Question

Medical researchers conducted a study to determine whether treadmill exercise could improve the walking ability of patients suffering from claudication, which is pain caused by insufficient blood flow to the muscles of the legs. A random sample of 48 patients walked on a treadmill for six minutes every day. After six months, the sample mean distance walked in six minutes was 348 meters, with a sample standard deviation of 80 meters. For a control group of 46 patients who did not walk on a treadmill, the sample mean distance was 309 meters with a sample standard deviation of 89 meters. Can you conclude that the population mean distance walked for patients using a treadmill is greater than that for those who do not walk on a treadmill? Test at the 5% level of significance

Answer #1

Sol:

Ho:mu1=mu2

Ha:mu1>mu2

alpha=0.05

test statistic

t=x1bar-x2bar/sqrt(s1^2/n1+s2^2/n2)

=(348-309)/sqrt((80^2/48+89^2/46))

t= 2.231199

df=n1+n2-2

=48+46-2

=92

p value in excel is

=T.DIST.2T(2.231199,92)

=0.028097625

p<0.05

Reject Ho

Accept Ha

Cocnclusion:

There is sufficient statistical evidence at 5% level of significance to conclude that

the population mean distance walked for patients using a treadmill is greater than that for those who do not walk on a treadmill.

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