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 4538 patients treated with the​ drug,145 developed the adverse reaction of nausea. Construct a 99% confidence interval for the proportion of adverse reactions.

A.) find the best point of the estimate of the population of portion p.

B.) Identify the value of the margin of error E.

C.) Construct the confidence interval. Round to 3 decimal places

D.) Write a statement that correctly interprets the confidence interval.

Answer 1

Solution :

Given that,

n = 4538

x = 145

A)

Point estimate = sample proportion = = x / n = 0.032

1 - = 0.968

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

B)

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

= 2.576 * (((0.032 * 0.968) / 4538)

= 0.007

C)

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.032 - 0.007 < p < 0.032 + 0.007

0.025 < p < 0.039

D)

A 99% confidence interval for the proportion of adverse reactions is between 0.025 to 0.039.