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

Interpreting Power For the sample data in Example 1 “Adult Sleep” from this section, Minitab and StatCrunch show that the hypothesis test has power of 0.4943 of supporting the claim that\(\mu < 7\)hours of sleep when the actual population mean is 6.0 hours of sleep. Interpret this value of the power, then identify the value of\(\beta \)and interpret that value. (For the t test in this section, a “noncentrality parameter” makes calculations of power much more complicated than the process described in Section 8-1, so software is recommended for power calculations.)

Short Answer

Expert verified

If the true population mean sleep time is equal to 6.0 hours, there is a 49.43% chance of making the correct conclusion of rejecting the false null hypothesis/supporting the stated claim.

The value of \(\beta \) is equal to 0.5057.

If the true population mean sleep time is equal to 6.0 hours, there is a 50.57% chance of making the incorrect conclusion of failing to reject the false null hypothesis/rejecting the stated claim.

Step by step solution

01

Given information

A sample of 12 adults is randomly selected. The actual population’s mean sleep time is equal to 6.0 hours. It is claimed that the mean sleep time is less than 7 hours.

02

Hypotheses

The appropriate hypotheses for testing the given claim are written as follows:

Null hypothesis: The mean sleep time is equal to 7.0 hours.

\({H_0}:\mu = 7\)

Alternative Hypothesis: The mean sleep time is less than 7.0 hours.

\({H_0}:\mu < 7\)

The test is left-tailed.

03

Power of the test

The power of the test refers to the probability of rejecting the null hypothesis when it is actually false. In simple words, the value of the power of the test explains the probability of making the correct conclusion of the claim.

Here, the value of the power of the test is equal to 0.4943.

The actual population’s mean sleep time is equal to 6.0 hours. It is claimed that the mean sleep time is less than 7 hours.

Thus, it means that there is a 49.43% probability of making the correct conclusion that the mean sleep time is less than 7 hours (rejecting the false null hypothesis) when the population’s true mean sleep time is equal to 6.0 hours.

04

Value of \(\beta \)

The value of\(\beta \)is computed using the following formula:

\(\begin{array}{c}{\rm{Power}} = 1 - \beta \\\beta = 1 - {\rm{Power}}\end{array}\)

Therefore, the value of\(\beta \)is equal to:

\(\begin{array}{c}\beta = 1 - {\rm{Power}}\\ = 1 - 0.4943\\ = 0.5057\end{array}\)

Thus, the value of \(\beta \) is equal to 0.5057.

05

Interpretation of the value of \(\beta \)

The value of\(\beta \)is the value of the Type II error which is the probability of failing to reject the null hypothesis when it is actually false.

In simple words,\(\beta \)refers to the probability of making an incorrect conclusion about the stated claim.

For the given claim, the value of\(\beta \)represents the probability of rejecting the claim that the mean sleep time is less than 7 hours when the true population’s mean sleep time is actually equal to 6.0 hours.

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

Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Mickey D’s In a study of the accuracy of fast food drive-through orders, McDonald’s had 33 orders that were not accurate among 362 orders observed (based on data from QSR magazine). Use a 0.05 significance level to test the claim that the rate of inaccurate orders is equal to 10%. Does the accuracy rate appear to be acceptable?

Finding P-values. In Exercises 5–8, either use technology to find the P-value or use Table A-3 to find a range of values for the P-value7. Old Faithful. The claim is that for the duration times (sec) of eruptions of the Old Faithful geyser, the mean is μ=240sec. The sample size is n = 6 and the test statistic is t = 1.340.

Interpreting Power Chantix (varenicline) tablets are used as an aid to help people stop smoking. In a clinical trial, 129 subjects were treated with Chantix twice a day for 12 weeks, and 16 subjects experienced abdominal pain (based on data from Pfizer, Inc.). If someone claims that more than 8% of Chantix users experience abdominal pain, that claim is supported with a hypothesis test conducted with a 0.05 significance level. Using 0.18 as an alternative value of p, the power of the test is 0.96. Interpret this value of the power of the test.

Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Testing Effectiveness of Nicotine Patches In one study of smokers who tried to quit smoking with nicotine patch therapy, 39 were smoking one year after the treatment and 32 were not smoking one year after the treatment (based on data from “High-Dose Nicotine Patch Therapy,” by Dale et al., Journal of the American Medical Association, Vol. 274, No. 17). Use a 0.05 significance level to test the claim that among smokers who try to quit with nicotine patch therapy, the majority are smoking one year after the treatment. Do these results suggest that the nicotine patch therapy is not effective?

Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Car Booster Seats The National Highway Traffic Safety Administration conducted crash tests of child booster seats for cars. Listed below are results from those tests, with the measurements given in hic (standard head injury condition units). The safety requirement is that the hic measurement should be less than 1000 hic. Use a 0.01 significance level to test the claim that the sample is from a population with a mean less than 1000 hic. Do the results suggest that all of the child booster seats meet the specified requirement?

774 649 1210 546 431 612

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