Elementary Statistics
Ch 9 Practice Test
Daphne Skipper
Chapter 9
1. Express the null and alternative hypotheses in symbolic form that would be used to test
each claim below. Use the correct symbol (µ or p) for the parameter. Identify which
hypothesis represents the claim.
(a) Claim: More than 20% of Yoga Journal subscribers choose a meat-free diet.
Solution:
H0 : p ≤ 0.2
Ha : p > 0.2 ← claim
(b) Claim: The average weight of an adult German shepherd dog is 45 pounds.
Solution:
H0 : µ = 45 ← claim
Ha : µ 6= 45
2. Suppose we perform a hypothesis test to test the claim that more than 20% of Yoga
Journal subscribers choose a meat-free diet. Express the conclusion of the test using a
complete sentence in non-technical language if...
(a) ...our analysis leads us to reject the null hypothesis.
Solution: There is sufficient evidence to support the claim that more than 20%
of Yoga Journal subscribers choose a meat-free diet.
(b) ...our analysis leads us to fail to reject the null hypothesis.
Solution: There is not sufficient evidence to support the claim that 20% of
Yoga Journal subscribers choose a meat-free diet.
3. A group of doctors is deciding whether or not to perform an operation. Suppose the null
hypothesis, H0 , is: the surgical procedure will go well.
(a) State the Type I and Type II errors in complete sentences.
Solution: Type 1: (Reject True Null) A type 1 error in this case would be
deciding that the surgical procedure would not go well, when in fact it would.
Type 2: (Fail to Reject False Null) A type 2 error would be deciding that the
surgical procedure would go well when, in fact it wouldn’t.
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Daphne Skipper
(b) Which error has the more serious consequences in this case? Explain.
Solution: A type 2 error would be more serious because it would encourage
the doctors to procede (unknowingly) with an unsafe operation.
4. A random sample of 40 high school basketball players is chosen. Their heights have a
sample mean 71 inches, and a sample standard deviation of 1.5 inches. Do the data
support the claim that the mean height of high school basketball players is less than 73
inches? Chapter 9 Practice 35 (edited).*
(a) Is this a test about population mean or population proportion?
Solution: population mean, µ
(b) State the null and alternative hypotheses.
Solution:
H0 : µ ≥ 73
Ha : µ < 73 ← claim
(c) Is this a right-tailed, left-tailed, or two-tailed test?
Solution: left-tailed
(d) Is the population standard deviation known and, if so, what is it?
Solution: unknown
(e) Find the following: x̄, s, n.
Solution: x̄ = 71 inches, s = 1.5 inches, and n = 40
(f) State the distribution to use for the hypothesis test: t or N (
missing values.
,
)? Fill in the
Solution: t39
(g) Which calculator function should be used: TTest, ZTest, or 1-PropZTest?
Solution: TTest
(h) Find the p-value (4 decimal places) and test statistic labeled as t or z (3 decimal
places).
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Daphne Skipper
Solution: p-value = 0.000, t = −8.433
(i) Explain the meaning of the p-value.
Solution: It is almost impossible for 40 randomly selected players to have an
average height of 71 inches or less, if the average height all players is 73 inches.
(j) At a significance level of α = 0.05, what is your decision about the null hypothesis?
Solution: Since p − value < α, we decide to reject the null hypothesis. (The
average height of these players is far below 73 inches (measuring in standard
deviations), invalidating the null hypothesis assumption that the average height
of all players is 73 inches.)
(k) Write your conclusion about the claim in a complete sentence.
Solution: There is sufficient evidence to support the claim that the mean height
of high school basketball players is less than 73 inches.
5. In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average,
an 80 hour week. Recently, many groups have been studying whether or not the women’s
movement has, in fact, resulted in an increase in the average work week for women
(combining employment and at-home work). Suppose a study was done to determine if
the mean work week has increased. 81 women were surveyed with the following results.
The sample mean was 83; the sample standard deviation was ten. Does it appear that
the mean work week has increased for women at the 5% significance level? Assume that
the population standard deviation for women’s work weeks is known to be twelve hours.
Chapter 9 Homework 79 (edited).*
(a) State the null and alternative hypotheses.
