Question

use the sample data and confidence level given below to complete parts (a) through (d).
a drug is used to help prevent blood clots in certain patients. in clinical trials, among 4105 patients treated with the drug, 111 developed the adverse reaction of
nausea. construct a 90% confidence interval for the proportion of adverse reactions.
a) find the best point estimate of the population proportion p.
0.027
(round to three decimal places as needed.)
b) identify the value of the margin of error e.
e = 0.004
(round to three decimal places as needed.)
c) construct the confidence interval.
□ < p < □
(round to three decimal places as needed.)

Explanation:

Step1: <Find best point estimate>

The best point estimate of the population proportion \(p\) is the sample proportion \(\hat{p}\).

$$ \hat{p}=\frac{x}{n} $$

where \(x = 111\) (number of successes, i.e., patients with adverse reaction) and \(n=4105\) (sample size).

$$ \hat{p}=\frac{111}{4105}\approx0.027 $$

Step2: <Construct confidence interval>

For a \(90\%\) confidence level, the critical value \(z_{\alpha/2}\) is \(z_{0.05}\). From the standard normal distribution table, \(z_{0.05} = 1.645\).
The formula for the confidence interval for a proportion is \(\hat{p}-EWe know \(\hat{p}=0.027\), \(n = 4105\), \(z_{\alpha/2}=1.645\)
First, calculate \(1-\hat{p}=1 - 0.027=0.973\)
Then \(\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.027\times0.973}{4105}}\)

$$ \frac{0.027\times0.973}{4105}=\frac{0.026271}{4105}\approx6.4\times10^{-6} $$

$$ \sqrt{6.4\times 10^{-6}}\approx0.00253 $$

\(E=z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=1.645\times0.00253\approx0.004\)
The confidence interval is \(\hat{p}-E=0.027 - 0.004=0.023\) and \(\hat{p}+E=0.027+0.004 = 0.031\)

Answer:

a) The best point estimate of the population proportion \(p\) is \(0.027\)
b) The value of the margin of error \(E = 0.004\)
c) The \(90\%\) confidence interval is \(0.023