Chapter Solutions
Solution 1
We use the five-step hypothesis testing procedure for the solution.
Step 1: State the null hypothesis and the alternate hypothesis.
Note that the testing company is attempting to show only that there is a difference in the time required to
affect relief. There is no attempt to show one tablet is “better than” or “worse than” the other. Thus, a
two-tailed test is applied.
H0:  1 =  2
H1:  1   2
Let 1 refer to the mean time to obtain relief using SINUS and 2 to the mean time to obtain relief using
ANTIDRIP.
Step 2: Select a level of significance.
The 0.05 significance level is to be used. The alternate hypothesis does not state a direction, so this is a
two-tailed test. The 0.05 significance level is divided equally into two tails of the standard normal
distribution. Hence, the area in the left tail is 0.0250 and the area in the right tail is 0.0250.
Step 3: Identify the test statistic.
Because both samples are large (greater than 30) the z distribution is used as the test statistic.
Step 4: Formulate a decision rule based on the
selected test statistic and level of significance.
The critical values that separate the two rejection
regions from the region of acceptance are 1.96 and
+1.96. To explain: if the area in a rejection region is
0.0250, the acceptance area is 0.4750, found by
0.5000  0.0250. The z value corresponding to an
area of 0.4750 is obtained by referring to the table of
areas of the normal curve (Appendix D). Search the
body of the table for a value as close to 0.4750 as possible and read the corresponding row and column
values. The area of 0.4750 is found in the row 1.9 and the column 0.06. Hence, the critical values are +
1.96 or 1.96.
This decision rule is shown on the following diagram.
Step 5: Make a decision to reject or not to reject the null hypothesis and interpret the results.
The computed value of z is 1.24, found by using formula [10-2]. Because the population standard
deviations are not known, the sample standard deviations are substituted.
z
X1  X 2
s12 s22

n1 n2

85.0  86.2
b6.0g b6.8g
2
100
177
2
 124
.
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The computed value of z is between 1.96 and +1.96. Thus H0 is not rejected. We conclude that there is
no difference in the mean time it takes SINUS and ANTIDRIP to bring relief. The difference of 1.2
minutes (85.0  86.2) can be attributed to sampling error (chance).
To determine the p-value we need to find the area to the left of 1.24 and add to it the area to the right of
1.24. We are concerned with both tails because H1 is two-tailed. The p-value is 0.2150, found by 2(0.5000
 0.3925). Since the p-value of 0.2150 is greater than the level of significance of 0.05, do not reject H0 .
Solution 2
The problem is to examine whether a higher proportion of working mothers of young children live on the
south side.
Step 1: State the null hypothesis and the alternate hypothesis.
The hypotheses are:
H 0 : 1   2
H1 :  1   2
Where:
1
refers to the proportion of working mothers on the south side.
2
refers to the proportion of working mothers on the east side.
Step 2: Select a level of significance.
The 0.05 significance level is stated in the problem.
Step 3: Identify the test statistic.
The standard normal distribution is the test statistic to be used. The z value is computed using formula
[10-3].
Step 4: Formulate a decision rule based on the selected test statistic and level of significance.
The alternate hypothesis indicates a direction, so this is a one-tailed test.
The critical value is 1.65 obtained from Appendix
D. The area in the upper tail of the curve is 0.05,
therefore the area between z = 0 and the critical
value is 0.4500, found by (0.5000  0.0500).
Search the body of the table for a value close to
0.4500. Since 1.64 is equal to 0.4495 and 1.65 is
equal to 0.4505, a value between 1.64 and 1.65 or
(1.645) could be used as the critical value. We
take the conservative approach and use 1.65. The
null hypothesis is rejected if the calculated z value
is greater than 1.65. This information is summarized in the diagram above.
Formula [10-3] for z is repeated below.
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p1  p2
z
pc 1  pc  pc 1  pc 

n1
n2
where pc is a pooled estimate of the population proportion and is computed using formula [10-4].
pc 
X1  X 2
n1  n2
In this problem X1, and X2 refer to the number of “successes” in each sample (number of working mothers
with children under 5 years), n1 and n2 refer to the number of housing units sampled in the south side and
east side, respectively. The pooled estimate of the population proportion is 0.4143, found as follows:
pc 
X1  X 2
88  57

