Solution to Problem 4

Suppose you want to construct a 90% confidence interval for the proportion of college students that want to work in the health care profession after graduation.  If you want a margin of error of no more than 5%, how many college students must you survey?

Solution

Since there is no preliminary estimate for the population proportion, we use the formula:

n = 1/4 * (z/E)^2

We have:

E  =  0.05

c  =  0.90

zc = 1.645

Plug these numbers into the formula to get:

        n = 1/4 * (1.645/0.05)^2 = 270.6025 

We always round up, so we can conclude that we must survey 271 college students.

Back to the handout

 

This site has had Hit Counter visitors since February 26, 2001