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 cinical trials, among 4779 patients treated win the drug, 126 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 (Round to three decimal places as needed.) b) Identity the value of the margin of error E. (Round to three decimal places as needed) c) Construct the confidence interval (Round to three decimal places as needed.)

Answer 1

a.

The point estimate of the population proportion P is,

I P = 126 4779 0.026

b.

Margin of error:

$\small \\ ME = z_{\alpha /2} \sqrt{\frac{\hat{p} \: (1-\hat{p})}{n}} \\\\\\ ME =2.58 \sqrt{\frac{0.026 (0.974)}{4779}} \\ \\ \\ ME =0.006$

c.

99% confidence interval,

$\small \\CI = \hat{p} \pm ME \\\\ CI = 0.026 \pm 0.006 \\\\ CI = (0.020 , 0.032)$

d.

One has 99% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.