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

2. Discuss the basic strategy for comparing the means of two populations based on a simple random paired sample.

Short Answer

Expert verified

The paired difference variable is normally distributed

Step by step solution

01

Step 1:Given information

The means of two populations based on a simple random paired sample.

02

Step 2:Concepts required

The basic strategy for comparing the means of two populations based on a simple random paired sample is given in the following steps:

1. Independently and randomly take a paired sample.

2. The variable analyzed is the difference between the paired values of the variable under study.

3. By the use of a paired sample, we can remove an extraneous source of variation.

4. The sampling error thus made in estimating the difference between the populations means will generally smaller and, therefore, more likely to detect difference between the population means when such differences exist.

03

Step 2:Explaination

Suppose that xis a variable on each of two populations whose members can be paired. For each pair, we calculate the difference between the values of the variable xon the members of the pair denoted by dand is called as paired-difference variable.

μd=μ1-μ2

The paired difference variable is normally distributed.

Test Statistic:

The test statistic for comparing two population means in a paired sample of size nis defined as follows:

t=d¯-μdsd/n~t(n-1)

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

In each of Exercises 10.75-10.80, we have provided summary statistics for independent simple random samples from two populations. In each case, use the non pooled t-test and the non pooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval,

x¯1=20,s1=4,n1=10,x¯2=23,s2=5,n2=15.

a. Left-tailed test, α=0.05.

b. 90%confidence interval.

Suppose that you want to perform a hypothesis test to compare the means of two populations, using independent simple random samples. Assume that the two distributions (one for each population) of the variable under consideration are normally distributed and have equal standard deviations. Answer the following questions and explain your answers.

a. Is it permissible to use the pooled t-test to perform the hypothesis test?

b. Is it permissible to use the Mann-Whitney test to perform the hypothesis test?

c. Which procedure is preferable, the pooled t-test or the Mann-Whitney test?

The intent is to employ the sample data to perform a hypothesis test to compare the means of the two populations from which the data were obtained. In each case, decide which of the procedures should be applied.

Paired: n=18.

Cooling Down. Cooling down with a cold drink before exercise in the heat is believed to help an athlete perform. Researcher 1. Dugas explored the difference between cooling down with an ice slurry (slushy) and with cold water in the article "lce Slurry Ingestion Increases Running Time in the Heat" (Clinical Journal of Sports Medicine, Vol. 21, No, 6, pp. 541-542). Ten male participants drank a flavored ice slurry and ran on a treadmill in a controlled hot and humid environment. Days later, the same participants drank cold water and ran on a treadmill in the same bot and humid environment. The following table shows the times, in minutes, it took to fatigue on the treadmill for both the ice slurry and the cold water.

At the 1%significance level, do the data provide sufficient evidence to conclude that, on average, cold water is less effective than ice slurry For optimizing athletic performance in the heat? (Note; The mean and standard deviation of the paired differences are -5.9minutes and 1.60minutes, respectively.)

H2:μ1>μ2

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