A survey published in the American Journal of Sports Medicin

A survey published in the American Journal of Sports Medicine^2 reported the number of meters (m) per week swum by two groups of swimmers-those who competed exclusively in breaststroke and those who competed in the individual medley (which includes breaststroke). The number of meters per week practicing the breaststroke was recorded for each swimmer, and the summary statistics are given below. Is there sufficient evidence to indicate that the average number of meters per week spent practicing breaststroke is greater for exclusive breaststrokers than it is for those swimming individual medley? State the null and alternative hypotheses. What is the appropriate rejection region for an alpha = .01 level test? Calculate the observed value of the appropriate test statistic. What is your conclusion? What is a practical reason for the conclusion you reached in part (d)?

Solution

Let mu1 be the mean for exclusively breaststroke

Let mu2 be the mean for individual medley

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(a) Ho: mu1=mu2 (i.e. null hypothesis)

Ha: mu1> mu2 (i.e. alternative hypothesis)

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(b) Since both sample sizes are grearter than n=30, we can use normal distriubtion.

It is a one-tailed test.

Given a=0.01, the critical value is Z(0.01) =2.33 (from standard normal table)

So the rejection region is Z>2.33

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(c) The test statistic is

Z=(xbar2-xbar2)/sqrt(s1^2/n1+s2^2/n2)

=(9017-5853)/sqrt(7162^2/130+1961^2/80)

=4.76

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(d) Since Z=4.76 is larger than 2.33, we reject the null hypothesis.

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(e) So we can conclude that there is sufficient evidence to indicate that the average number of meter per week spent practicing breaststroke is greater for exclusive breaststrokers than it is for those swimming individual medley