HON 180 Review Questions Brief Answers
1. A distribution table is show below. The table gives the distribution of the degree of astigmatism
in 1033 young men, aged 18-22, who were accepted for military service in Great Britain.
Astigmatism is a condition were light rays coming into the eye are not focused on a single point,
causing blurry eyesight. The degree of astigmatism is measured in diopters. The class intervals
include the left endpoint, but exclude the right one.
Degree of Astigmatism (diopters) Percent
0 – 0.15
44
0.15 – 0.35
26
0.35 – 0.55
15
0.55 – 6.00
15
a. Make a histogram for this data. Make sure you label the axes correctly.
b. What is the 70th percentile for the degree of astigmatism in the sample of 1033 men?
0.35 diopters
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2. A hundred draws are made at random with replacement from the box
0
0
0
1
2
a. Calculate the probability that the sum of the one hundred draws will be greater than 65.
Answer. Normalcdf(65, ∞, 0.6, 8) ≈ 0.266
b. Calculate the probability that 1 comes up between 19 and 21 times out of the one hundred draws.
Answer. Normalcdf(19, 21,20,4) ≈ 0.197
3. Los Angeles has about four times as many registered voters as San Diego. A simple random sample of
registered voters is taken in each city, to estimate the percentage who will vote for school bonds. Other
things being equal, a sample of 4000 taken in Los Angeles will be about
(i) four times as accurate
(ii) twice as accurate
(iii) as accurate
as a sample of 1000, taken in San Diego. Choose one option and say why.
Answer. It will be twice as accurate (option ii). The new SE will be
4 N (SD of box)
4 N (SD of box) 2 N (SD of box)


4N
4N
4N
1 N (SD of box) 1

 (original SE)
2
N
2
2
4. A confidence interval calculated from a simple random sample of 4000 will be approximately
(i) one fourth the width
(ii) one half the width
(iii) the same width
as a sample of 1000. Choose one option and say why.
Answer. It will be twice as accurate (option ii). The new SE will be
4 N (SD of box)
4 N (SD of box) 2 N (SD of box)


