CHAPTER 19 SOLUTIONS AND MINI-PROJECT NOTES
CHAPTER 19
THE DIVERSITY OF SAMPLES FROM THE SAME POPULATION
EXERCISE SOLUTIONS
19.1
For condition 1, there exists a population (students at the college) with a fixed proportion with a certain trait
(left-handed). For condition 2, a random sample of students is selected. For condition 3, the sample size of
200 is large enough that we are indeed likely to see at least five right-handed and five left-handed students.
19.2
The picture is bell-shaped with a mean of 0.12 and standard deviation of 0.023 ( (.12)(. 88) ).
200
19.3
a. There exists an actual population of adults, a fixed proportion of whom have fallen asleep at the wheel
in the last year (and technically, are willing to admit it when asked!).
b. The standard deviation is the square root of (.4)(.6)/1027, or about 0.015. The interval in which about
95% of sample proportions should fall is thus about .4  .03, or .37 to .43, equivalent to 37% to 43%. So
the observed proportion of 37% is just covered by the interval.
c. It would be reasonable to conclude that they differ, since 25% is very far from the interval computed in
part b.
19.4
a. The picture is bell-shaped with a mean of 0.50 and standard deviation of the square root of
(0.5)(0.5)/800 = 0.0177 or about 0.02.
b. Yes, it would be very unlikely. The picture does not extend that far. In fact, it is virtually impossible. It
represents a standardized score of over 10.
c. Optimism or overconfidence. People think they are likely to be better than average on most things.
19.5
Possible sample means are bell-shaped with a mean of 25 and standard deviation of 1/ 9 = 1/3 or about
0.3. So 68% are between 24.7 and 25.3, 95% are between 24.4 and 25.6 and almost all are between 24.1
and 25.9.
19.6
Now the picture is even more compact. The mean is still 25 but the standard deviation is only 1/10 = 0.1.
19.7
This will differ for each student.
19.8
Using the Rule for Sample Means, we would expect the possible sample means to be bell-shaped, with a
mean of 105 and a standard deviation of 15/ 100 = 15/10 = 1.5. Therefore, we would find them: between
103.5 and 106.5 about 68% of the time, between 102 and 108 about 95% of the time, between 100.5 and
109.5 about 99.7% of the time.
19.9
Yes. The Rule for Sample Proportions tells us that the sample proportions for this situation would be bellshaped with a mean of 0.60 and standard deviation of the square root of (0.6)(0.4)/100 = 0.049. Therefore,
95% of the time, the sample proportion would be within 2 standard deviations (0.098) of 0.60, or between
0.502 and 0.698. So there is a very small probability (less than 2.5%) that the sample proportion would fall
below 50%. As long as the sample proportion fell above 50%, there would be convincing evidence that the
quarter system is favored. Another way to see this is to compute the standardized score for a proportion of
0.50. It would be (0.5 − 0.6)/0.049 = −2.04. That lies at the 2nd percentile, meaning that a value below
50% would occur only with probability of 0.02 or 2%.
19.10
a. The standard deviation for the possible sample proportions is approximately the square root of
(.09)(.91)/8709, or about .003. Almost all sample proportions are likely to be within three standard
deviations of the true proportion, which is about .009, or just under 1%.
b. The proportion found in 1976 is likely to be within 1% of the true proportion for that time period,
indicating that it was between 8% and 10%. The value of 25% found in 1997 far exceeds this range, so it is
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CHAPTER 19 SOLUTIONS AND MINI-PROJECT NOTES
likely to represent a real increase in the proportion of all adults who believe in reincarnation.
19.11
a. The frequency curve would be bell-shaped with a mean of 0.2 and a standard deviation of 0.008.
b. No, it is not in danger. We can be almost certain that the sample proportion will be between 17.6% and
22.4%. So it is not impossible, but quite unlikely. The standardized score is (.17−.2)/.008 = −3.75.
19.12
The Rule for Sample Means says that the standard deviation of the frequency curve for the samples will be
the population standard deviation divided by the square root of the sample size. So the larger the sample
size the smaller the standard deviation, and hence the better the estimate.
19.13
a. The picture is bell-shaped with mean 6.95 hours and standard deviation of 0.15 hours.
b. Yes, a mean of 7.1 hours is about one standard deviation above the mean of the possible values for
sample means based on 190 people, so it is well within the range of what is expected.
c. The sample is unlikely to be representative of all adults. There are many ways in which college students
differ from all adults in factors affecting sleep, such as age range, flexibility of schedule, perceived need
and desire to stay up all night, and so on.
19.14
a. This random sample does not meet the criteria of condition 3, because the sample size is not large
enough. Even if the sample did show exactly 10% of the cars did not meet emissions standards, this would
be only 3, not the required minimum of 5.
b. This meets all the conditions.
c. This sample would violate condition 1. It would not be random, and the relative frequency of rain or
snow changes from the winter months to the summer months.
d. This sample meets all the conditions.
19.15
a. It does not apply because the sample is not random; he biased it by using only heart attack patients.
b. Again this is probably a biased sample, so the rule does not apply. People who leave during that time
may be different in their responses from people who work late.
c. Incomes are not bell-shaped but this is a large sample so the rule applies (as long as the university is
careful to contact the particular people chosen to be in the sample). The population under consideration is
alumni and the measurement of interest is income.
d. The rule would probably not apply here because sales prices are not likely to be bell-shaped and the
sample is small.
19.16
a. 68% of the time the sample proportion should be between 0.548 and 0.572, 95% of the time it should
be between 0.536 and 0.584, and almost all the time it should be between 0.524 and 0.596.
b. No, it is outside of the expected range.
c. The standard deviation would be 0.025, so the standardized score for 0.61 would be (0.61 0.56)/0.025, or about 2. According to Table 8.1 this is about the 97.5th percentile. It is possible that all the
people sampled were telling the truth.
19.17
It is bell-shaped with a mean of 3.1 and standard deviation of 0.05.
19.18
They want three standard deviations to equal 1, where a single standard deviation is 5/n (n is the size of
the sample). So they need 15/n= 1, or n = 152 = 225.
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CHAPTER 19 SOLUTIONS AND MINI-PROJECT NOTES
NOTES ABOUT MINI-PROJECTS FOR CHAPTER 19
Mini-Project 19.1
Make sure the method is one for which you put all of the individuals back in place before the start of each sample,
and for which the proportions are correct. As an example of how to do this, one student used two different colors of
cotton swabs, counting out 240 blue and 160 yellow, and placing them in a container. He picked 15, counted the
proportion of yellow, then put them all back and started over. The Rule for Sample Proportions tells us that the
sample proportions based on a true proportion of 0.4 and a sample size of 15 are approximately bell-shaped with a
mean of 0.40 and standard deviation of 0.126. So we should see about 68% of our samples (13 or 14 of them)
between 0.274 and 0.526, about 95% (19 of them) between .148 and .652 and all 20 between 0.022 and 0.778. You
should compare the percentages you found in each of the three intervals to what the rule tells us to expect.
Mini-Project 19.2
Make sure the method is one for which everyone goes back into the population after each sample of 9 is drawn. The
Rule for Sample Means tells us to expect the sample means to be bell-shaped with a mean of 110 and standard
deviation of 5/3 = 1.67. Therefore, about 68% of the sample means should fall between 108.33 and 111.67 (about
13 or 14 of the sample means), 95% should fall between 106.67 and 113.33 (about 19 of the sample means) and all
20 should fall within 5 of the population mean, or between 105 and 115. Compare your results to these anticipated
percentages.
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