Question

Is college worth it? Among a simple random sample of 344 American adults who do not have a four-year college degree and are not currently enrolled in school, 166 said they decided not to go to college because they could not afford school.

1. Calculate a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it, and interpret the interval in context. Round to 4 decimal places.

2. Suppose we wanted the margin of error for the 90% confidence level to be about 2.25%. What is the smallest sample size we could take to achieve this? Note: For consistency's sake, round your z* value to 3 decimal places before calculating the necessary sample size.

choose n=

Answer 1

Solution :

Given that,

n = 344

x = 166

Point estimate = sample proportion = = x / n = 166 / 344 = 0.4826

1 - = 1 - 0.4826 = 0.5174

1) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_0.05 = 1.645

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

= 1.645 (((0.4826 * 0.5174) / 344)

= 0.0443

A 90% confidence interval for population proportion p is ,

± E

= 0.4826  ± 0.0443

= ( 0.4383, 0.5269 )

2) Margin of error = E = 2.25% = 0.0225

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645 / 0.0225)² * 0.4826 * 0.5174

= 1334.69

sample size = n = 1335