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 4628 patients treated with
the drug, 153 developed the adverse reaction of nausea. construct a 95% confidence interval for the proportion of
adverse reactions.

a) find the best point estimate of the population proportion p.
.033
(round to three decimal places as needed.)
b) identify the value of the margin of error e.
e =
(round to three decimal places as needed.)

Explanation:

Step1: Calculate \(\hat{p}\) and \(\hat{q}\)

The best point - estimate of the population proportion \(p\) is \(\hat{p}=\frac{x}{n}\), where \(x = 153\) and \(n=4628\). So \(\hat{p}=\frac{153}{4628}\approx0.033\). Then \(\hat{q}=1 - \hat{p}=1 - 0.033 = 0.967\)

Step2: Find \(z_{\alpha/2}\) for a 95% confidence level

For a 95% confidence level, \(\alpha=1 - 0.95=0.05\), and \(\alpha/2=0.025\). From the standard normal distribution table, \(z_{\alpha/2}=z_{0.025}=1.96\)

Step3: Calculate the margin of error \(E\)

The formula for the margin of error \(E\) for a proportion is \(E = z_{\alpha/2}\sqrt{\frac{\hat{p}\hat{q}}{n}}\)
Substitute \(\hat{p}=0.033\), \(\hat{q}=0.967\), \(n = 4628\), and \(z_{\alpha/2}=1.96\) into the formula:

$$
LATEXBLOCK0
$$

Rounding to three decimal places, \(E = 0.005\)

Answer:

\(E = 0.005\)