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 4829 patients treated with
the drug, 115 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.
0 024
(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 sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 115$ (number of patients with nausea) and $n=4829$ (total number of patients).
$\hat{p}=\frac{115}{4829}\approx0.024$ (already given for part a).

Step2: Find $z -$ score for 90% confidence level

For a 90% confidence level, the significance level $\alpha=1 - 0.90=0.10$. Then $\alpha/2=0.05$.
From the standard normal distribution table, $z_{\alpha/2}=z_{0.05}\approx1.645$.

Step3: Calculate margin of error formula

The formula for the margin of error $E$ for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
Substitute $\hat{p}=0.024$, $n = 4829$, and $z_{\alpha/2}=1.645$ into the formula.
First, calculate $1-\hat{p}=1 - 0.024=0.976$.
Then $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.024\times0.976}{4829}=\frac{0.023424}{4829}\approx4.85\times10^{-6}$.
$\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\approx\sqrt{4.85\times10^{-6}}\approx0.0022$.
$E=1.645\times0.0022\approx0.004$.

Answer:

$E = 0.004$