Download Ch. 9 Review: Testing a Claim Given α = 0.05, which of the following

Ch. 9 Review: Testing a Claim
1. Given α = 0.05, which of the following is true?
(A) P(Type II error) = 0.95
(B) The power of the test is 0.95.
(C) P(rejecting H0 when H0 is true) = 0.05
(D) P(failing to reject H0 when H0 is false) = 0.05
(E) The probability of a Type II error is independent of the value of α.
2. The mayor of a large city will run for governor if he believes that more than 30% of the
voters in the state already support him. He will have a survey firm ask a random sample
of n voters whether or not they support him. He will use a large sample test for
proportions to test the null hypothesis that the proportion of all voters who support him is
30% or less against the alternative that the percentage is higher than 30%. Suppose that
35% of all voters in the state actually support him. In which of the following situations
would the power for the test be the highest?
(A) The mayor uses a significance level of 0.01 and n=250 voters.
(B) The mayor uses a significance level of 0.01 and n= 500 voters.
(C) The mayor uses a significance level of 0.01 and n= 1,000 voters.
(D) The mayor uses a significance level of 0.05 and n= 500 voters.
(E) The mayor uses a significance level of 0.05 and n= 1,000 voters.
3. When performing a test of significance for a null hypothesis, Ho, against an alternative Ha,
the p-value is
(A) The probability that Ho is true.
(B) The probability that Ha is true.
(C) The probability that Ho is false.
(D) The probability of observing a value of a test statistic at least as extreme as that
observed in the sample if Ho is true.
(E) The probability of observing a value of a test statistic at least as extreme as that
observed in the sample if Ha is true.
4. An opinion poll asks a random sample of adults whether they favor banning ownership of
handguns by private citizens. A commentator believes that more than half of all adults favor
such a ban. The null and alternative hypotheses you would use to test this claim are:
(A) H 0 : pˆ = 0.5; H a : pˆ > 0.5
(B) H 0 : pˆ = 0.5; H a : pˆ ¹ 0.5
(C) H0 : p = 0.5;Ha : p > 0.5
(D) H 0 : p = 0.5; H a : p ¹ 0.5
(E) H 0 : p = 0; H a : p > 0
5. A significance test gives a P-value of 0.29. From this we can
(A) Accept H0 at the 5% significance level
(B) Fail to reject H0 at the 5% significance level
(C) Reject H0 at the 5% significance level
(D) Say that the probability that H0 is false is 0.29
(E) Say that the probability that H0 is true is 0.29
6. In a significance test of the hypotheses H0: μ = 15 vs. Ha: μ > 15, what is the P-value if the
test statistic z is 1.2? (MAKE A SKETCH OF THE SAMPLING DISTRIBUTION AND SHADE
THE AREA UNDER THE CURVE THAT IS THE P-VALUE.)
7. What is the P-value if the test statistic z is the same as in part #6 but the alternative
hypothesis is Ha: μ ≠ 15? (MAKE A SKETCH OF THE SAMPLING DISTRIBUTION AND
SHADE THE AREA UNDER THE CURVE THAT IS THE P-VALUE.)
8. Suppose the people of Sleepy Eye, MN have always believed that the mean height of their
eighth grade girls was 60 inches. A sample of the heights of 23 randomly selected eighth
grade girls gave a mean of 62.35. A one-sample t-test with a one-sided alternative was
done and the resulting P-value was 0.0498. Interpret the P-value. (In other words,
explain what 0.0498 is the probability of in the context of the problem. You are not being
asked to state whether you would reject or fail to reject H0.)
9.
A random sample of 15 cigarettes of a certain brand was tested for nicotine content. The
average content of these 15 cigarettes was found to be 20.3 mg. The standard deviation
for the nicotine of all cigarettes of this brand is 3.0 mg. A 95% confidence interval for the
mean nicotine content for all cigarettes of this brand is calculated and found to be 18.782
to 21.818 mg.
The manufacturer claims that the mean nicotine content in its cigarettes is 18.3 mg. Based
on the confidence interval above, is there evidence at the 5% level that the mean nicotine
content in the manufacturer’s cigarettes is different than 18.3 mg?
10. Statistics can help decide the authorship of literary works. Researchers have reviewed the
sonnets of Sir Edward Dyer and calculated the number of new words (words not used in
any of his other works) in each one. The number of new words in the sonnets of Sir
Edward Dyer is Normally distributed with a mean of 𝜇 = 6.9 and a standard deviation of
𝜎 = 2.7 words. Now a new manuscript has come to light with many new sonnets, and
scholars are debating whether it is the poet’s work. They take a simple random sample of
five sonnets from the new manuscript and count the number of new words in each one.
We expect poems by another author to contain more new words than found in the
Elizabethan poet’s poems.
When you examine the five new works, you find that the mean number of new words is 𝑥̅
= 9.2. Below is a dot plot showing the results of simulating 200 samples of size 5 from a
Normal distribution with a mean of 6.9 and a standard deviation of 2.7, and calculating the
mean for each sample. Use the results of this simulation study to estimate the P-value of
this test, and draw an appropriate conclusion using significance level of α = 0.05.
11. When a law firm represents a group of people in a class action lawsuit and wins that
lawsuit, the firm receives a percentage of the group’s monetary settlement. That
settlement amount is based on the total number of people in the group – the larger the
group and the larger the settlement, the more money the firm will receive.
A law firm is trying to decide whether to represent car owners in a class action lawsuit against
the manufacturer of a certain make and model for a particular defect. If 5 percent or less of the
cars of this make and model have the defect, the firm will not recover its expenses. Therefore,
the firm will handle the lawsuit only if it is convinced that more than 5 percent of cars of this
make and model have the defect. The firm plans to take a random sample of 1,000 people who
bought this car and ask them if they experienced this defect in their cars.
(a) Define the parameter of interest and state the null and alternative hypotheses that the law
firm should test.
(b) In the context of this situation, describe Type I and Type II errors and describe the
consequences of each of these for the law firm.
(c) Give two ways to reduce the probability of a Type II error.
(d) Give two ways to increase the power of the test.
12. In recent years, the mean yield of corn in the United States has been about 120 bushels per
acre. A survey of 41 farmers this year gives a sample mean yield of 123.8 bushels per acre
and a standard deviation of 10 bushels per acre. The farmers surveyed are a SRS from the
population of all commercial corn growers. We want to know whether the sample result is
good evidence that the national mean this year is different from 120 bushels per acre. Use
a = 0.01. Follow the 4-step process.