/*! 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. 9.46 In each of Exercises 9.41-9.46 ,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In each of Exercises 9.41-9.46 ,determine the critical values for a one-mean z-test. For each exercise, draw a graph that illustrates your answer

A two-tailed test withα=0.05

Short Answer

Expert verified

Here we set the α=0.05value level of the given test that describes z with a two-tailed view. For a single z-definition test, under the null hypothesis, z-test statistics have a fairly common distribution.

z-check. In each test Two double-tailed tail tests, we have a drop point on both sides of the regular curve.

Therefore the value of two z points divides the area below the normal curve into the central area of 0.95 and the two areas outside of 0.025

That means key values ±z0.025

Used for areas under the standard curve table, ±z0.025=±1.96.

Step by step solution

01

Step 1. Given 

A two -tailed test withα=0.05

02

Step 2. Graph 

Here we set the α=0.05value level of the given test that describes z with a two-tailed view. For a single z-definition test, under the null hypothesis, z-test statistics have a fairly common distribution.

z-check. In each test Two double-tailed tail tests, we have a drop point on both sides of the regular curve.

Therefore the value of two z points divides the area below the normal curve into the central area of 0.95 and the two areas outside of 0.025

That means key values ±z0.025

Used for areas under the standard curve table, ±z0.025=±1.96

The rejection position is therefore found as in the following graph

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

9.90 Hotels and Motels. The daily charges, in dollars, for a sample of 15 hotels and motels operating in South Carolina are provided on the WeissStats site. The data were found in the report South Caroline Statistical Abstract, sponsored by the South Carolina Budget and Control Board.
a. Use the one-mean z-test to decide, at the 5%significance level, whether the data provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than \(75. Assume a population standard deviation of \)22.40.

b. Obtain a normal probability plot, boxplot, histogram, and stem-and-leaf diagram of the data.
c. Remove the outliers (if any) from the data and then repeat part (a).
d. Comment on the advisability of using the z-test here.

Dementia is the loss of the in actual and social abilities severe enough to interfere with judging behavior and daily functioning. Alzheimer's disease is the most common type of dementia. In the article "Living with Early Onsite dementia: Exploring the Experience and Developing Evidence Guidelines for Practice", P Harris and J Keady explored the experiment struggles of people diagnosed with dementia and their familiar simple random sample \(21\) people with early-onset dementia the following data on age at diagnosis in years.

At the \(1%\) significance level, do the data provide sufficient evidence to conclude that the mean age at diagnosis of all people with early onset dementia is less than \(55\) years old? Assume that the population is standard deviation is \(6.8\) years.

We have been provided a sample mean, sample size, and population standard deviation. In the given case, use the one-mean z-test to perform the required hypothesis test at the 5% significance level.

x¯=24,n=15,σ=4,H0:μ=22,Ha:μ>22

In Exercise 8.146 on page 345, we introduced one-sided one-mean t-intervals. The following relationship holds between hypothesis tests and confidence intervals for one-mean t-procedures: For a left-tailed hypothesis test at the significance level α, the null hypothesis H0:μ=μ0will be rejected in favor of the alternative hypothesis Ha:μ<μ0if and only if μ0is greater than or equal to the 1-α- level upper confidence bound for μ. In each case, illustrate the preceding relationship by obtaining the appropriate upper confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.117

Part (b): Exercise 9.118

9.89 Job Gains and Losses. In the article "Business Employment Dynamics: New Data on Gross Job Gains and Losses" (Monthly Labor Review, Vol. 127. Issue 4. pp. 29-42). J. Spletzer et al. examined gross job gains and losses as a percentage of the average of previous and current employment figures. A simple random sample of 20 quarters provided the net percentage gains (losses are negative gains) for jobs as presented on the WeissStats site. Use the technology of your choice to do the following.
a. Decide whether, on average, the net percentage gain for jobs exceeds 0.2. Assume a population standard deviation of 0.42. Apply the one-mean z-test with a 5% significance level.
b. Obtain a normal probability plot, boxplot, histogram, and stem-and-leaf diagram of the data.
c. Remove the outliers (if any) from the data and then repeat part (a).
d. Comment on the advisability of using the z-test here.

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.