22 CHAPTER SOLUTIONS AND MINI-PROJECT NOTES
CHAPTER 22
REJECTING CHANCE TESTING HYPOTHESES IN RESEARCH
EXERCISE SOLUTIONS
22.1
Null: Mothers who smoke have the same average level of education as those who do not smoke.
Alternative: Mothers who smoke do not have the same average level of education as those who do not
smoke.
22.2
If there were no difference in average education level for the two populations (mothers who smoke while
pregnant and those who do no), we would expect to see a sample difference as large or larger than the one
we saw in about 1% of all samples of this size.
22.3
a. Null: Calcium does not have an impact on the severity of premenstrual symptoms. Alternative:
Calcium reduces the severity of premenstrual symptoms.
b. Type 1, concluding calcium has an impact when it does not.
c. The consequence of a type 1 is that some women may take calcium because they have been told it
would reduce premenstrual symptoms, but in fact it doesn't. The consequence of a type 2 error is that
women are told calcium is not effective in reducing premenstrual symptoms, and thus do not take it, when
in fact it would help.
22.4
a. Null: The calcium and placebo-treated participants came from populations with equally severe PMS
mood swings. Alternative: The calcium and placebo-treated participants came from populations with PMS
mood swings of different severity. Conclusion: The null hypothesis is not rejected, we cannot conclude
that the two sets of participants come from different populations.
b. Null: At the third cycle after treatment with calcium and treatment with placebo, mood swings would
be equally severe among women similar to those in this study. Alternative: At the third cycle after
treatment with calcium, mood swings would be less severe than after similar treatment with a placebo.
Conclusion: Reject the null hypothesis and accept the alternative, treatment with calcium significantly
reduces mood swings when compared to treatment with a placebo.
c. Null: At the third cycle after treatment with calcium and treatment with placebo, insomnia would be
equally severe among women similar to those in this study. Alternative: At the third cycle after treatment
with calcium, insomnia would be less severe than after similar treatment with a placebo. Conclusion: Do
not reject the null hypothesis, treatment with calcium does not significantly reduce insomnia when
compared to treatment with a placebo.
22.5
a. Null: Psychotherapy and desipramine are equally effective in treating cocaine use. Alternative: One
method is more effective than the other.
b. A type 1 error would occur if there is no difference in the treatments, but the conclusion is that they do
differ. A type 2 error would occur if there is a more effective treatment but the study fails to make that
conclusion.
c. Type 1; concluding there is a difference when there isn't.
22.6
a. Null: Working 5 hours a day or more at a computer does not contribute to deteriorating eyesight.
Alternative: Working 5 hours a day or more at a computer does contribute to deteriorating eyesight.
b. Null: Placing infants in incubators does not lead to claustrophobia. Alternative: Placing infants in
incubators leads to claustrophobia.
c. Null: Placing plants in an office does not reduce number of sick days. Alternative: Placing plants in
an office reduces the number of sick days taken by the person in that office.
22.7
a. Type 1 error is that working 5 or more hours at a computer does not harm eyesight but the research
concludes that it is harmful. Millions of workers would have needless concern. Type 2 is that there really
is a harmful effect but it goes undetected. Millions of workers who should be warned about this problem
would not learn about it.
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CHAPTER 22 SOLUTIONS AND MINI-PROJECT NOTES
b. Type 1 error is that incubators do not cause claustrophobia but the research finds that they do. As a
consequence, incubator use might be reduced or counseling offered to people who were really okay. Type
2 is that incubators do cause a problem but it goes undetected. Those who were put in incubators would
wonder why they had claustrophobia and would not be told there was a reason, knowledge that might help
them overcome it.
c. Type 1 error is that plants have no impact but the research finds that they do; as a consequence
companies would needlessly offer plants to their employees. Type 2 error is that plants do help reduce
sick days, but the link goes undetected. A useful strategy would go unused.
22.8
We do not specify a particular value for the population when we state the alternative hypothesis, so there is
no way to determine what to expect by chance as there is when we assume the null hypothesis is true.
22.9
The standard deviation is the square root of (0.344)(0.656)/355 = 0.025; a 95% confidence interval is
0.344 2(0.025) or 0.294 to 0.394 or 29.4% to 39.4%.
22.10
A type 1 error would be that no ESP is present but the study concludes it is being used. A type 2 error
would be that ESP is indeed present but the study does not detect it.
22.11
a. You could say either of the following. "Do not reject the null hypothesis" or "There is no statistically
significant difference (relationship, effect, etc.)."
b. You could say any of the following. "Reject the null hypothesis" or "Accept the alternative
hypothesis" or "Conclude that there is a statistically significant difference (relationship, effect, etc.)."