Solution:
H0 : µ = 80
Ha : µ > 80 ← claim
(b) In words, clearly state what your random variable X̄ or P 0 represents.
Solution: X̄ represents the sample mean work week (in hours) of a random
sample of 81 women.
(c) State the distribution to use for the test: t or N (
Fill in the missing values.
,
)? (Hint: Is σ known?)
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Solution: X̄ ∼ N (80, √1281 )
(d) What is the test statistic? Label it as z or t.
Solution: z = 2.5
(e) What is the p-value (use four decimal places)? Explain in a sentence what the
p-value means for this problem.
Solution: p-value = 0.0122. There is about a 0.01 chance (1 in 100) of randomly
selecting a sample of 81 women who have a mean work week of 83 hours or
more, if the average work week for all women is 80 hours, as assumed in the null
hypothesis.
(f) Indicate the correct decision (“reject” or “do not reject” the null hypothesis) and
the reason for it, including a sentence explaining the reason.
Solution: We reject the null hypothesis because the p-value < α. The sample mean was enough above 80 hours (measuring in standard deviations) to
invalidate the assumption that the mean work week is 80 hours.
(g) Write an appropriate conclusion using complete sentences.
Solution: The sample data leads us to conclude that the work week has increased for 25 year-old mothers of three since 1955.
6. Toastmasters International cites a report by Gallop Poll that 40% of Americans fear
public speaking. A student believes that less than 40% of students at her school fear
public speaking. She randomly surveys 361 schoolmates and finds that 135 report they
fear public speaking. Conduct a hypothesis test to determine if the percent at her school
is less than 40%. Chapter 9 Homework 79 (edited).*
(a) State the null and alternative hypotheses.
Solution:
H0 : p = 0.4
Ha : p < 0.4 ← claim
(b) What is the test statistic? Label it as z or t and use two decimal places.
Solution: z = −1.01
(c) What is the p-value (use four decimal places)? Explain in a sentence what the
p-value means for this problem.
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Daphne Skipper
Solution: p-value = 0.1563. There is about a 16% chance of random selecting
361 students in which 37% or less fear public speaking, if 40% of the student
population fears public speaking.
(d) Indicate the correct decision (“reject” or “do not reject” the null hypothesis) and
the reason for it, including a sentence explaining the reasoning.
Solution: We do not reject the null hypothesis because the p-value > α. This
sample outcome is close enough to the assumed value of 0.4 (or 40%) (measured
in standard devations) that the 0.4 seems like a reasonable estimate for the
proportion of all students who fear public speaking.
(e) Write an appropriate conclusion using complete sentences.
Solution: There is insufficient evidence to conclude that fewer than 40% of the
students at this school fear public speaking.
7. According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample
of a college math class resulted in the family sizes listed below. At α = 0.05 level, is the
class’ mean family size greater than the national average? Chapter 9 Homework 115.*
5 4 5 4 4 3 6 4 3 3
5 5 6 3 3 2 7 4 5 2
2 2 3 2
(a) List the null and alternative hypotheses.
Solution:
H0 : µ = 3.18
Ha : µ > 3.18 ← claim
(b) List the test statistic and p-value. For the test statistic use 2 decimal places and
label it as z or t. Use 4 decimal places for the p-value.
Solution: t = 2.23, p-value = 0.0178
(c) Explain the meaning of the p-value in the context of this problem.
Solution: There is about a 2% chance of randomly selecting 24 student from
the class that have a mean family size of 3.83 or more, if the entire class has a
mean family size of 3.18.
(d) Indicate the correct decision (“reject” or “do not reject” the null hypothesis) and
the reason for it, including a sentence explaining the reasoning.
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Daphne Skipper
Solution: We reject the null hypothesis because the p-value < α. The sample
mean of 3.83 is far enough from the assumed value of 3.18 to invalidate the
assumption.
(e) Write an appropriate conclusion using complete sentences.
Solution: The sample data leads us to conclude that this class has a population
mean family size that is greater than the national average.
Many of these problems are from Barbara Illowsky & Susan Dean. “Introductory Statistics.” OpenStax College, 2013. iBooks.
https://itun.es/us/kFeL1.l