Question

Is college worth it? Among a simple random sample of 324 American adults who do not have a four-year college degree and are not currently enrolled in school, 143 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 3.25%. 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 = 324

x = 143

Point estimate = sample proportion = = x / n = 143 / 324 =0.4414

1 - = 1 - 0.4414 = 0.5586

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 * ((( 0.4414 * 0.5586) / 324 )

= 0.0454

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.4414 - 0.0454 < p < 0.4414 + 0.0454

0.3960 < p < 0.4868

( 0.3960 , 0.4868 )

The 90% confidence interval for the population proportion p is : - ( 0.3960 , 0.4868 )

( 2 )

Given that,

= 143 / 324 = 0.441

1 - = 1 - 0.441 = 0.559

margin of error = E = 3.25% = 0.0325

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 = 0.10

/2 = 0.05

Z/2 = Z0.05  = 1.645

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

= ( 1.645 / 0.0325 )2 * 0.441 * 0.559

= 631.56

Sample size = n = 632