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, 155 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. ( 0.4383 , 0.5269 )

2. Suppose we wanted the margin of error for the 90% confidence level to be about 3.5%. 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 = 1335

Answer 1

Solution :

Given that,

n = 344

x = 155

Point estimate = sample proportion = = x / n = 155 / 344 = 0.4505

1 - = 1 - 0.4505=5494

1) 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 (((0.4505 * (1-0.4505) / 344)

= 0.0441

A 90% confidence interval for population proportion p is ,

± E

= 0.4505 ± 0.0441

= ( 0.4065, 0.4974 )

Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is ( 0.4065, 0.4974)

That means, we are 90% confident that the proportion of Americans who decide to not to go to college because they cannot afford it is between 0.4065 and 0.4974

2) margin of error ( E ) = 3.5% = 0.035

we want to find, the sample size ( n ),

Case 1) if we use above proportion p = 0.4505 then required sample size is,

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

= (1.645 / 0.035)2 * 0.4505 * (1-0.4505)

=546.8552

sample size = n = 547

Case 2) if we don't use above proportion then we assume p = 0.5 then required sample size is,

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

= (1.645 / 0.035)2 * 0.5 * (1-0.5)

=552.25

sample size = n = 552