2%.
11.27
a. Ho: p = .50 (or Ho: p  .50)
Ha: p >.50
(more than one-half do)
p = proportion of all adult American Catholics who favor allowing women to be priests.
Usually the “yes” answer to the research question is the alternative hypothesis, and that is what has been done here.
b. The sample was randomly selected, which is one of the necessary conditions. The sample size is large enough,
which is the other necessary condition. Both np0 and n(1  p 0 ) are greater than 10, as they should be to use a zstatistic. Here, n = 507 and p 0 =.50.
c. z 
Sample estimate - Null value

Null standard error
pˆ  p 0

.59  .50

.09
 4.05
.0222
p 0 (1  p 0 )
.5(1  .5)
507
n
d. p-value  .00003. This is the probability (area) to the right of z = 4.05. To find the p-value exactly, use software or
a calculator that can give normal curve probabilities. Table A.1 can be used to approximate the p-value. Due to
symmetry, the area to the right of z = 4.05 equals the area to the left of z = 4.05. Near the bottom of left-side page of
Table A.1, probabilities (areas) to the left of 3.72 and 4.25 are given as .0001 and .00001respectively. The correct
p-value is between these two values. Typically, this information may be stated as p-value < .0001.
e. Reject the null hypothesis and decide in favor of the alternative hypothesis. The p-value is smaller than 0.05. This is
evidence that more than one-half (p >.50) of all adult American Catholics favor allowing women to be priests.
pˆ (1  pˆ )
.58 (1  .58 )

 .022 .
n
507
An approximate 95% confidence interval is p̂  2  standard error, which is .59 (2) (.022), or .55 to .63. All values
in this confidence interval are greater than .50. This confirms the conclusion that a majority of adult American
Catholics favor allowing women priests.
f. Standard error = s.e.( pˆ ) 
11.29
The p-value would be 2.04 = .08. As long as the p-value is less than 1/2, the p-value for a two-sided test is twice the
p-value for a one-sided test of the same proportion. We're assuming that the null value p 0 is the same for both the onesided and two-sided tests.
11.31
Step 1: H0: p = .5
Ha: p  .5
where p = the proportion of college students whose feet are the same length.
Step 2: The necessary conditions for using the z-statistic are present. This is a convenience sample, but the assumption
is that the sample represents all college students in the United States, which is reasonable for a question about foot
lengths. The sample size is large enough so that both np0 and n(1  p 0 ) are greater than 10, as they should be to use a
z-statistic. Here, n = 215 and p 0 = .5.
Test statistic is z 
Sample estimate - Null value

Null standard error
pˆ  p 0
p 0 (1  p 0 )
n
103
 .4791
215
pˆ  p 0
.4791  .5  .0209


 0.61
.0341
p 0 (1  p 0 )
.5(1  .5)
215
n
Sample estimate = pˆ 
z
Step 3: p-value  .54. Table A.1 gives the information that P(z < 0.61) =.2709. The alternative hypothesis is twosided so the p-value is 2 .2709 = .5418.
Step 4: Cannot reject the null hypothesis. The result is not statistically significant because the p-value is not less than
.05 (the usual standard for significance).
Step 5: We cannot conclude that the proportion of United States college students whose feet are the same length differs
from one-half.
Note: Minitab can be used to find the z-statistic and corresponding p-value. The p-value differs slightly from the one in
Step 3 because Minitab uses more significant digits for the calculations. See the Minitab Tip on page 373 of the text
for guidance. The output for this exercise is
Output for Exercise 11.31
Test of p = 0.5 vs p not = 0.5
Sample
X
1
103
11.61.
N Sample p
215 0.479070
95.0% CI
(0.412294, 0.545845)
Z-Value
-0.61
P-Value
0.539
a. .p-value = .1470. This is the combined probability to the right of z = 1.45 and left of z = 1.45. Table A.1 can be
used to find that P(Z < 1.45) = .0735. By symmetry, the area to the right of z = 1.45 is also .0735. So, the p-value = 2
 .0735 = .1470.
Figure for Exercise 11.61a
Stat 250.3
November 5, 2003
b. p-value = .0375. This is the probability (area) to the left of z = 1.78. Table A.1 can be used to find P(Z < 1.78) =
.0375.
Figure for Exercise 11.61b
c. p-value = .9625 This is the probability (area) to the left of z = 1.78. Table A.1 can be used to find P(Z < 1.78) =
.9625.
Figure for Exercise 11.61c
d. p-value = .0122. This is the probability (area) to right of z = 2.25. By the symmetry of the normal curve, this
probability equals the area to the left of z = 2.25. Table A.1 can be used to find P(Z< 2.25)=.0122.
Figure for Exercise 11.61d
11.62
The rejection region rules are given on page 372 of the text.
a. Rejection region rule is to reject H0 if | z | >1.96.
Cannot reject H0. z-statistic = 1.45 is not greater than 1.96.
b. Rejection region rule is to reject H0 if z < 1.645.
Reject H0. z-statistic = 1.78 is less than 1.645.
c. Rejection region rule is to reject H0 if z < 1.645.
Cannot reject H0. z-statistic = 1.78 is not less than 1.645.
d. Rejection region rule is to reject H0 if z >1.645.
Reject H0. z-statistic = 2.25 is greater than 1.645.