# Solved (Free): The following data were collected in a clinical trial to compare a new drug to a placebo for its effectiveness in lowering total serum cholesterol

#### ByDr. Raju Chaudhari

Mar 13, 2021

The following data were collected in a clinical trial to compare a new drug to a placebo for its effectiveness in lowering total serum cholesterol. Generate a 95% confidence interval for the difference in mean total cholesterol levels between treatments.

. New Drug (n=75) Placebo (n = 75) Total Sample (n = 150)
Mean (SD) Total Serum Cholesterol 185.0 (24.5) 204.3 (21.8) 194.7 (23.2)
% of patients with Total Cholesterol < 200 78.0% 65.0 % 71.5 %

#### Solution

Given that $n_1 = 75$, $\overline{X}_1 =185$, $s_1 = 24.5$, $n_2 =75$, $\overline{X}_2 =204.3$ and $s_2 = 21.8$.

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

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

Given information

Given that $n_1 = 75$, $\overline{X}_1= 185$, $s_1 = 24.5$.

$n_2 = 75$, $\overline{X}_2= 204.3$, $s_2 = 21.8$.

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 $t_{\alpha/2,n_1+n_2-2} = t_{0.025,148} = 1.976$.

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{24.5^2}{75}+\frac{21.8^2}{75}}\\ & = 7.483. \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\\ (185-204.3) - 7.483 & \leq (\mu_1-\mu_2) \leq (185-204.3) + 7.483\\ -26.783 & \leq (\mu_1-\mu_2) \leq -11.817. \end{aligned}
Thus, $95$% confidence interval for the difference in mean total cholesterol levels between treatments $(\mu_1-\mu_2)$ is $(-26.783,-11.817)$.