Question

2. 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 4691 patients treated with the​ drug, 158 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.

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

E =

​c) Construct the confidence interval.

??? <p< ???

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

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

B. 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.

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

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

Answer 1

Solution :

Given that,

n = 4691

x = 158

a)Point estimate = sample proportion = = x / n = 158 / 4691 = 0.034

1 - = 1 - 0.034 = 0.966

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z _0.05 = 1.645

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

= 1.645 * (((0.034*0.966) / 4691)

= 0.004

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

- E < p < + E

0.034 - 0.004 < p < 0.034 + 0.004

0.030 < p < 0.038

The 90% confidence interval for the population proportion p is : ( 0.030 , 0.038 )

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

bound.