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 4432 patients treated with the drug, 104 developed the adverse reaction of nausea. Construct a 95% 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. EN (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. O A. There is a 95% chance that the true value of the population proportion will fall between the lower bound and the upper bound O B. One has 95% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion OC. One has 95% confidence that the sample proportion is equal to the population proportion. OD. 95% of sample proportions will fall between the lower bound and the upper bound.

Answer 1

Answer:

C=95%, x=104, n=4432

a) Point estimate is

$\hat p =\frac{x}{n} =\frac{104}{4432} = 0.023465704$

Point estimate =0.023

b) Formula for margin of error is

р* (1 —р) п

Where Zc is the z critical value for C=95%

Zc=1.96

$1.96 * \sqrt{\frac{0.023*(1-0.023)}{4432}}$

Margin of error (E)= 0.0044133401

Margin of error (E)= 0.004

c) Formula for confidence interval is

р*(1— р) V p Zc п

$0.023\pm1.96 * \sqrt{\frac{0.023*(1-0.023)}{4432}}$

0.023 $\pm$ 0.004

Thus we get confidence interval as 0.019 < p < 0.027

d)

b) one has 95% confident that the interval from lower bound to upper bound actually does contain the true value of population proportion.