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 4630 patients treated with the​ drug, 130 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.

​(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 sample proportion is equal to the population proportion.

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

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

D.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.

Answer 1

Solution :

Given that,

n = 4630

x = 130

Point estimate = sample proportion = = x / n = 130 / 4630 = 0.028

a) The best point estimate of the population proportion p : 0.028

1 - = 1 - 0.028 = 0.972

At 99% confidence level

= 1 - 99%

=1 - 0.99 = 0.01

/2 = 0.005

Z/2 = Z_0.005 = 2.576

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

= 2.576 * ((( 0.028 * 0.972) ) / 4630)

= 0.006

b) The value of the margin of error E = 0.006

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.028 - 0.006 < p < 0.028 + 0.006

0.022 < p < 0.034

c) The 99% confidence interval for the population proportion p is : 0.022 < p < 0.034

d) Answer :

D.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.