Question

A random sample of 200 physicians indicates that 36 of them make at least $400,000 a year. Construct a 99% confidence interval estimate of the population proportion of physicians who make at least $400,000 a year. State the conclusion.

Answer 1

Solution:
Given:

Sample size = n = 200

x = number of physicians make at least $400,000 a year = 36

We have construct a 99% confidence interval estimate of the population proportion of physicians who make at least $400,000 a year.

Formula:

where

and

Z_c is z critical value for c = 0.99 confidence level.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area and find corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and 0.9951 and both are at same distance from 0.9950, Hence corresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Z_c = 2.575

Thus

Thus

Conclusion:
We are 99% confident that the true population proportion of physicians who make at least $400,000 a year is between (0.11 , 0.25 )