22.12
a. Null: There is no link between vertex baldness and heart attacks in men. Alternative: There is a link
between vertex baldness and heart attacks men.
b. in There is no link between the two but the study concludes that there is.
c. There is a link between vertex baldness and heart attacks but the evidence in the sample wasn't strong
enough to conclusively detect it.
22.13
a. Null: Level of education has no relationship to developing Alzheimer's disease. Alternative: Highly
educated people have a lower chance of developing the disease.
b. It is implying that a study found a statistically significant relationship. "A new study found that the
chances of highly educated people developing Alzheimer's disease were statistically significantly lower
than the chances for others." (Notice that a causal conclusion is not justified.)
c. It was a one-sided test. The researchers specified a direction in the alternative hypothesis, namely, that
if there was a relationship between education and developing Alzheimers disease, it would be that those
with more education were less likely to develop the disease.
22.14
a. (0.29 0.25)/0.02 = 2.0.
b. From Table 8.1, it is between 97.5 and 98, so about the 97.75th percentile. Using Excel, it is 97.725 th
percentile.
c. The corresponding p-value is about 0.023, or to be exact, 0.02275.
d. The p-value is 0.023, which is less than 0.05, so the null hypothesis is rejected. We can conclude that
the population proportion is higher than 0.25.
e. The alternative hypothesis contains a whole range of values, from just above 0.25 to 1. Therefore we
would not know what to use as the mean and standard deviation.
22.15
a. No! There will be one or two false negatives per 100 infected people tested, and because there is no
way to know who they are they would not be retested. Their contaminated blood would end up in the
blood supply.
b. Of the 100 infected, 99 would test positive. Of the 9,900 not infected, 1% or 99 would test positive.
So the probability of actually being infected given a positive test is 99/(99+99) = 0.5 or one-half.
22.16
a. Minor disease with serious treatment, like tonsillitis where the treatment is surgery.
b. Being infected with HIV or having a fast-spreading cancer.
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CHAPTER 22 SOLUTIONS AND MINI-PROJECT NOTES
22.17
They should use a higher p-value, thus increasing the probability of making a type 1 error and reducing the
probability of a type 2 error.
22.18
Type 1 error would be that there is no relationship between aspirin and heart attack but the study finds one.
People would start taking aspirin needlessly. Type 2 error is that aspirin does help prevent heart attacks
but the study does not discover the relationship, so people who could be helped by aspirin are not told to
take it.
22.19
a. Null: Listening to the sound of a heartbeat in the first few days of life does not influence weight gain
of infants. Alternative: Infants who hear the sound of a heartbeat in the first few days of life gain more
weight than those who do not.
b. This was a one-sided test. Dr. Salks research hypothesis was that listening to the heartbeat sound
would result in infants gaining more weight.
c. A type 1 error could occur if there really is no benefit from listening to a heartbeat, but the study finds
that there is. A type 2 error could occur if there really is a benefit from listening to the heartbeat but the
study fails to detect it.
d. Because Salk rejected the null hypothesis, he could have only committed a type 1 error. A type 2
error can be committed only when the null hypothesis is not rejected.
e. It would have been useful to see a confidence interval for the difference in average weight gain under
the two conditions.
22.20
a. The results probably apply to female capuchin monkeys, but possibly only to those who are kept in
captivity and interact with humans.
b. The alternative hypothesis is that for the population of similar monkeys, the proportion of times they
would cooperate is lower after observing another monkey receiving a free grape than it is after observing
another monkey trading a token for a piece of cucumber.
c. It is less than 0.05.
d. They are using 0.05 as the level of significance.
e. Reject the null hypothesis. We can conclude that for the population of monkeys the proportion willing
to cooperate would be lower after observing another monkey receiving a free grape than after watching
another monkey do what they were asked to do, namely, give up a token for a piece of cucumber.
22.21
Step 1: The null hypothesis is that the population proportion of college students who experienced at least
one hangover symptom two or fewer times in the past year is 0.5, and the alternative hypothesis is that the
proportion is less than 0.5.
Step 2: The standard deviation is the square root of (.5)(.5)/1216 = 0.014. The test statistic is standardized
score = (0.4 0.5)/0.014 = 6.97.
Step 3: The p-value is extremely small. From Table 8.1 we can see that it is less than 0.000000001.
Step 4: Reject the null hypothesis. We can conclude that less than half of college students would report
experiencing hangover symptoms two or fewer times. In other words, the majority of college students
would report experiencing hangover symptoms more than twice in the past year.
22.22
For women only, the sample size is 749 and the proportion is 0.14 (0 times) + 0.28 (1-2 times) = 0.42. The
hypotheses are the same, except that they now refer to the population of college women only. The standard
deviation is the square root of (.5)(.5)/749 = 0.018 and the test statistic is 5.47. The conclusion is the
same as it was for all students, that less than half of college women would report experiencing hangover
symptoms two or fewer times in the past year.
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