4. Suppose that the variable under consideration is normally distributed on each of the two populations and that you are going to use independent simple random samples to compare the population means. Fill in the blank and explain your answer: Unless you are quite sure that the are equal, the nonpooled t-procedures should be used instead of the pooled t-procedures.

Short Answer

Expert verified

The two distributions are normal

Step by step solution

01

Step 1:Given information

The variable under consideration is normally distributed on each of the two populations

02

Step 2:Explaination

Assumption for pooled t-test is given below:

- The selected sample should be a simple random sample from two populations.

- They are independent of one another.

- Population is approximately normal.

- Equal population standard deviations.

Assumption for nonpooled t-test is given below:

- The selected sample should be a simple random sample from two populations.

- They are independent of one another.

- Population is approximately normal.

- Sample size is large.

ExplanatIion:

Frpm the above situation we can say that, the two distributions are normal and use the independent simple random samples to compare the population means.

Unless you are quite sure that the population standard deviations are equal, the non pooled t-procedures should be used instead of the pooled t-procedures.

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Most popular questions from this chapter

Two-Tailed Hypothesis Tests and CIs. As we mentioned on page 413, the following relationship holds between hypothesis tests and confidence intervals: For a two-tailed hypothesis test at the significance level α, the null hypothesis H0:μ1=μ2will be rejected in favor of the alternative hypothesis Ha:μ1μ2if and only if the ( 1-α)-level confidence interval for μ1-μ2does not contain 0. In each case, illustrate the preceding relationship by comparing the reults of the hypothesis test and confidence interval in the specified xercises.

a. Exercises 10.48 and 10.54.

b. Exercises 10.49 and 10.55.

Identify the assumption for using the two means ztest and the two mean zinterval procedure that renders those procedures generally impractical.

Wing Length. Refer to Exercise 10.86 and find a 999 con fidence interval for the difference between the mean wing lengths o the two subspecies.

Right-Tailed Hypothesis Tests and CIs. If the assumptions for a pooled t-interval are satisfied, the formula for a (1-α)-level lower confidence bound for the difference, μ1-μ2, between two population means is

x¯1-x^2-ta·Sp1/n1+1/n2

For a right-tailed hypothesis test at the significance level α,

the null hypothesis H0:μ1=μ2will be rejected in favor of the alternative hypothesis Ha:μ1>μ2if and only if the (1-α)-level lower confidence bound for μ1-μ2is greater than or equal to 0. In each case, illustrate the preceding relationship by obtaining the appropriate lower confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

a. Exercise 10.47

b. Exercise 10.50

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.)

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