Old Exam Questions - Solutions
Hypothesis Testing (Chapter 7)
1. First note that this is a claim about a population PROPORTION. Thus we will be using the
symbol p. Note that the claim can be written symbolically as
p 1
4
The opposite of the claim is
p≤ 1
4
Since the claim does NOT contain equality, it becomes the alternative hypothesis. The
hypothesis always has  in it. Thus we have
H0 : p  1
4
1
H1 : p 
4
2. First note that the claim, null & alternative hypotheses are
Claim:  ≠ 98. 6
H 0 :   98. 6
H 1 :  ≠ 98. 6
Since the p value (0.045) is less than  (0.05), the conclusion about the null hypothesis
would be: Reject the null hypothesis. Using Figure 7-7 on page 346, the final wording
about the claim would be: "The sample data support the claim that the mean body
temperature of adult humans is not 98.6 degrees."
3. First note that for the sample data we have
n  23, x̄  3. 5 years, and s  1. 9 years
We do NOT know the population standard deviation and we are told to assume the
population distribution is normal. Thus we will use the t distribution to find critical values.
Further, note that the claim, null & alternative hypotheses are
Claim:   4
H0:   4
H1:   4
Thus we have a LEFT-tail test (we will need to make sure the critical value is negative!).
We are told that   0. 1 and note that the degrees of freedom  n - 1  23 - 1  22. Thus
look in Table A-3 with   0. 1 (in ONE tail) and df  22. We find that our critical value
is -1.321. Note that we had to include the negative sign since this was a LEFT-tail test.
Note that the decision test will be that we will reject the null hypothesis if our test statistic
is less than -1.321 (T.S.  -1.321).
4. Note we are testing a claim about a population standard deviation. Thus critical values will
come from Table A-4 (chi-square distribution). We have a left tail test. Since   0. 05,
the area to the RIGHT of the critical value must be 1 - .05  .95. With df  24the critical
value is 13.848. The correct choice is (b).
5. Note that this is a claim about a population PROPORTION. We are given that n  603 and
p̂  0.43.
Claim: p  0.5
H 0 : p  0.5
H 1 : p ≠ 0. 5
We have a two-tail test.
Critical Values: 1. 96 (from Table A-2)
Decision Test: Reject the null hypothesis if T.S.  -1.96 or if T.S.  1.96.
Test Statistic:
p̂ − p
z
 0. 43 − 0. 5 ≈ −3. 44
pq
n
0.50.5
603
Null hypothesis Conclusion: Reject H 0
Non-technical restatement: There is sufficient evidence to warrant the rejection of the
claim that McDonald’s is preferred by half of all children.
6. To compute the p value for #5, we find the area under the standard normal curve to the left
of the test statistic and double it since we have a two-tail test (see Figure 7-6 on page 344).
The test statistic was -3.44. Using Table A-2, the area to the left of the test statistic is
0.0003. Thus the p value is 2(0.0003)  0.0006.
7. This is a test of a population MEAN. We have
n  35, x̄  6. 3 hours, and s  2. 1 hours
Since n  30 and the population standard deviation is unknown, we have a t test.
Claim:   7
H0:   7
H1:   7
We have a left-tail test. Thus the critical value is -2.441 (From Table A-3 with df  34 and
  0. 01 in one tail). Hence the decision test is to reject the null hypothesis if the test
statistic is less than -2.441. The test statistics is
x̄ − 
t
 6. 3 − 7 ≈ −1. 97
s
n
2.1
35
Since the test statistic is not in the critical region, the conclusion is: Failure to reject the
null hypothesis. Using Figure 7-7 on page 346 (and in your formula card), the wording of
the final conclusion is: "There is NOT sufficient sample evidence to support the claim that
statistics students spend less than 7 hours studying per week."
8. We are testing a claim about a population mean. We have n  7 and
x̄ ≈ 9915. 6 and s ≈ 693. 4
Since we are assuming that the population is normally distributed, we have a t-test.
Claim:   9, 000
H 0 :   9, 000
H 1 :   9, 000
We have a right-tail test.
Critical Value: 1.943 (t-test with 6 degrees of freedom and   0. 05 in one tail)
Decision test: Reject the null hypothesis if T.S.  1.943.
Test Statistic:
x̄ − 
t
 9915. 6 − 9000 ≈ 3. 494
s
n
693.4
7
H 0 Conclusion: Reject H 0
Non-technical restatement: The sample data SUPPORT the claim that the mean annual
consumption amount is more than 9,000 kWh.
9. We are testing a claim about a population standard deviation. We have
n  22 and s  0. 0089
The claim is   0. 01 so we have
H 0 :   0. 01
H 1 :   0. 01
This is a left-tail test.
Critical Value: 11.591 (Table A-4 with df  21 and   0. 05 so that area to right is 1 0.05  0.95)
Decision Test: Reject H 0 if TS  11.591.
Test Statistic:
n − 1s 2
22 − 10. 0089 2

≈ 16. 634
2 
2
0. 01 2
H 0 Conclusion: Fail to reject H 0
Non-technical restatement: There is NOT sufficient sample evidence to support the claim
that the standard deviation is less than 0.01 millimeters.