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 4413 patients treated with the​ drug, 173 developed the adverse reaction of nausea. Construct a 95​% confidence interval for the proportion of adverse reactions.

Answer 1

Solution :

Given that,

n = 4413

x = 173

Point estimate = sample proportion = = x / n = 173/4413=0.039

1 -   = 1- 0.039 =0.961

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.039*0.961) /4413 )

E = 0.0057

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.039-0.0057 < p < 0.039+ 0.0057

0.0333< p < 0.0447

The 95% confidence interval for the population proportion p is : 0.0333,0.0447