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 4330 patients treated with the​ drug, 121 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.

​(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.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.

B.One has 99​% confidence that the sample proportion is equal to the population proportion.

C. 99​% of sample proportions will fall between the lower bound and the upper bound.

D.There is a 99​% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

Answer 1

Solution :

Given that,

n = 4330

x = 121

a) Point estimate = sample proportion = = x / n = 121 / 4330 = 0.028

1 - = 1 - 0.028 = 0.972

b) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.960

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

= 1.96 (((0.028 * 0.972) / 4330 )

= 0.005

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

- E < p < + E

0.028 - 0.005 < p < 0.028 + 0.005

( 0.023 < p < 0.033 )

A.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.