Chapter 10.2, Problem 48E

(a)

To determine

To explain why the distribution of pulse rate in each group is not normal.

Explanation of Solution

The population distribution is not normal because:

  $\begin{array}{l}{n}_1=30\\ n_2=30\end{array}$

Central limit theorem is if the sample size is large then the sampling distribution of the sample mean is approximately normal.

Since the sample size of $30\text{ is at least 30}$ we can use the central limit theorem and thus the sampling distribution of the sample means is approximately Normal.

The sampling distribution of the difference between the sample means is then also approximately Normal.

(b)

To determine

To construct and interpret $99\%$ confidence interval for the difference in mean pulse rate for patients like these who receive a beta blocker or a placebo.

Answer to Problem 48E

There is $99\%$ confidence that the mean pulse rate for patients that receive a beta blocker is between $10.8311\text{ lower and 0}\text{.6311 higher}$ than the mean pulse rate for patients that receive the placebo.

Explanation of Solution

Given:

  $\begin{array}{l}{\overline{x}}_1=65.2\\ {\overline{x}}_2=70.3\\ n_1=30\\ n_2=30\\ s_1=7.8\\ s_2=8.3\\ c=0.99=99\%\end{array}$

Calculating confidence interval:

Now we will be calculating the t -value for this we need to find out the degree of freedom. Thus, the degree of freedom will be:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (30-1,30-1)\\ =29\end{array}$

Then the t -value will be as:

  $t*=2.756$

Thus the confidence interval be:

  $\begin{array}{l}({\overline{x}}_1-{\overline{x}}_2)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(65.2-70.3)-2.756\times \sqrt{\frac{{7.8}^2}{30}+\frac{{8.3}^2}{30}}\\ =-10.8311\\ ({\overline{x}}_1-{\overline{x}}_2)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(65.2-70.3)+2.756\times \sqrt{\frac{{7.8}^2}{30}+\frac{{8.3}^2}{30}}\\ =0.6311\end{array}$

Thus, we conclude that there is $99\%$ confidence that the mean pulse rate for patients that receive a beta blocker is between $10.8311\text{ lower and 0}\text{.6311 higher}$ than the mean pulse rate for patients that receive the placebo.

(c)

To determine

To interpret the $99\%$ confidence level in the context of this study.

Explanation of Solution

The $99\%$ confidence interval means that on average $99\%$ of all possible samples will have a $99\%$ confidence interval that contains the true difference in pulse rates of the patients who receive the beta blockers and the mean pulse rates of the patients who receive a placebo.