Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 7: Q96E (page 439)

a. List three factors that will increase the power of a test.

b. What is the relationship between b, the probability of committing a Type II error, and the power of a test?

Short Answer

Expert verified

a. Three factors:

  • Increase the value of α
  • Sample size increase
  • Distance between increase μα(that is,actual value) as well as the null hypothesis value.

b. Power of the test = 1-β, where β is the probability of committing type II error.

Step by step solution

01

(a) Three factors of increase the power test

The power of a test is the probability that testing will result in the null hypothesis being rejected for a given average value in the alternative hypothesis, and it is equivalent

to 1-β.

Increasing the α, decreasing the β, as well as increasing the sample size n are three parameters that will boost the power of a test.

  1. Increase the value of α
  2. Sample size increase
  3. Distance between increase μα(that is,actual value) as well as the null hypothesis value.
02

Step 2:(b) Relationship between b and the probability of committing a type II

Power of the test =1-β, whereβis the probability of committing type II error.The probability of making a type II error is one minus the test's power, commonly known as beta. With a larger sample size, the power of the test can be enhanced, lowering the chance of making a type II mistake.

The link between β and test power is 1-β

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.

Start your free trial

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

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

Feminized faces in TV commercials. Television commercials most often employ females or “feminized” males to pitch a company’s product. Research published in Nature (August 27 1998) revealed that people are, in fact, more attracted to “feminized” faces, regardless of gender. In one experiment, 50 human subjects viewed both a Japanese female face and a Caucasian male face on a computer. Using special computer graphics, each subject could morph the faces (by making them more feminine or more masculine) until they attained the “most attractive” face. The level of feminization x (measured as a percentage) was measured.

a. For the Japanese female face, x = 10.2% and s = 31.3%. The researchers used this sample information to test the null hypothesis of a mean level of feminization equal to 0%. Verify that the test statistic is equal to 2.3.

b. Refer to part a. The researchers reported the p-value of the test as p = .021. Verify and interpret this result.

c. For the Caucasian male face, x = 15.0% and s = 25.1%. The researchers reported the test statistic (for the test of the null hypothesis stated in part a) as 4.23 with an associated p-value of approximately 0. Verify and interpret these results.

An analyst tested the null hypothesis that μ ≥ 20against the alternative hypothesis that μ <20. The analyst reported a p-value of .06. What is the smallest value ofαfor which the null hypothesis would be rejected?

In a test of \({H_0}:\mu = 100\) against \({H_a}:\mu \ne 100\), the sample data yielded the test statistic z = 2.17. Find the p-value for the test.

Manufacturers that practice sole sourcing. If a manufacturer (the vendee) buys all items of a particular type from a particular vendor, the manufacturer is practicing sole sourcing (Schonberger and Knod, Operations Management, 2001). As part of a sole-sourcing arrangement, a vendor agrees to periodically supply its vendee with sample data from its production process. The vendee uses the data to investigate whether the mean length of rods produced by the vendor's production process is truly 5.0 millimetres (mm) or more, as claimed by the vendor and desired by the vendee.

a. If the production process has a standard deviation of .01 mm, the vendor supplies n = 100 items to the vendee, and the vendee uses a = .05 in testing H0: m = 5.0 mm against Ha: m < 5.0 mm, what is the probability that the vendee's test will fail to reject the null hypothesis when in fact m = 4.9975 mm? What is the name given to this Type of error?

b. Refer to part a. What is the probability that the vendee's test will reject the null hypothesis when m = 5.0? What is the name given to this Type of error?

c. What is the power of the test to detect a departure of .0025 mm below the specified mean rod length of 5.0 mm?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation
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