Question

Aspirin prevents blood from clotting and so helps prevent strokes. The Second European Stroke Prevention Study asked whether adding another anti-clotting drug, named dipyridamole, would be more effective for patients who had already had a stroke. Here are the data on strokes and deaths during the two years of the study.

$\begin{array}{lcc}& \begin{array}{c}\text{Number of}\\ \text{patients}\end{array}& \begin{array}{c}\text{Number of}\\ \text{strokes}\end{array}\\ \text{Aspirin alone}& 1649& 206\\ \text{Aspirin + dipyridamole}& 1650& 157\\ & & \end{array}$

The study was a randomized comparative experiment.

(a) Is there a significant difference in the proportion of strokes between these two treatments? Carry out an appropriate test to help answer this question.

(b) Describe a Type I and a Type II error in this setting. Which is more serious? Explain

Short answer

(a) There is sufficient evidence to support the claim of a difference between the population proportions.

(b) Type I error is more serious, as a type I error could be detrimental to the health of people.

Step-by-step solution

Part(a) Step 1: Given Information

Given

$x_1=206$

$n_1=1649$

$x_2=157$

$n_2=1650$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2\ne 0$

Part(a) Step 2: Explanation

The sample proportion is the number of successes divided by the sample size:

${\hat{p}}_1=\frac{x_1}{n_1}=\frac{206}{1649}\approx 0.125$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{157}{1650}\approx 0.095$

${\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}=\frac{206+157}{1649+1650}=\frac{363}{3299}\approx 0.110$

Determine the value of the test statistic:

localid="1650451069873" $$\begin{gathered} z=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \\ =\frac{0.125-0.095}{\sqrt{0.110(1-0.110)}\sqrt{\frac{1}{1649}+\frac{1}{1650}}} \\ \approx 2.75 \end{gathered}$$

The $p$-value is the probability of obtaining the value of the test statistic, or a value more extreme. Determine the $p$-value using table A:

localid="1650451084455" $$\begin{gathered} P=P(Z<-2.75\text{or}Z>2.75) \\ =2\times P(Z<-2.75) \\ =2\times 0.0030 \\ =0.0060 \end{gathered}$$

If the $p$-value is smaller than the significance level, reject the null hypothesis:

$P<0.05\Rightarrow \text{Reject}{H}_0$

Part(b) Step 1: Given Information

Given

$x_1=206$

$n_1=1649$

$x_2=157$

$n_2=1650$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2\ne 0$

Part(b) Step 2: Explanation

Type I error: Rejecting the null hypothesis $H_0$, when $H_0$is true

Interpretation: The significance test indicates that the proportions are significantly different, while they are actually the same. This then implies that we assume that the treatment is more effective when in reality the treatment is not more effective and thus this might be detrimental to the health of people.

Type II error: Failing to reject the null hypothesis $H_0$, when $H_0$is false

Interpretation: The significance test indicates that the proportions are the same, while they are actually different. This then implies that we assume that the treatment is not effective when in reality the treatment is effective and thus we might be missing a useful treatment.

We then note that a type I error is more serious, as a type I error could be detrimental to the health of people.