Question

(3 points) Is college worth it? Among a simple random sample of 304 American adults who do not have a four-year college degree and are not currently enrolled in school, 137 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

(1)

n = 304

p = 137/304 = 0.4507

q = 1 - p = 0.5493

SE =

Vpq/n-v/0.4507 x 0.5493/3040.0285 0.4507 x0.5493/304-0.0285

$\alpha$ = 0.10

From Table critical values of Z = 1.645

Confidence interval:

0.4507 $\pm$ (1.645 X 0.0285)

= 0.4507 $\pm$ 0.0469

= ( 0.4038 ,0.4976)

Confidence Interval:

0.4038 < P < 0.4976

(b)

Smallest sample size (n) is given by:

nーで/2p(1-p)

Given:

$\alpha$ = 0.10

From Table, critical values of Z^* = $\pm$ 1.645

p = 0.4507

e = 0.0225

Substituting, we get:

1.6452 × 0.4507 × (1-0.4507) 0.02252 rl = 1324

So,

Answer is:

1324