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Chapter 10: Q. 10.117 (page 441)

Ha:μ1≠μ2

Short Answer

Expert verified

Since the value of test statistic is fall in the rejection region.

Thus, the nullhypothesis is rejected.

Step by step solution

01

Given Information

The hypothesis is

H0:μ1=μ2,Ha:μ1≠μ2

and the level of significance is 0.1.

02

Explanation

We have to find the sample mean as follow:

d¯=∑dn

=187

=2.571

Let's compute the standard deviation:

sd=∑di2-∑di2nn-1

=76-(18)277-1

=2.2254

The test statistics are,

t=d¯sdnn

Replace the given values in the equation above.

z=2.57122047

=3.06

Compute the degree of freedom as follow

df=n-1

=7-1=6

The critical value for a level of significance of 0.1is 1.943.

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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-fest and the non pooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.

x~1=20,s1=2,n1=30,x2=18,s2=5,n2=40.

a. Right-tailed test, α=0.05

b. 90%confidence interval.

In this section, we introduced the pooled t-test, which provides a method for comparing two population means. In deriving the pooled f-test, we stated that the variable

z=f^1-x^2-μ1-μ2σ1/n1+1/n2

cannot be used as a basis for the required test statistic because σ is unknown. Why can't that variable be used as a basis for the required test statistic?

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.

Doing Time. Refer to Exercise 10.45 and obtain a 90% confidence interval for the difference between the mean times served by prisoners in the fraud and firearms offense categories.

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.)
See all solutions

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