Question

Critical Thinking. In Exercises 17–28, use the data and confidence level to construct a confidence interval estimate of p, then address the given question.

Lipitor In clinical trials of the drug Lipitor (atorvastatin), 270 subjects were given a placebo, and 7 of them had allergic reactions. Among 863 subjects treated with 10 mg of the drug, 8 experienced allergic reactions. Construct the two 95% confidence interval estimates of the percentages of allergic reactions. Compare the results. What do you conclude?

Short answer

The 95% confidence interval estimate of the proportion of subjects who were given a placebo and developed allergic reactions is equal to (0.70%, 4.49%).

The 95% confidence interval estimate of the proportion of subjects who were treated with the drug and developed allergic reactions is equal to (0.29%, 1.57%).

The drug Lipitor does not significantly result in the occurrence of allergic reactions.

Step-by-step solution

Given information

In a sample of 270 subjects who were given a placebo, 7 developed allergic reactions. In another sample of 863 subjects who were treated with the drug, 8 developed allergic reactions.

Step 2:Calculation of the sample proportions

The sample proportion of subjects who were given a placebo and developed allergic reactions is computed below:

$$\begin{gathered} {\hat{p}}_1=\frac{7}{270} \\ =0.0259 \end{gathered}$$

Hence,

$$\begin{gathered} {\hat{q}}_1=1-{\hat{p}}_1 \\ =1-0.0259 \\ =0.9741 \end{gathered}$$

The sample proportion of subjects who were treated with the drug and developed allergic reactions is computed below:

$$\begin{gathered} {\hat{p}}_2=\frac{8}{863} \\ =0.0093 \end{gathered}$$

Hence,

$$\begin{gathered} {\hat{q}}_2=1-{\hat{p}}_2 \\ =1-0.0093 \\ =0.9907 \end{gathered}$$

Calculation of the margins of error

The given level of significance is 0.05.

Therefore, the value of$z_{\frac{\alpha }{2}}$ from the standard normal table is equal to 1.96.

The margin of error corresponding to the placebo group is computed below:

$$\begin{gathered} E_1=z_{\frac{\alpha }{2}}\times \sqrt{\frac{{\hat{p}}_1{\hat{q}}_1}{n_1}} \\ =1.96\times \sqrt{\frac{0.0259\times 0.9741}{270}} \\ =0.0190 \end{gathered}$$

Therefore, the margin of error corresponding to the placebo group is equal to 0.0190.

The margin of error corresponding to the treatment group is computed below:

$$\begin{gathered} E_2=z_{\frac{\alpha }{2}}\times \sqrt{\frac{{\hat{p}}_1{\hat{q}}_1}{n_2}} \\ =1.96\times \sqrt{\frac{0.0093\times 0.9907}{863}} \\ =0.0064 \end{gathered}$$

Therefore, the margin of error corresponding to the treatment group is equal to 0.0064.

Calculation of the confidence intervals

The 95% confidence interval estimate of the proportion of subjects who were given a placebo and developed allergic reactions is computed below:

$$\begin{gathered} {\hat{p}}_1-E_1<p<{\hat{p}}_1+E_1 \\ 0.0259-0.0190<p<0.0259+0.0190 \\ 0.0070<p<0.0449 \\ 0.70\%<p<4.49\% \end{gathered}$$

Thus, the 95% confidence interval estimate of the proportion of subjects who were given a placebo and developed allergic reactions is equal to (0.70%, 4.49%).

The 95% confidence interval estimate of the proportion of subjects who were treated with the drug and developed allergic reactions is computed below:

$$\begin{gathered} {\hat{p}}_2-E_2<p<{\hat{p}}_2+E_2 \\ 0.0093-0.0064<p<0.0093+0.0064 \\ 0.0029<p<0.0157 \\ 0.29\%<p<1.57\% \end{gathered}$$

Thus, the 95% confidence interval estimate of the proportion of subjects who were treated with the drug and developed allergic reactions is equal to (0.29%, 1.57%).

Comparison

Upon observing the confidence intervals, it can be said that the percentage of subjects who were given a placebo and developed allergic reactions is comparatively more than the percentage of subjects who were treated with the drug and developed allergic reactions.

Therefore, there is no particular instance of the occurrence of allergic reactions associated with the drug.