Question

A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 N Mean StDev SE Mean men and 74 women to participate in the study. Each subject was required Men 54 112.6 11.9 1.6 to step up and down a 6-inch platform. The pulse of each subject was Women 74 118.3 14.9 1.7 then recorded. The following results were obtained. 99% CI for mu Men -mu Women ( - 11.97, 0.57) Two sample T for Men vs Women T-Test mu Men=mu Women (vs < ) T= -2.40 P = 0.0089 DF = 124

What is the 99​% confidence interval for the mean difference in pulse rates of men versus​ women? Identify the lower bound and the upper bound. Round to two decimal places as needed.

Answer 1

In general, (1-$\alpha$)% confidence interval for the population mean difference is obtained as follows :
$(\bar{x}_{M}-\bar{x}_{F}\mp t_{\alpha/2,n_{1}+n_{2}-2}*S*\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}})$
Here, t ($\alpha$/2, n1+n2-2) = t (0.005, 126) = 2.6154   (from t tables)

S =  $\sqrt{\frac{(n_{1}-1)*S_{1}^{2}+(n_{2}-1)*S_{2}^{2}}{n_{1}+n_{2}-2}}$ = 13.7183

Thus, 99% confidence interval for the population mean difference is :
$(112.6-118.3-2.6154*13.7183*\sqrt{\frac{1}{54}+\frac{1}{74}},112.6-118.3+2.6154*13.7183*\sqrt{\frac{1}{54}+\frac{1}{74}})$
$\Rightarrow$ ( -12.1214, 0.7214 )
$\Rightarrow$ ( -12.12, 0.72 )

Hope this answers your query!