Chapter 10.1, Problem 22E

(a)

To determine

To test whether is there any significant difference in proportion of stokes between these two treatments.

Explanation of Solution

Given the following information:

  $\begin{array}{l}{n}_1=1649n_2=1650\\ x_1=206x_2=157\end{array}$

  $x_1$ , $x_2$ are the number stokes in the two treatments respectively.

The test procedure:

The null and alternative hypothesis is given by

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_1:p_1\ne p_2\end{array}$

The test statistic is given by,

  $Z=\frac{p_1-p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}}~N(0,1)$

Reject the null hypothesis if $\left|Z\right|>Z_{\alpha /2}$

Calculation:

On substituting the given values the test statistic is given by

  $\begin{array}{l}Z=\frac{p_1-p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}}\\ Where{p}_1=\frac{x_1}{n_1}=\frac{206}{1649}=0.1249\\ p_2=\frac{x_2}{n_2}=\frac{157}{1650}=0.0952\\ \therefore Z=\frac{0.1249-0.0952}{\sqrt{\frac{0.1249(1-0.1249)}{1649}+\frac{0.0952(1-0.0952)}{1650}}}\\ \text{ =}\frac{0.0297}{\sqrt{0.00007+0.00005}}\\ \text{ =2}\text{.7112}\end{array}$

The critical value at 0.05 level of significance is $Z_{0.025}=1.96$

Since $\left|Z\right|>1.96$ , we reject the null hypothesis and conclude that there is a significant difference between two treatments.

(b)

To determine

To describe the Type 1 error and Type 2 error in this problem.

Explanation of Solution

Type 1 error is the mistakes in which we reject the proportion stoke in the treatment Aspirin and Aspirin with dipyridamole are same even though it is true.

Type 2 error is the mistakes in which we do not reject the proportion of the proportion stoke in the treatment Aspirin and Aspirin with dipyridamole are same even though it is not true.

Type 1 error is more dangerous since the chance that a null hypothesis which is not true will not be rejected. Therefore Type 1 error should be maintain under control.