Question

Constructing and Interpreting Confidence Intervals. In Exercises 13–16, use the given sample data and confidence level. In each case, (a) find the best point estimate of the population proportion p; (b) identify the value of the margin of error E; (c) construct the confidence interval; (d) write a statement that correctly interprets the confidence interval.

Eliquis The drug Eliquis (apixaban) is used to help prevent blood clots in certain patients. In clinical trials, among 5924 patients treated with Eliquis, 153 developed the adverse reaction of nausea (based on data from Bristol-Myers Squibb Co.). Construct a 99% confidence interval for the proportion of adverse reactions.

Short answer

(a)Thebest point estimate of the proportion ofpatients who took Eliquis and developed nauseais equal to 0.026.

(b)the margin of error is equal to 0.0053.

(c)The 99% confidence interval estimate of the population proportion ofpatients who took Eliquis and developed nauseais equal to (0.021, 0.031).

(d) There is 99% confidence that the true proportion of patients who took Eliquis and developed nausea will lie between the values 0.021 and 0.031.

Step-by-step solution

Given information

In a sample of 5924 patients treated with Eliquis drug, 153 patients suffered from the adverse reaction of nausea.

Compute the sample proportion

(a)

The best point estimate of the proportion of patients who took Eliquis and developed nausea is computed below:

$$\begin{gathered} \hat{p}=\frac{153}{5924} \\ =0.026 \end{gathered}$$

Thus, the sample proportion equal to 0.026 is the best point estimate of the proportion of patients who took Eliquis and developed nausea.

Compute the margin of error

(b)

The confidence level is equal to 99%. Thus, the corresponding level of significance is equal to 0.01.

From the standard normal distribution table, the right-tailed value of $z_{\frac{\alpha }{2}}$for $\alpha =0.01$is equal to 2.5758.

The margin of error is calculated below:

$$\begin{gathered} E=2.5758\times \sqrt{\frac{0.026\times 0.9741}{5924}} \\ =0.0053 \end{gathered}$$

Thus, the margin of error is equal to 0.0053.

Compute the confidence interval

(c)

The formula for computing the confidence interval estimate of the population proportion is written below:

$CI=\left(\hat{p}-E,\hat{p}+E\right)$

The 95% confidence interval becomes equal to:

$$\begin{gathered} CI=\left(0.026-0.0053,0.026+0.0053\right) \\ =\left(0.021,0.031\right) \end{gathered}$$

Therefore, the 99% confidence interval estimate of the population proportion of patients who took Eliquis and developed nausea is equal to (0.021, 0.031).

Interpretation of the confidence interval

(d)

There is 99% confidence that the true proportion ofpatients whotook Eliquis anddeveloped nauseawill lie between the values, 0.021 and 0.031.