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 4778 patients treated with the​ drug, 104 developed the adverse reaction of nausea. Construct a 99​% confidence interval for the proportion of adverse reactions. ​

a) Find the best point estimate of the population proportion p.

0.022 ​(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

Solution :

Given that,

n = 4778

x = 104

= x / n = 104 / 4778 = 0.022

a) Point estimate = 0.022

1 - = 1 - 0.022 = 0.978

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.576 * (((0.022 * 0.978) / 4778)

b) Margin of error = 0.005

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.022 - 0.005 < p < 0.022 + 0.005

0.017 < p < 0.027

(0.017,0.017)

c) The 99% confidence interval is 0.017 to 0.027