# Solved (Free): A study is run comparing HDL cholesterol levels between men who exercise regularly and those who do not

#### ByDr. Raju Chaudhari

Mar 13, 2021

A study is run comparing HDL cholesterol levels between men who exercise regularly and those who do not. The data are shown below.

Regular Exercise N Mean Std Dev
Yes 35 48.5 12.5
No 120 56.9 11.9

Generate a 95 % confidence interval for the difference in mean HDL levels between men who exercise regularly and those who do not.

#### Solution

Given that $n_1 = 35$, $\overline{X}_1 =48.5$, $s_1 = 12.5$, $n_2 =120$, $\overline{X}_2 =56.9$ and $s_2 = 11.9$.

Specify the confidence level $(1-\alpha)$

The confidence level is $1-\alpha = 0.95$, thus $\alpha = 0.05$.

Given information

Given that $n_1 = 35$, $\overline{X}_1= 48.5$, $s_1 = 12.5$.

$n_2 = 120$, $\overline{X}_2= 56.9$, $s_2 = 11.9$.

Specify the formula

$100(1-\alpha)$% confidence interval estimate for the difference $(\mu_1-\mu_2)$ is

 \begin{aligned} (\overline{X} -\overline{Y})- E \leq (\mu_1-\mu_2) \leq (\overline{X} -\overline{Y}) + E. \end{aligned}

where $E = t_{\alpha/2,n_1+n_2-2} \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$.

Determine the critical value

The critical value $t_{\alpha/2,n_1+n_2-2} = t_{0.025,153} = 1.976$.

Compute the margin of error

The margin of error for proportions is

 \begin{aligned} E & = t_{\alpha/2,n_1+n_2-2} \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ & = 1.976 \sqrt{\frac{12.5^2}{35}+\frac{11.9^2}{120}}\\ & = 4.695. \end{aligned}

Determine the confidence interval

$95$% confidence interval estimate for the difference $(\mu_1-\mu_2)$ is

 \begin{aligned} (\overline{X} -\overline{Y})- E & \leq (\mu_1-\mu_2) \leq (\overline{X} -\overline{Y}) + E\\ (48.5-56.9) - 4.695 & \leq (\mu_1-\mu_2) \leq (48.5-56.9) + 4.695\\ -13.095 & \leq (\mu_1-\mu_2) \leq -3.705. \end{aligned}

Thus, $95$% confidence interval for the difference in mean HDL levels between men who exercise regularly and those who do not is $(-13.095,-3.705)$.