Chapter 10.1, Problem 26E

(a)

To determine

To construct and interpret a $95\%$ confidence interval for the difference between the true proportions and assume that the conditions for inference are met.

Answer to Problem 26E

There is $95\%$ confidence that the true death rate of people who take Prachachol is between $0.0129\text{ and 0}\text{.0651}$ higher than the true death rate for people who take Lipitor.

Explanation of Solution

Referring to the previous exercise we have,

Given:

  $\begin{array}{l}{n}_1=2063\\ n_2=2099\\ {\widehat{p}}_1=0.263\\ {\widehat{p}}_2=0.224\\ \alpha =0.05\end{array}$

Conditions to be satisfied:

There are three conditions to be satisfied:

Random: It is satisfied because the volunteers are independent random samples assigned to the treatment groups.

Independent: It is satisfied because the $2063$ people with heart disease are less than $10\%$ of all people with heart disease and $2099$ people with heart disease are less than $10\%$ of all people with heart disease.

Normal: It is satisfied because

  $\begin{array}{l}{n}_1{\widehat{p}}_1=2063(0.263)\\ =542.569\\ n_1(1-{\widehat{p}}_1)=2063(1-0.263)\\ =2063(0.737)\\ =1520.431\\ n_2{\widehat{p}}_2=2099(0.224)\\ =470.176\\ n_2(1-{\widehat{p}}_2)=2099(1-0.224)\\ =2099(0.776)\\ =1628.824\end{array}$

are all at least ten.

Thus, all the conditions are satisfied.

Calculation:

Now, for the confidence interval $1-\alpha =0.95$ , we have to find out the value of $z_{\alpha /2}=z_{0.025}$ using the table. Thus the z score will be:

  $z_{\alpha /2}=1.96$

Thus, the confidence interval will be:

  $\begin{array}{l}({\widehat{p}}_1-{\widehat{p}}_2)-z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}\\ =(0.263-0.224)-1.96\times \sqrt{\frac{0.263(1-0.263)}{2063}+\frac{0.224(1-0.224)}{2099}}\\ \approx 0.0129\\ ({\widehat{p}}_1-{\widehat{p}}_2)+z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}\\ =(0.263-0.224)+1.96\times \sqrt{\frac{0.263(1-0.263)}{2063}+\frac{0.224(1-0.224)}{2099}}\\ \approx 0.0651\end{array}$

Thus, we conclude that, we are $95\%$ confident that the true death rate of people who take Prachachol is between $0.0129\text{ and 0}\text{.0651}$ higher than the true death rate for people who take Lipitor.

(b)

To determine

To explain how the confidence interval provides more information than the test in previous exercise.

Explanation of Solution

The confidence interval gives more information than a significance test because the confidence interval gives us a range of possible values for the difference between the true proportions and the significance test only checks the one possible value for the difference between the proportions.