Solutions to Worksheet on Hypothesis Tests
1. A production line produces rulers that are supposed to be 12 inches long. A sample of 49 of the rulers
had a mean of 12.1 and a standard deviation of .5 inches. The quality control specialist responsible for
the production line decides to do a hypothesis test at the 90 percent significance level to determine
whether the production line is really producing rulers that are 12 inches long or not.
a. What is the null hypothesis?
Solution:
H0 : µ = 12
b. What is the alternative hypothesis?
Solution:
H1 : µ ≠ 12
c. What is the value of the test statistic?
Solution:
x−µ
12.1 − 12
z = s 0 = .5
= 1.4
n
49
d. What is the rejection region (with its numerical value)?
Solution:
z > zα = z.05 = 1.645
2
e. What conclusion do you draw?
Solution:
Do not reject H0 since | 1.4| is not greater than 1.645.
f. What does this mean in terms of the problem situation?
Solution:
We cannot conclude that the mean length of the population of rulers from the production line is not
equal to 12. We should continue with our working assumption that it is equal to 12.
2. A sample of 20 OU freshmen had a mean GPA of 2.8 over all their courses taken in their first semester at
OU. This had a variance of .25. Perform a hypothesis test at the 95 percent level to determine if the first
semester GPA of all OU freshmen is less than a B (3.0).
a. What is the null hypothesis?
Solution:
H0 : µ = 3.0
b. What is the alternative hypothesis?
Solution:
H1 : µ < 3.0 (one-sided test, "less than"
c. What is the value of the test statistic?
Solution:
x −µ
2.8 − 3.0
t = s 0 = .5
≈ −1.79 (small sample)
n
20
d. What is the rejection region (with its numerical value)?
Solution:
t < −t n−1,α = −t19,.05 = −1.729
e. What conclusion do you draw?
Solution:
Reject H0 since -1.79 < -1.729.
f. What does this mean in terms of the problem situation?
Solution:
We must reject our working hypothesis that the mean GPA is 3.0 or more and conclude that it is
less than 3.0.
3. A researcher decides to look at the variance of the production line in Problem 1 She decides to do a
hypothesis test at the 90 percent significance level to determine if the variance is actually less than .25.
a. What is the null hypothesis?
Solution:
H0 : σ 2 = .25
b. What is the alternative hypothesis?
Solution:
H1 : σ 2 < .25
c. What is the value of the test statistic?
Solution:
χ2 =
(n − 1)s2
σ 02
=
( 49 − 1)(.5)2
.25
= 48
d. What is the rejection region (with its numerical value)?
Solution:
χ 2 < χ 2n−1,1 −α = χ 248,.90 ≈ 29.050
e. What conclusion do you draw?
Solution:
We cannot reject H0 since 48 is not less than 29.05
f. What does this mean in terms of the problem situation?
Solution:
We must continue with our working assumption that the true variance is .25 or more and cannot
assume that it is less than .25.
4. A math teacher for Math 4753 wants to determine if a new book proposed for the course is better than
the old book that was used before. The math teacher decides that the two classes might have different
math ability coming into the course. He gives a test at the beginning of the semester to measure the
math ability the students have when they come into the course. He uses this test to create a matched pair
design for 30 students from each class. The scores he gets are as follows:
Student Pair
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
New Book Class
96
95
94
92
92
90
90
89
88
88
86
86
84
84
84
83
83
81
79
78
77
77
72
70
68
67
65
64
62
62
Old Book Class
97
94
94
95
90
91
90
85
87
92
85
81
89
83
85
80
72
89
76
78
75
76
74
81
71
66
65
68
65
58
Perform a hypothesis test at the 95 percent level for the matched pairs design in Problem 8 to see if the
scores for the new book are higher than those for the old book.
a. What is the null hypothesis?
Solution:
H0 : µ D = µ1 −µ 2 = 0 ( µ1 is the mean for the new book)
b. What is the alternative hypothesis?
Solution:
H1 : µ D = µ1 −µ 2 > 0
c. What is the value of the test statistic?
Solution:
First we need to find the differences, their mean and their variance:
Student Pair
New Book Class Old Book Class Differences
1
96
97
-1
2
95
94
1
3
94
94
0
4
92
95
-3
5
92
90
2
6
90
91
-1
7
90
90
0
8
89
85
4
9
88
87
1
10
88
92
-4
11
86
85
1
12
86
81
5
13
84
89
-5
14
84
83
1
15
84
85
-1
16
83
80
3
17
83
72
11
18
81
89
-8
19
79
76
3
20
78
78
0
21
77
75
2
22
77
76
1
23
72
74
-2
24
70
81
-11
25
68
71
-3
26
67
66
1
27
65
65
0
28
64
68
-4
29
62
65
-3
30
62
58
4
-0.2
d =
16.7172414
sd2 =
The value of the test statistic is then
d − δ0
−.2 − 0
−.2
t= s
=
≈
≈ −0.273
16.1724
d
.7342
30
n
d. What is the rejection region (with its numerical value)?
Solution:
t > tn −1,α = t 29,.05 = 1.699
e. What conclusion do you draw?
Solution:
We cannot reject H0 since -0.273 is not greater than 1.699.
f. What does this mean in terms of the problem situation?
Solution:
We must continue to use the working assumption that the means are not different and cannot
conclude that the means for the new book are higher than for the old book.
5. Five engineering students and 5 computer science students are arranged in a matched pair design on the
basis of their GPA prior to the last semester of their senior year. Their GPAs for their last semester are
as follows:
Pair Number
1
2
3
4
5
Engineering Major
3.89
3.61
2.80
2.75
3.41
CS Major
375
3.87
3.02
2.92
1.28
Perform a 90 percent hypothesis test to determine whether the means for CS majors are less than the
means for Engineering majors.
a. What is the null hypothesis?
Solution:
H0 : µ D = µ1 −µ 2 = 0 (This is Engineers minus CS majors)
b. What is the alternative hypothesis?
Solution:
H1 : µ D = µ1 −µ 2 > 0 (Since this is Engineering majors minus CS majors and we are interested
in whether the means for CS majors are less, we can answer the question by seeing if the
Engineering majors are more and that's what we are doing since we have already found the
differences as Engineering majors minus CS majors and not the other way around.)
c. What is the value of the test statistic?
Solution:
First we need to find the differences, their mean and their variance:
Pair Number
1
2
3
4
5
Engineering
CS Major
Major
3.89
3.61
2.8
2.75
3.41
d=
sd2 =
So
d − δ0
.324 − 0
.324
t= s
=
≈
≈ 0.709
1.04413
d
.4570
5
n
d. What is the rejection region (with its numerical value)?
Solution:
t > tn −1,α = t 4, .1 = 1.533
Differences
3.75
3.87
3.02
2.92
1.28
0.14
-0.26
-0.22
-0.17
2.13
0.324
1.04413
e. What conclusion do you draw?
Solution:
We cannot reject H0 since .709 is not greater than 1.533.
f. What does this mean in terms of the problem situation?
Solution:
We must continue to assume that there is no difference between the grades for the two majors and
cannot conclude that CS majors have lower grades.
6. The teacher in Problem 5 is still not sure about the differences in Problem 5 between CS and Engineering
majors. He takes independent samples of size 10 from each major at random. He does not use their
previous GPA to make matched pairs. The 10 Engineering majors have a mean last-semester GPA of
2.89 with a variance of .5. The 10 CS majors have a mean last semester GPA of 2.66 with a variance of
.3. He decides to do a hypothesis test to see if the GPAs are different for the two populations. He does
the test at the 90 percent level.
a. What is the null hypothesis?
Solution:
H0 : µ1 − µ 2 = 0
b. What is the alternative hypothesis?
Solution:
H1 : µ1 − µ 2 ≠ 0
c. What is the value of the test statistic?
Solution:
x − y − δ0
2.89 − 2.66 − 0
0.23
=
≈
≈ 1.247
1
1
1
1  0.1844

