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 4469 patients treated with the​ drug, 190 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.

​(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.)

​c) Construct the confidence interval.

_ < p < _

​(Round to three decimal places as​ needed.)

​d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below.

A. 90​% of sample proportions will fall between the lower bound and the upper bound.

B.There is a 90% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

C.One has 90​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

D.One has 90​% confidence that the sample proportion is equal to the population proportion.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 190 / 4469 = 0.043

Z_/2 = 1.645

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

= 1.645 * (((0.043 * 0.957) / 4469)

= 0.005

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.043 - 0.005 < p < 0.043 + 0.005

0.038 < p < 0.048

C.One has 90​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.