Question

A random sample of 400 Michigan State University (MSU) students were surveyed recently to determine an estimate for the proportion of all MSU students who had attended at least three football games. The estimate revealed that between .372 and .458 of all MSU students attended. Given this information, we can determine that the confidence coefficient was approximately: a. .92 b. .95 c. .88 d. .90 e. .99

Answer 1

Answer:

a. .92

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 of the interval is:

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

The lower bound is the point estimate $\pi$ subtracted by the margin of error.

The upper bound is the point estimate $\pi$ added to the margin of error.

Point estimate:

The confidence interval is symmetric, so it is the mean between the two bounds.

In this problem:

$\pi = \frac{0.372 + 0.458}{2} = 0.415$

Sample of 400, which means that $n = 400$

Margin of error is the estimate subtracted by the lower bound. So $M = 0.415 - 0.372 = 0.043$

We have to find z.

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

$0.043 = z\sqrt{\frac{0.415*0.585}{400}}$

$z = \frac{0.043\sqrt{400}}{\sqrt{0.415*0.585}}$

$z = 1.745$

$z = 1.745$ has a pvalue of 0.96.

This means that:

$1 - \frac{\alpha}{2} = 0.96$

$\frac{\alpha}{2} = 1 - 0.96$

$\frac{\alpha}{2} = 0.04$

$\alpha = 0.08$

Confidence level:

$1 - \alpha = 1 - 0.08 = 0.92$

So the correct answer is:

a. .92