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Lady bird [3.3K]
3 years ago
9

In a study of government financial aid for college​ students, it becomes necessary to estimate the percentage of​ full-time coll

ege students who earn a​ bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.04 margin of error and use a confidence level of 99​%. Complete parts​ below.
​(a) Assume that nothing is known about the percentage to be estimated.
(b) Assume prior studies have shown that about 55​% of​ full-time students earn​ bachelor's degrees in four years or less.
Mathematics
1 answer:
ICE Princess25 [194]3 years ago
6 0

Answer:

a) n = 1037.

b) n = 1026.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

​(a) Assume that nothing is known about the percentage to be estimated.

We need to find n when M = 0.04.

We dont know the percentage to be estimated, so we use \pi = 0.5, which is when we are going to need the largest sample size.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 2.575\sqrt{\frac{0.5*0.5}{n}}

0.04\sqrt{n} = 2.575*0.5

(\sqrt{n}) = \frac{2.575*0.5}{0.04}

(\sqrt{n})^{2} = (\frac{2.575*0.5}{0.04})^{2}

n = 1036.03

Rounding up

n = 1037.

(b) Assume prior studies have shown that about 55​% of​ full-time students earn​ bachelor's degrees in four years or less.

\pi = 0.55

So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 2.575\sqrt{\frac{0.55*0.45}{n}}

0.04\sqrt{n} = 2.575*\sqrt{0.55*0.45}

(\sqrt{n}) = \frac{2.575*\sqrt{0.55*0.45}}{0.04}

(\sqrt{n})^{2} = (\frac{2.575*\sqrt{0.55*0.45}}{0.04})^{2}

n = 1025.7

Rounding up

n = 1026.

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