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 4318 patients treated with the​ drug,188 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.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.There is a 90​% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

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

Answer 1

Solution :

Given that,

n = 4318

x = 188

a) Point estimate = sample proportion = = x / n = 188 / 4318 = 0.044

1 - = 1 - 0.044 = 0.956

b) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

E = 1.645 (((0.044 * 0.956) / 4318)

E  = 0.005

c) A 90% confidence interval for population proportion p is ,

- E < p <   + E

0.044 - 0.005 < p < 0.044 + 0.005

( 0.039 < p < 0.049 )

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.