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

4884

patients treated with the​ drug,

117

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.

nothing

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

​b) Identify the value of the margin of error E.

Eequals=nothing

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

​c) Construct the confidence interval.

nothing less than p less than<p<nothing

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

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

A.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.

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

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

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

Answer 1

Solution :

Given that,

a) Point estimate = sample proportion = = x / n = 117 / 4884 = 0.024

1 - = 1 - 0.024 = 0.976

Z/2 = Z0.05 = 1.645

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

= 1.645 (((0.024 * 0.976) / 4884)

= 0.004

c) A 95% confidence interval for population proportion p is ,

- E < p < + E

0.024 - 0.004 < p < 0.024 + 0.004

( 0.020 < p < 0.028 )

d) A.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.