Question

Suppose a 90% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$38,737, $50,463]. The population standard deviation used for the analysis is known to be $14,300.
a. What is the point estimate of the mean salary for all college graduates in this town?
Point estimate
b. Determine the sample size used for the analysis.
Sample size

Answer 1

Answer:

a. The point estimate was of $44,600.

b. The sample size was of 16.

Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

The margin of error is the difference between the two bounds, divided by 2.

a. What is the point estimate of the mean salary for all college graduates in this town?

Mean of the bounds, so:

(38737+50463)/2 = 44600.

The point estimate was of $44,600.

b. Determine the sample size used for the analysis.

First we need to find the margin of error, so:

$M = \frac{50463-38737}{2} = 5863$

Relating the margin of error with the sample size:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$.

That is z with a pvalue of $1 - 0.05 = 0.95$, so Z = 1.64.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

For this problem, we have that $\sigma = 14300, M = 5863$. So

$M = z\frac{\sigma}{\sqrt{n}}$

$5863 = 1.645\frac{14300}{\sqrt{n}}$

$5863\sqrt{n} = 1.645*14300$

$\sqrt{n} = \frac{1.645*14300}{5863}$

$(\sqrt{n})^2 = (\frac{1.645*14300}{5863})^2$

$n = 16$

The sample size was of 16.