Question

From a random sample of 500 Americans, it was discovered that 300 of them have some level of college education. (Source: The Environmental Protection Agency)

*What is the sample proportion, p^, of Americans who have some level of college education.

*Construct and interpret a 90% confidence interval for the proportion of Americans who have some level of college education.

* Construct and interpret a 99% confidence for the proportion of Americans who have some level of college education.

Answer 1

Solution :

Given that,

n = 500

x = 300

a) Point estimate = sample proportion = = x / n = 300 / 500 = 0.60

1 - = 1 - 0.60 = 0.40

b) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 (((0.60 * 0.40) / 500 )

= 0.036

A 90% confidence interval for population proportion p is ,

± E   

= 0.60 ± 0.036

= ( 0.564, 0.636 )

We are 90% confident that the true proportion of Americans who have some level of college education between 0.564 and 0.636.

c) At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576 (((0.60 * 0.40) / 500 )

= 0.056

A 99% confidence interval for population proportion p is ,

± E   

= 0.60 ± 0.056

= ( 0.544, 0.656 )

We are 99% confident that the true proportion of Americans who have some level of college education between 0.544 and 0.656.