Solved (Free): A clinical trial is run to compare the effectiveness of an experimental drug in reducing preterm delivery to a drug considered standard care and to piacebo

ByDr. Raju Chaudhari

Mar 13, 2021

A clinical trial is run to compare the effectiveness of an experimental drug in reducing preterm delivery to a drug considered standard care and to piacebo. Pregnant women are enrolled and randomly assigned to receive the experimental drug, the standard drug or placebo. Women are followed through delivery and classified as delivering preterm (< 37 weeks) or not. The resulting data are shown below.

Preterm Delivery Experimental Drug Standard Drug Placebo
Yes 17 23 35
No 83 77 65

Previous studies have shown that approximately 32% of women deliver prematurely without treatment. Is the proportion of women delivering prematurely significantly higher in the placebo group? Run the test at a 5 % level of significance.

Solution

Given that $n = 100$, $X= 35$.

The sample proportion is

$$\hat{p}=\frac{X}{n}=\frac{35}{100}=0.35$$.

Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : p = 0.32$ against $H_1 : p > 0.32$ ($\text{right-tailed}$)

Test Statistic

The test statistic for testing above hypothesis testing problem is

\begin{aligned} Z & = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \end{aligned}
which follows $N(0,1)$ distribution.

Significance Level

The significance level is $\alpha = 0.05$.

Critical values

As the alternative hypothesis is $\textit{right-tailed}$, the critical value of $Z$ $\text{ is }$ $\text{1.64}$.

The rejection region (i.e. critical region) for the hypothesis testing problem is $\text{Z > 1.64}$.

Computation

The test statistic is

\begin{aligned} Z & = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\\ &= \frac{0.35-0.32}{\sqrt{\frac{0.32* (1-0.32)}{100}}}\\ & =0.643 \end{aligned}

Decision

The test statistic is $Z =0.643$ which falls $outside$ the critical region, we $\text{fail to reject}$ the null hypothesis.

$p$-value approach:

This is a $\text{right-tailed}$ test, so the p-value is the area to the left of the test statistic ($Z=0.643$). Thus the $p$-value = $P(Z < 0.643) =0.2601$.

The p-value is $0.2601$ which is $\text{greater than}$ the significance level of $\alpha = 0.05$, we $\text{fail to reject}$ the null hypothesis.

We conclude that the proportion of women delivering prematurely is not significantly higher in the placebo group at 0.05 level of significance.