s
+
(.17) + 
n1 n2
10 10
2
(n1 − 1)s1 + ( n2 − 1)s22 (10 − 1)(.5)2 + (10 − 1)(.3)2 3.06
2
=
=
= 0.17
where s =
n1 + n2 − 2
10 + 10 − 2
18
t=
d. What is the rejection region (with its numerical value)?
Solution:
t > tn 1 +n 2−2 ,α = t18,.05 = 1.734
2
e. What conclusion do you draw?
Solution:
We cannot reject H0 because 1.247 is not greater than 1.734.
f. What does this mean in terms of the problem situation?
Solution:
We cannot conclude that the two means are different since we cannot reject the null hypothesis.
7. Suppose in the situation in Problem 6, the teacher had the GPAs for a sample of 10 Engineering majors
but only 8 CS majors but otherwise the data were the same. Suppose he decided to do a 90 percent
hypothesis test for the difference in GPAs for the two populations of majors.
a. What is the null hypothesis?
Solution
H0 : µ1 − µ 2 = 0
b. What is the alternative hypothesis?
Solution
H1 : µ1 − µ 2 ≠ 0
c. What is the value of the test statistic?
Solution
t=
x − y − δ0 2.89 − 2.66 − 0
0.23
=
≈
≈ 1.208
2
2
2
2
0.1904
s1 s1
.5) (.3)
(
+
+
n1 n2
10
8
d. What is the rejection region (with its numerical value)?
Solution
t > tν, α = t15,.05 = 1.753
2
2
2
 (.5)2 (.3)2 
 s12 s22 


