/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q.85 The P-value for a two-sided test... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The P-value for a two-sided test of the null hypothesis H0:μ=10is 0.06

(a) Does the95%confidence interval forMinclude 10? Why or why not?

(b) Does the 90%confidence interval forMinclude 10? Why or why not

Short Answer

Expert verified

(a) 95%confidence interval for μwill includes 10

(b)90%confidence interval for μdoes not includes 10

Step by step solution

01

Part (a) Step 1: Given information

Given in the question that, The P-value for a two-sided test of the null hypothesis H0:μ=10is 0.06.

we need to find that Whether the95%confidence interval for μinclude 10.

02

Part (a) Step 2: Explanation

H0:μ=10

P-value =0.06

Reject, H0, if level of significance(α)<P-value .

The null and alternative hypotheses could be written as:

H0:μ=10

Ha:μ≠10

The level of significance is0.05

Here,

p-value (0.06)>α(0.05)

Thus, the null hypothesis is not rejected. So, it could be concluded that 95%confidence interval for μwill includes 10

03

Part (b) Step 1: Given information

Given in the question that, The P-value for a two-sided test of the null hypothesis H0:μ=10is 0.06.

We need to find that Whether the90%confidence interval for μinclude10.

04

Part (b) Step 2: Explanation

Here,

H0:μ=10

P-value=0.06

Reject, H0, if level of significance (α)<P-value

The null and alternative hypotheses could be written as:

H0:μ=10

Ha:μ≠10

The level of significance is 0.10.

Here, p-value (0.06)<α(0.10)

Thus, the null hypothesis is rejected. So, it could not be concluded that 90%confidence interval for μ will include10.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Refer to Exercise 1. In Simon’s SRS, 16 of the students were left-handed. A significance test yields a P-value of 0.2184.

(a) Interpret this result in context.

(b) Do the data provide convincing evidence against the null hypothesis? Explain.

We hear that listening to Mozart improves students’ performance on tests. Maybe pleasant odors have a similar effect. To test this idea, 21subjects worked two different but roughly equivalent paper-and-pencil mazes while wearing a mask. The mask was either unscented or carried a floral scent. Each subject used both masks, in a random order. The table below gives the subjects’ times with both masks.

Bullies in middle school A University of Illinois study on aggressive behavior surveyed a random sample of 558 middle school students. When asked to describe their behavior in the last 30 days, 445 students said their behavior included physical aggression, social ridicule, teasing, name-calling, and issuing threats. This behavior was not defined as bullying in the questionnaire. Is this evidence that more than three-quarters of the students at that middle school engage in bullying behavior? To find out, Maurice decides to perform a significance test. Unfortunately, he made a few errors along the way. Your job is to spot the mistakes and correct them.

H0:p=0.75Ha:p->0.797

where p= the true mean proportion of middle school students who engaged in bullying.

- A random sample of 558 middle school students was surveyed.

- 558(0.797)=444.73 is at least 10.

z=0.75−0.7970.797(0.203)445=−2.46;P-value=2(0.0069)=0.0138

The probability that the null hypothesis is true is only 0.0138, so we reject H0. This proves that more than three-quarters of the school engaged in bullying behavior.

A Gallup Poll report on a national survey of 1028 teenagers revealed that 72% of teens said they seldom or never argue with their friends. Yvonne wonders whether this national result would be true in her large high school. So she surveys a

a random sample of 150 students at her school.

Strong chairs? A company that manufactures classroom chairs for high school students claims that the mean breaking strength of the chairs that they make is 300pounds. One of the chairs collapsed beneath a 220-pound student last week. You wonder whether the manufacturer is exaggerating the breaking strength of the chairs.

(a) State null and alternative hypotheses in words and symbols.

(b) Describe a Type I error and a Type II error in this situation, and give the consequences of each.

(c) Would you recommend a significance level of0.01,0.05, or 0.10for this test? Justify your choice.

(d) The power of this test to detect μ=294is 0.71. Explain what this means to someone who knows little statistics.

(e) Explain two ways that you could increase the power of the test from (d).

- If conditions are met, conduct a significance test about a population proportion.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.