For each of the following rejection regions, sketch the sampling distribution for z and indicate the location of the rejection region.

a. \({H_0}:\mu \le {\mu _0}\) and \({H_a}:\mu > {\mu _0};\alpha = 0.1\)

b. \({H_0}:\mu \le {\mu _0}\) and \({H_a}:\mu > {\mu _0};\alpha = 0.05\)

c. \({H_0}:\mu \ge {\mu _0}\) and \({H_a}:\mu < {\mu _0};\alpha = 0.01\)

d. \({H_0}:\mu = {\mu _0}\) and \({H_a}:\mu \ne {\mu _0};\alpha = 0.05\)

e. \({H_0}:\mu = {\mu _0}\) and \({H_a}:\mu \ne {\mu _0};\alpha = 0.1\)

f. \({H_0}:\mu = {\mu _0}\) and \({H_a}:\mu \ne {\mu _0};\alpha = 0.01\)

g. For each rejection region specified in parts a–f, state the probability notation in z and its respective Type I error value.

Short Answer

Expert verified

a)

b)

c)

d)

e)

f)

g)

Step by step solution

01

Given information

For each part, the null and the alternative hypotheses are given with a specific significance level

02

(a) Sketching the sampling distribution of z for H0:μ⩽μ0 and Ha:μ>μ0;α=0.1 

Consider,

\({H_0}:\mu \le {\mu _0}\)

\({H_a}:\mu > {\mu _0}\)

The given alternative hypothesis is right-tailed.

Also,

\(\alpha = 0.1\)

Therefore, from the table of Standard Normal Distribution z score valueis 1.282.

The sampling distribution of Z and the rejection region's location is shown in the following diagram.

03

(b) Sketching the sampling distribution of z for H0:μ⩽μ0and Ha:μ>μ0;α=0.05 

Consider,

\({H_0}:\mu \le {\mu _0}\)

\({H_a}:\mu > {\mu _0}\)

The given alternative hypothesis is right-tailed.

Also,

\(\alpha = 0.05\)

Therefore, from the table of Standard Normal Distribution z score value is 1.96.

The sampling distribution of Z and the rejection region's location is shown in the following diagram.

04

(c) Sketching the sampling distribution of z for H0:μ⩾μ0 and Ha:μ<μ0;α=0.01 

Consider,

\({H_0}:\mu \ge {\mu _0}\)

\({H_a}:\mu < {\mu _0}\)

Given alternative hypothesis is left-tailed.

Also,

\(\alpha = 0.01\)

Therefore, from the table of Standard Normal Distribution z score value is -2.326.

The sampling distribution of Z and the location of the rejection region at \(Z < - 2.326\) is shown in the following diagram.

05

(d) Sketching the sampling distribution of z for H0:μ=μ0 and Ha:μ≠μ0;α=0.05    

Consider,

\({H_0}:\mu = {\mu _0}\)

\({H_a}:\mu \ne {\mu _0}\)

The given alternative hypothesis is two-tailed.

Also,

\(\alpha = 0.05\)

Therefore, from the table of Standard Normal Distribution z score value is 1.96.

The sampling distribution of Z and the rejection region's location is shown in the following diagram.

06

(e) Sketching the sampling distribution of z for H0:μ=μ0 and Ha:μ≠μ0;α=0.1

Consider,

\({H_0}:\mu = {\mu _0}\)

\({H_a}:\mu \ne {\mu _0}\)

The given alternative hypothesis is two-tailed.

Also,

\(\alpha = 0.1\)

Therefore, from the Standard Normal distribution table, z score values are -1.282 and 1.282.

The sampling distribution of Z and the rejection region's location is shown in the following diagram.

Consider,

\({H_0}:\mu = {\mu _0}\)

\({H_a}:\mu \ne {\mu _0}\)

The given alternative hypothesis is two-tailed.

Also,

\(\alpha = 0.1\)

Therefore, from the Standard Normal distribution table, z score values are -1.282 and 1.282.

07

(f) Sketching the sampling distribution of z for H0:μ=μ0 and Ha:μ≠μ0;α=0.01

Consider,

H0:μ=μ0Ha:μμ0

The given alternative hypothesis is two-tailed.

Also,

α=0.01

Therefore, from the Standard Normal distribution table, z score values are -2.326 and 2.326.

The sampling distribution of Z and the rejection region's location is shown in the following diagram.

08

(g) Stating the probabilities of Type I errors.

Since Type I error is the probability of rejecting the null hypothesis when it is true.

Now, the required table is as follows:

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 Vaia!

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

Libor interest rate. The interest rate at which London banks lend money to one another is called the London interbank offered rate, or Libor. The British Bankers Association regularly surveys international banks for the Libor rate. One recent report (Bankrate.com, March 16, 2016) had the average Libor rate at 1.2% for 1-year loans—a value considered high by many Western banks. Set up the null and alternative hypotheses for testing the reported value.

Question:Accounting and Machiavellianism. Refer to the Behavioral Research in Accounting (January 2008) study of Machiavellian traits in accountants, Exercise 6.19 (p. 341). A Mach rating score was determined for each in a random sample of 122 purchasing managers, with the following results: = 99.6 s = 12.6. Recall that a director of purchasing at a major firm claims that the true mean Mach rating score of all purchasing managers is 85.

a. Suppose you want to test the director’s claim. Specify the null and alternative hypotheses for the test.

b. Give the rejection region for the test using α = 0.10.

c. Find the value of the test statistic.

d. Use the result, part c, to make the appropriate conclusion.

“Streaming” of television programs is trending upward. According to The Harris Poll (August 26, 2013), over one-third of American’s qualify as “subscription streamers,” i.e., those who watch streamed TV programs through a subscription service such as Netflix, Hulu Plus, or Amazon Prime. The poll included 2,242 adult TV viewers, of which 785 are subscription streamers. On the basis of this result, can you conclude that the true fraction of adult TV viewers who are subscription streamers differs from one-third? Carry out the test using a Type I error rate of α=.10. Be sure to give the null and alternative hypotheses tested, test statistic value, rejection region or p-value, and conclusion.

Accidents at construction sites. In a study published in the Business & Economics Research Journal (April 2015), occupational accidents at three construction sites in Turkey were monitored. The total numbers of accidents at the three randomly selected sites were 51, 104, and 37.

Summary statistics for these three sites are:\(\bar x = 64\)and s = 35.3. Suppose an occupational safety inspector claims that the average number of occupational accidents at all Turkish construction sites is less than 70

a. Set up the null and alternative hypotheses for the test.

b. Find the rejection region for the test using\(\alpha = .01\)

c. Compute the test statistic.

d. Give the appropriate conclusion for the test.

e. What conditions are required for the test results to be valid?

For each of the following situations, determine the p-value and make the appropriate conclusion.

a.\({H_0}:\mu \le 25\),\({H_a}:\mu > 25\),\(\alpha = 0.01\),\(z = 2.02\)

b.\({H_0}:\mu \ge 6\),\({H_a}:\mu < 6\),\(\alpha = 0.05\),\(z = - 1.78\)

c.\({H_0}:\mu = 110\),\({H_a}:\mu \ne 110\),\(\alpha = 0.1\),\(z = - 1.93\)

d. \({H_0}:\mu = 10\), \({H_a}:\mu \ne 10\), \(\alpha = 0.05\), \(z = 1.96\)

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.

Sign-up for free