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.
(round to three decimal places as needed.)
b) identify the value of the margin of error e.
e = 005
(round to three decimal places as needed.)
c) construct the confidence interval.
<p<
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion $\hat{p}$

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 153$ (number of successes) and $n=4628$ (sample size). So, $\hat{p}=\frac{153}{4628}\approx0.033$.

Step2: Use the formula for the confidence interval

The confidence interval for a proportion is $\hat{p}-E < p<\hat{p} + E$. We know $\hat{p}\approx0.033$ and $E = 0.005$ (given).

Step3: Calculate the lower and upper bounds

Lower bound: $\hat{p}-E=0.033 - 0.005=0.028$.
Upper bound: $\hat{p}+E=0.033+ 0.005=0.038$.

Answer:

$0.028 < p<0.038$