Question

A random sample of 500 nursing school applications were selected and it was noted if the applicant
was a male
or female. The results are noted below:
Female 432
Male 68
a.    Using 99% confidence, compute the margin of error for the proportion of men applying to the
nursing
program.

b.    Construct a 99% confidence interval for the proportion of all nursing school applicants that
are men.

c.    Write a statement explaining your confidence intervals.

Answer 1

Solution:

Given:

n = 500

x = number of men = 68

Part a) Using 99% confidence, compute the margin of error for the proportion of men applying to the nursing program.

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

Part b) Construct a 99% confidence interval for the proportion of all nursing school applicants that are men.

Part c) Write a statement explaining your confidence intervals.

We are 99% confident that the true population proportion of all nursing school applicants that are men is between 0.0965 and 0.1755.