 0.4143
n1  n2
200  150
Inserting the pooled estimate of 0.4143 in the formula and solving for z in formula [10-3] gives 1.13.
z
p1  p2
0.44  0.38

 113
.
b g p b1  p g b0.4143gb1  0.4143g b0.4143gb1  0.4143g
pc 1  pc
n1
c
c
200
n2
150
Step 5: Make a decision to reject or not to reject the null hypothesis.
The computed value: z = 1.13, is less than the critical value of 1.65 so the null hypothesis is not rejected.
The difference can be attributed to sampling error (chance). To put it another way, the proportion of
mothers who work and have children under 5 on the south side is not significantly greater than the east
side. The p-value is 0.1292, found by (0.5000  0.3708). So, the probability of finding a value of the test
statistic this large or larger is 0.1292.
Solution 3
The null hypothesis is that accounting majors earn the same or less than general business majors. The
alternate hypothesis is that accounting majors earn more. They are written as follows:
H0: 1   2
H1: 1 >  2
Where:
1 refers to accounting majors (graduates).
2 refers to general business majors (graduates).
The required assumptions are:
1. The samples are independent.
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2. The two populations follow the normal distribution.
3. The population standard deviations are equal.
The t distribution is the test statistic. There are 16 degrees of freedom, found by (n1 + n2  2) = (10 + 8
 2) = 16. The alternate hypothesis is a one-tailed test with the rejection region in the upper tail. From
Appendix F, the critical value is 1.746. Hence, H0 is rejected if the computed value of the test statistic
exceeds 1.746.
The first step is to pool the variances, using formula [10-5].
s 2p
s b
n  1g
s
bn  1gdi
di
b10  1gb2,000g b8  1gb1,500g 3,234,375


2
1
1
2
2
2
2
n1  n2  2
2
10  8  2
Next, the value of t is computed, using formula [10-6].
t
X1  X 2
s 2p

$30,000  $29,000
1 1I
F
F
1
1I
3,234,375g
G
 J b
G
H10  8 J
K
Hn n K
1
1,000


b g
3,234,375 0.225
1,000
 117
.
853.073
2
Because the computed value of t (1.17) is less than the critical value of 1.746, H0 is not rejected. The
sample evidence does not suggest a difference in the mean salaries of the two groups.
We determine the p-value by using Appendix F. Move down the left column to the 16 degrees of freedom
row. Move across the row until you locate a value larger than 1.17—the computed value of t. The value
is in the first column. It is 1.337. Note that this column has 0.10 as a heading. Since this is a one-tailed
test, we conclude that the p-value is greater than 0.10.
Solution 4
Let d be the mean difference between the fall and spring semester grades for all business students at
Kingsport U. in their senior year. Since we want to explore whether grades decrease, a one-tailed test is
appropriate.
H0:  d  0
H1:  d > 0
There are six paired observations; therefore, there are (n  1) = (6  1) = 5 degrees of freedom. Using
Appendix F with 5 degrees of freedom, the 0.05 significance level and a one-tailed test, the critical value
of t is 2.015. H0 is rejected if the computed value of the test statistic exceeds 2.015.
The value of the test statistic is determined from formula [10-7]. t 
d
sd / n
Where:
d is the mean of the differences between fall and spring GPAs.
sd is the standard deviation of those differences.
n is the number of paired observations.
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First, subtract Spring semester grades from fall semester grades. If this difference is positive, then a
decline has occurred. The sample data is shown below and the values of d and sd computed:
Student
Fall
Spring
d
d2
A
B
C
D
E
F
2.7
3.4
3.5
3.0
2.1
2.7
3.1
3.3
3.3
2.9
1.8
2.4
0.4
0.1
0.2
0.1
0.3
0.3
0.6
0.16
0.01
0.04
0.01
0.09
0.09
0.40
The t statistic is computed by:
t
d
d 0.6

 010
.
N
6
bd g
2
sd 
d 2 
n 1
n
b0.6g
2

0.40 
6 1
6
 0.2608
d
010
.
010
.


 0.94
.
sd / n 0.2608 / 6 01065
Since the computed value of t (0.94) is less than the critical value of 2.015, H0 is not rejected. The
evidence does not suggest a reduction in grades from the fall to the spring semester. The decrease in
GPAs can be attributed to chance. The p-value is greater than 0.10.
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