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

According to sleep researchers, if you are between the ages of 12and 18years old, you need 9hours of sleep to function well. A simple random sample of 28students was chosen from a large high school, and these students were asked how much sleep they got the previous night. The mean of the responses was 7.9hours with a standard deviation of 2.1hours. If we are interested in whether students at this high school are getting too little sleep, which of the following represents the appropriate null and alternative hypotheses ?

  1. H0:μ=7.9and Ha:μ<7.9
  2. H0:μ=7.9and Ha:μ7.9
  3. H0:μ=9and Ha:μ9
  4. H0:μ=9and width="69" height="24" role="math">Ha:μ<9
  5. H0:μ9andHa:μ9

Short Answer

Expert verified

Option (d) is correct :H0:μ=9andHa:μ<9

Step by step solution

01

Given Information

We are given following information:

Mean of responses = 7.9

Standard deviation =2.1

02

Explanation

Remember that the null is identical to the hypothesized value while constructing the null and alternative.

The sample mean is 7.9, which differs from the hypothesized value. So the hypothesized value is 9.

Null Hypothesis H0:μ=9

Whether the test is a left or right tail, or both, is determined by the alternative hypothesis.

We're curious if kids are receiving less sleep than we will go for the left tail test. It would be a right tail test if we were interested in pupils sleeping too much. Because the researchers are deliberately looking for too little sleep. So researchers will go for left tail test and alternative hypothesis will be less than null hypothesis.

Alternative Hypothesis Ha:μ<9

Hence, option (d) is correct.

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

Children make choices Refer to Exercise 15.

a. Explain why the sample results give some evidence for the alternative hypothesis.

b. Calculate the standardized test statistic and P-value.

c. What conclusion would you make?

Broken crackers We don’t like to find broken crackers when we open the package. How can makers reduce breaking? One idea is to microwave the crackers for 30seconds right after baking them. Randomly assign 65newly baked crackers to the microwave and another 65to a control group that is not microwaved. After 1day, none of the microwave group were broken and 16of the control group were broken. Let p1be the true proportions of crackers like these that would break if baked in the microwave and p2be the true proportions of crackers like these that would break if not microwaved. Check if the conditions for calculating a confidence interval forp1-p2met.

A researcher wants to determine whether or not a 5-week crash diet is effective over a long period of time. A random sample of 15five-week crash dieters is selected. Each person’s weight (in pounds) is recorded before starting the diet and 1year after it is concluded. Do the data provide convincing evidence that 5-week crash dieters weigh less, on average, 1year after finishing the diet?

An SRS of size 100is taken from Population A with proportion 0.8of successes. An independent SRS of size 400is taken from Population B with proportion 0.5of successes. The sampling distribution of the difference (A − B) in sample proportions has what mean and standard deviation?

a. mean=0.3; standard deviation =1.3

b. mean=0.3; standard deviation =0.40

c. mean=0.3; standard deviation =0.047

d. mean=0.3; standard deviation =0.0022

e. mean=0.3; standard deviation =0.0002

In an experiment to learn whether substance M can help restore memory, the brains of 20rats were treated to damage their memories. First, the rats were trained to run a maze. After a day, 10rats (determined at random) were given substance M and 7of them succeeded in the maze. Only 2of the 10control rats were successful. The two-sample z test for the difference in the true proportions

a. gives z=2.25,P<0.02 .

b. gives z=2.60,P<0.005 .

c. gives z=2.25,P<0.04 but not<0.02

d. should not be used because the Random condition is violated.

e. should not be used because the Large Counts condition is violated.

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