4N
4N
4N
1 N (SD of box) 1

 (original SE)
2
N
2
5. One hospital has 218 live births during the month of January. Another has 536. Which is likelier to
have 55% or more male births? Or is it equally likely? (There is about a 51% chance for a live-born infant
to be male.)
Answer. The hospital with 218 live births is more likely to 55% or more male births because by the law of
average the proportion of live births will be close the theoretical 51% for a large sample.
6. a. There are about 25000 high schools in the United States; each high school has a principal. As part
of a national survey of education, a simple random sample of 225 high schools is chosen. In 202 of the
sample high schools, the principal has an advanced degree.
(a) If possible, find an approximate 95% confidence interval for the percentage of all 25000 high school
principals who have advanced degrees, showing your work. If this is impossible, explain why.
23
225  202
202
225  225
which is 85.7% to 93.8%.
2
225
225
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The 25000 high school in the United States employ a total of about one million teachers. As it turned
out, the 225 sample high schools employed a total of 10,000 teachers, of whom 5,010 had advanced
degrees.
(b) If possible, find an approximate 95% confidence interval for the percentage of all one million high
school teachers with advanced degrees, showing your work. If this is impossible, explain why.
Answer. It can’t be done with our methods. This is a cluster sample not simple random sample.
7. A survey organization takes a simple random sample of 625 households from a city of 80,000
households. On the average, there are 2.3 persons per sample household, and the SD is 1.75. Say
whether each of the following statements is true or false and explain.
(a) The 2.30 is 0.07 or so off the average number of persons per household in the whole city.
True. The SE is estimated as √625∙1.75/625 ≈ 0.7. A sample average is the expected average give or
take random error. The expected error is the average number of persons per household in the whole
city. The random error is estimated by the SE. So the sample average and the average number of
persons per household for the whole city are within an SE of each other usually.
(b) A 95%-confidence interval for the average household size in the sample is 2.16 to 2.44.
False. There is no such thing as a “confidence interval for the sample average.” We know the sample
average exactly. Confidence intervals are for the average of the entire population.
(c) A 95% confidence interval for the average house size in the city is 2.16 to 2.44.
True. 2.3 ± 2∙0.7 is 2.16 to 2.44.
(d) 95% of the households in the city contain between 2.16 and 2.44 persons.
False. A confidence interval does not give us information about the large population. It estimates the
average of the entire population.
(e) Household size in the city follows the normal curve.
False. Based on the sample, the mean is about 2.3 persons per household with a SD of 1.75. Since the
SD is so large, the population is not normal. If it were, there would have to be some households with
negative persons per household, which is impossible.
(f) The 95% confidence interval is about right because household size follows the normal curve.
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False. For one thing , household size doesn’t follow the normal curve. The 95% confidence interval is
about right because by the Central Limit Theorem, sample averages of size 225 drawn from this
population will follow the normal curve.
8. When Gregor Mendel conducted his famous hybridization experiments with peas, one such
experiment resulted in offspring consisting of 428 peas with green pods and 152 peas with yellow pods.
According to Mendel’s theory, ¼ of the offspring peas should have yellow pods. More than ¼ of
Mendel’s experimental plants had yellow pods. Do the results contradict the theory? Or can they be
explained by random chance? Conduct a hypothesis test to find out. Make sure you carefully state your
null hypothesis, your alternative hypothesis, your p-value, whether you reject your null hypothesis or
not, and your conclusion.
Answer.
The null hypothesis is that percentage of peas with yellow pods is 25%. The alternative hypothesis is
that more than 25% of the peas have yellow pods. The SE is
580
152
580
580
428
 580
 1.82% . The p-value is normalcdf(152/580, ∞, 0.25, 0.0182) =0.254. We cannot
reject the null hypothesis because the p-value is so high. We conclude that these results do not
contradict Mendel’s theory; they in fact support Mendel’s theory.
9. A researcher takes a simple random sample of registered voters in Illinois and among other things
asks about their age. In a sample of 1500 voters, the average age was 47.2 and the SD of the sample
was 13.9 years. Is there evidence that the average age of all voters in Illinois is greater than 45? Use a
hypothesis test. Make sure you carefully state your null hypothesis, your alternative hypothesis, your pvalue, whether you reject your null hypothesis or not, and your conclusion.
Answer.
The null hypothesis is that average age of all voters in Illinois is 45. The alternative hypothesis is that the
average age of all voters in Illinois is greater than 45. The SE is
1500 13.9
 0.359 . The p-value is normalcdf(47.2, ∞, 45, 0.359)  0. We reject the null hypothesis
1500
because the p-value is extremely low. We conclude that there is strong evidence that the average age
of voters in Illinois is greater than 45.
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10. Consider the following box of tickets.
0
0
1
1
3
1
a. If you draw 3 tickets without replacement what is the chance of getting a 1 on the first draw?
Answer. 3/6 = 0.5.
b. If you draw 3 tickets without replacement what is the chance of getting a 1 on the second draw?
Answer. Also 3/6 = 0.5.
c. If you draw 3 tickets with replacement what is the chance of getting a 1 on the second draw?
Answer. Also 3/6 = 0.5.
d. If you draw 2 tickets without replacement and you get draw a 0 on the first draw, what is the chance
of drawing a 1 on the second draw?
Answer. 3/5 = 0.6.
e. When drawing 2 tickets without replacement, what is the chance of drawing a 0 followed by a 1?
Answer. 2/63/5 = 0.2
f. When drawing 2 tickets without replacement, what is the chance of drawing a 1 followed by a 0?
Answer. 3/62/5 = 0.2
g. When drawing 2 tickets with replacement, what is the chance of drawing a 0 followed by a 1?
Answer. 2/63/6 = 1/6.
h. When drawing 2 tickets with replacement, what is the chance of drawing a 1 followed by a 0?
Answer. 3/62/6 = 1/6.
i. If you draw 1000 tickets from this box with replacement and add up the tickets, what is the chance
that the sum will be between 975 and 1025?
The average of the box is 1. The SD for the box is 1. The answer is normalcdf(975,1025, 1000,√1000∙1)
≈ 0.571.
j. If you draw 50 tickets from this box with replacement and add up the tickets, approximately what is
the probability of getting a sum of exactly 50?
Answer. normalcdf(49.5, 50.5, 50,√50∙1) ≈ 0.056.
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