Beta-blockers In a study of heart surgery, one issue was the effect of drugs called beta-blockers on the pulse rate of patients during surgery. The available subjects were randomly assigned into two groups. One group received a beta-blocker; the other group received a placebo. The pulse rate of each patient at a critical point during the operation was recorded. Here are the data in summary form:

a. The distribution of pulse rate in each group is not Normal. The use of two-sample t procedures is still justified. Why?

b. Construct and interpret a $99\%$confidence interval for the difference in mean pulse rates for patients like these who receive a beta-blocker or a placebo.

c. Interpret the $99\%$confidence level in the context of this study.

The population distribution is not normal for the following reasons:

$$\begin{gathered} n_1=30 \\ {n}_2=30 \end{gathered}$$

If the sample size is large, the sampling distribution of the sample mean is approximately normal, according to the central limit theorem.

We can use the central limit theorem because the sample size of 30 is at least$30$, and thus the sampling distribution of the sample means is approximately Normal.

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

$$\begin{gathered} {\overline{x}}_1=65.2 \\ {\overline{x}}_2=70.3 \\ {n}_1=30 \\ {n}_2=30 \\ {s}_1=7.8 \\ {s}_2=8.3 \\ c=0.99=99\% \end{gathered}$$

Confidence interval calculation:

Now we'll calculate the $t$-value, and to do so, we'll need to know how many degrees of freedom there are. As a result, the degree of liberty will be:

$$\begin{gathered} dfmin(n_1-1,n_2-1) \\ =min(30-1,30-1) \\ =29 \end{gathered}$$

Then the $t$-value will be as:

$t^{*}=2.756$

As a result, the confidence interval will be:

role="math" localid="1654743888820" $$\begin{gathered} ({\overline{x}}_1-{\overline{x}}_2)-t_{\frac{\alpha }{2}}\times \sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}} \\ =(65.2-70.3)-2.756\times \sqrt{\frac{7.8^2}{30}+\frac{8.3^2}{30}} \\ =-10.8311 \end{gathered}$$

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

Therefore,, we conclude that there is $99\%$the confidence that the mean pulse rate for patients that receive a beta-blocker is between$10.8311$ lower and$0.6311$ higher than the mean pulse rate for patients that receive the placebo.

The $99\%$confidence interval means that $99\%$of all possible samples will have a $99\%$ confidence interval that includes the true difference in pulse rates between beta-blocker patients and placebo patients.