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Q.12

Expert-verified

The Practice of Statistics for AP

Found in: Page 546

Book edition 4th

Author(s) David Moore,Daren Starnes,Dan Yates

Pages 809 pages

ISBN 9781319113339

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Short Answer

Refer to Exercise 2. For Yvonne's survey, 96 students in the sample said they rarely or never argue with friends. A significance test yields a P-value of 0.0291.
(a) Interpret this result in context.
(b) Do the data provide convincing evidence against the null hypothesis? Explain.

a. Chance that students in the sample said rarely or never argue with friends is $0.0291$

b. Yes, the data provide convincing evidence against the null hypothesis

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: introduction

Given,

sample size is $150$

population value is $72 %$

new population value is 0.0291

Step 2: explanation part (a)

Calculating the null and alternative hypotheses,

we get,

$\begin{aligned} H_{o} : p = 0.72 \\ H_{a} : p \neq 0.72 \end{aligned}$

The chance that students in the sample said rarely or never argue with friends is $0.0291$ if the population proportion is $0.72$ .

Step 3: explanation part (b)

The null hypothesis is not accepted as the value of p is less than the significant level of $5 %$ or $0.05$

But $p = 0.0291$ which is less than $0.05$ .

Yes, the data provide convincing evidence against the null hypothesis

Most popular questions for Math Textbooks

Study more! The significance test in Exercise $76$ yields a P-value of $0.0622$ .

(a) Describe a Type I and a Type II error in this setting. Which type of error could you have made in Exercise $76$ ? Why?

(b) Which of the following changes would give the test a higher power to detect $μ = 120$ minutes: using $α = 0.01$ or $α = 0.10$ ? Explain.

Which of the following $95 %$ confidence intervals would lead us to reject H0 : $p = 0.30$ in favor of

Ha :p not equal to $0.30$ at the $5 %$ significance level?

(a) (0.29, 0.38) (c) (0.27, 0.31) (e) N o n e o f t h e s e (b) (0.19, 0.27) (d) (0.24, 0.30)

Refer to Exercise 80. Construct and interpret a $95 %$ confidence interval for the population mean $M$ . What additional information does the confidence interval provide?

Paying high prices? A retailer entered into an exclusive agreement with a supplier who guaranteed to provide all products at competitive prices. The retailer eventually began to purchase supplies from other vendors who offered better prices. The original supplier filed a lawsuit claiming violation of the agreement. In defense, the retailer had an audit performed on a random sample of $25$ invoices. For each audited invoice, all purchases made from other suppliers were examined and compared with those offered by the original supplier. The percent of purchases on each invoice for which an alternative supplier offered a lower price than the original supplier was recorded. $26$ For example, a data value of $38$ means that the price would be lower with a different supplier for $38 %$ of the items on the invoice. A histogram and some computer output

for these data are shown below. Explain why we should not carry out a one-sample t-test in this setting.

To determine the reliability of experts who interpret lie detector tests in criminal investigations, a random sample of $280$ such cases was studied. The results were

(a) $15 / 280$ .

(b) $9 / 280$ .

(c) $15 / 140$ .

(d) $9 / 140$ .

(e) $15 / 146$ .

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