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Q. 11

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The Practice of Statistics for AP
Found in: Page 735
The Practice of Statistics for AP

The Practice of Statistics for AP

Book edition 4th
Author(s) David Moore,Daren Starnes,Dan Yates
Pages 809 pages
ISBN 9781319113339

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Short Answer

A large distributor of gasoline claims that 60%all cars stopping at their service stations choose regular unleaded gas and that premium and supreme are each selected 20%of the time. To investigate this claim, researchers collected data from a random sample of drivers who put gas in their vehicles at the distributor's service stations in a large city. The results were as follows:

Carry out a significance test of the distributor's claim. Use a 5%significance level.

There is sufficient evidence to reject the distributor's claim.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given Information

Need to find whether there is sufficient evidence to reject the distributor's claim.

Step 2: Explanation

Determine the observed frequencies and the chi-square subtotals:

The value of the test statistic is thus:

χ2=1.8675+10.5125+0.8 =13.15

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table C containing the t-value in the row

d f =c-1 =3-1 =2

0.001<P<0.0025

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

localid="1650541589569" P<0.05=5% ⇒ Reject H0

There is sufficient evidence to reject the distributor's claim.

Most popular questions for Math Textbooks

Which hypotheses would be appropriate for performing a chi-square test?

(a) The null hypothesis is that the closer students get to graduation, the less likely they are to be opposed to tuition increases. The alternative is that how close students are to graduation makes no difference in their opinion.

(b) The null hypothesis is that the mean number of students who are strongly opposed is the same for each of the four years. The alternative is that the mean is different for at least two of the four years.

(c) The null hypothesis is that the distribution of student opinion about the proposed tuition increase is the same for each of the four years at this university. The alternative is that the distribution is different for at least two of the four years.

(d) The null hypothesis is that year in school and student opinion about the tuition increase in the sample are independent. The alternative is that these variables are dependent.

(e) The null hypothesis is that there is an association between a year in school and opinion about the tuition increase at this university. The alternative hypothesis is that these variables are not associated.

The expected count of cases of lymphoma in homes with an HCC is

(a) 79·31215.

(b) 10·21215.

(c) 79·3110.

(d) 136·31215.

(e) None of these.

Refer to Exercises 1 and 3.

(a) Confirm that the expected counts are large enough to use a chi-square distribution. Which distribution (specify the degrees of freedom) should you use?

(b) Sketch a graph like Figure 11.4 (page 683) that shows the P-value.

(c) Use Table C to find the P-value. Then use your calculator’s C2cdf command

(d) What conclusion would you draw about the company’s claimed distribution for its deluxe mixed nuts? Justify your answer.

The General Social Survey asked a random sample of adults, “Do you favour or oppose the death penalty for persons convicted of murder?” The following table gives the responses of people whose highest education was a high school degree and of people with a bachelor’s degree:

We can test the hypothesis of “no difference” in support for the death penalty among people in these educational categories in two ways: with a two-sample z test or with a chi-square test.

(a) Minitab output for a chi-square test is shown below. State appropriate hypotheses and interpret the P-value in context. What conclusion would you draw? Chi-Square Test: C1, C2 Expected counts are printed below-observed counts Chi-Square contributions are printed below expected counts

(b) Minitab output for a two-sample z test is shown below. Explain how these results are consistent with the test in part (a).

Calculate the expected counts. Show your work

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