+ 
+
 n1 n2 
 10
8 
(.03625)2
where ν =
=
≈ 15.01 ≈ 15
2
2 =
2
2 2
 s12 
 s22 
 (.5)2 
8.7525 × 10 −5

(
)
.3

 
 




 n1 
 n2 
 10 
 8 
+
+
n1 − 1 n2 − 1
10 − 1
8 −1
e. What conclusion do you draw?
Solution
We cannot reject H0 because 1.208 is not greater than 1.753.
f. What does this mean in terms of the problem situation?
Solution
We cannot conclude that the means are different and must continue to assume they are equal.
8. The math teacher for Math 4753 wants to determine if the new book proposed for the course is better
than the old book that was used before. He teaches two classes that are each a random sample of 42
people. The class using the new book got an average of 88 on the final exam given by the math teacher
and had a standard deviation of 3 points. The class using the old book had an average of 90 with a
standard deviation of 2 points. Suppose the teacher decides to do a hypothesis test at the 98 percent
level to see if the variances for Engineering majors is lower than the variances for CS majors.
a. What is the null hypothesis?
Solution:
H0 : σ 12 ≤ σ 22
b. What is the alternative hypothesis?
Solution:
H1 : σ 11 > σ 22
c. What is the value of the test statistic?
Solution:
s12 (3) 2
F= 2 =
= 2.25
s2 (2 )2
d. What is the rejection region (with its numerical value)?
Solution:
F > fn1 −1, n2 −1,α = f41,41,.01 ≈ 2.11 using the values for 40,40 rather than for 41,41.
e. What conclusion do you draw?
Solution:
We can reject H0 since 2.25 is greater than 2.11.
f. What does this mean in terms of the problem situation?
Solution:
We conclude that the variance for the new books is higher than the variance for the old books.
9. A survey of 4000 people in the US finds that 2856 of them believe that daily weather reports are totally
useless because meteorology is not really a science. Given this data perform a 95 percent hypothesis test
to see if more than half of the people in the US believe that weather reports are useless.
a. What is the null hypothesis?
Solution:
H0 : p ≤ .5
b. What is the alternative hypothesis?
Solution:
H1 : p > .5
c. What is the value of the test statistic?
Solution:
z=
pˆ − p0 .714 − .5 .214
=
≈
≈ 27.09
p0 q0
.5 × .5 .0079
n
4000
d. What is the rejection region (with its numerical value)?
Solution:
z > zα = z.05 ≈ 1.645
e. What conclusion do you draw?
Solution:
We can reject H0 since 27.09 > 1.645.
f. What does this mean in terms of the problem situation?
Solution:
We can conclude that it is likely (at the 95% confidence level) that more than half of the people in
the US believe that weather reports are useless.
10. A survey of 200 regular viewers of Channel 5 in Oklahoma City show that 68 believe that Gary
England is God. A survey of 100 regular viewers of Channel 9 in Oklahoma City show that 68 of them
also believe that Gary England is God. Given these data perform a 90 percent hypothesis to determine if
the proportion of Channel 9 viewers who believe that Gary England is God is greater than the
proportion of Channel 5 viewers who believe it.
a. What is the null hypothesis?
Solution:
H0 : p2 − p1 ≤ δ 0 = 0 (Population 2 are the regular Channel 9 viewers)
b. What is the alternative hypothesis?
Solution:
H1 : p2 − p1 > δ 0 = 0
c. What is the value of the test statistic?
Solution:
pˆ 2 − pˆ 1
.68 − .34
.34
=
≈
≈ 5.58
1  .06097
1
 1
1

.4533 × .5467
+
pˆ qˆ  + 
200 100 
 n1 n2 
x +y
68 + 68
where pˆ =
=
≈ 0.4533 and we use this form of the test statistic because
n1 + n2 200 + 100
δ0 = 0 (see Equation 9.12).
z=
d. What is the rejection region (with its numerical value)?
Solution:
z > zα = z.1 ≈ 1.285
e. What conclusion do you draw?
Solution:
Reject H0 because 5.58 > 1.285.
f. What does this mean in terms of the problem situation?
Solution:
This means that we can conclude that H1 is true and that p2 − p1 > 0 or, in other words, p2 > p1
(a greater proportion of regular Channel 9 viewers believe that Gary England is God than the
proportion of regular Channel 5 viewers that believe that Gary England is God. [Oh, please, Gary,
forgive the Channel 5 viewers for they know not what they do]).