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 4194 patients treated with the drug, 197 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 047 (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)

Answer 1

Given:

n =4194, X = 197, $\alpha$ = 0.10

a)

Point estimate = X/n = 197/4194 = 0.047

b)

Critical value:

Z$\alpha$/2 = $\pm$ 1.645            ........Using standard Normal table

Margin of error (ME):

$ME =Z_{\alpha /2} * \sqrt{\frac{pq}{n}}$

$ME =1.64 * \sqrt{\frac{0.047*0.953}{4194}}$

ME = 0.005