Question

In a study of government financial aid for college​ students, it becomes necessary to estimate the percentage of​ full-time college students who earn a​ bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.02 margin of error and use a confidence level of 99​%. Complete parts​ (a) through​ (c) below.
a. Assume that nothing is known about the percentage to be estimated.n = ________b. Assume prior studies have shown that about 55% of​ full-time students earn​ bachelor's degrees in four years or less.n = _______c. Does the added knowledge in part​ (b) have much of an effect on the sample​ size?

Answer 1

Answer:

(a) The sample size required is 2401.

(b) The sample size required is 2377.

(c) Yes, on increasing the proportion value the sample size decreased.

Step-by-step explanation:

The confidence interval for population proportion p is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hatp(1-\hat p)}{n}}$

The margin of error in this interval is:

$MOE=z_{\alpha/2}\sqrt{\frac{\hatp(1-\hat p)}{n}}$

The information provided is:

MOE = 0.02

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

(a)

Assume that the proportion value is 0.50.

Compute the value of n as follows:

$MOE=z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\n=\frac{1.96^{2}\times0.50(1-0.50)}{0.02^{2}}\\=2401$

Thus, the sample size required is 2401.

(b)

Given that the proportion value is 0.55.

Compute the value of n as follows:

$MOE=z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.55(1-0.55)}{n}}\\n=\frac{1.96^{2}\times0.55(1-0.55)}{0.02^{2}}\\=2376.99\\\approx2377$

Thus, the sample size required is 2377.

(c)

On increasing the proportion value the sample size decreased.