Question

In a study of government financial aid for college students, researchers needed to estimate the proportion of full-time college students who earn a bachelor's degree in 4 years or less. Assuming a confidence level of 90%, find the sample size needed to estimate that proportion with a 0.03 margin of error in two cases: (1) no assumptions are made about the value of the sample proportion, and (2) prior studies have shown that roughly 60% of full-time students earn a bachelor's degree in four years or less. Round your answers upward to the next higher integer.

(1) If no assumptions are made about the sample proportion, the required sample size required to ensure a margin of error of 0.03 is n = .

(2) If it is assumed that roughly 60% of full-time students earn a bachelor's degree in four years or less, the required sample size required to ensure a margin of error of 0.03 is n = .

Answer 1

Solution,

Given that,

1) $\hat p$ =  1 - $\hat p$ = 0.5

margin of error = E = 0.03

At 90% confidence level

$\alpha$ = 1 - 90%  

$\alpha$ = 1 - 0.90 =0.10

$\alpha$/2 = 0.05

Z$\alpha$/2 = Z0.05  = 1.645

sample size = n = (Z$\alpha$ / 2 / E )2 * $\hat p$ * (1 - $\hat p$ )

= (1.645 / 0.03 )2 * 0.5 * 0.5

= 751.67

sample size = n = 752

2) $\hat p$ = 0.60

1 - $\hat p$ = 1 - 0.60 = 0.40

sample size = n = (Z$\alpha$ / 2 / E )2 * $\hat p$ * (1 - $\hat p$ )

= (1.645 / 0.03 )2 * 0.60 * 0.40

= 721.60

sample size = n = 722