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 4691 patients treated with the​ drug, 146 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.

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

c) ​Construct the confidence interval. ???<p<???

Answer 1

Solution :

Given that,

n = 4691

x = 146

Point estimate = sample proportion = = x / n = 146 / 4691 = 0.031

( a ) The best point estimate of the population proportion p = 0.031

1 - = 1 - 0.031 = 0.969

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.960

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

= 1.960 ((( 0.031 * 0.969 / 4691 )

= 0.005

( b ) The value of the margin of error E. E = 0.005

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.031 - 0.005 < p < 0.031 - 0.005

0.026 < p < 0.036

( 0.026 , 0.036 )

( c ) The 95% confidence interval for the population proportion p is : - ( 0.026 < p < 0.036 )