Suppose a 95% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$36,080, $43,920]. The population standard deviation used for the analysis is known to be $12,000.

a. What is the point estimator of the mean salary of all college graduates in this town?
b. Determine the sample size used for the analysis.

a. Well, the point estimator for the mean salary of all college graduates in the town would be the midpoint of the confidence interval. So, let me do the math here... *calculating sounds*... Ah! The point estimator is $40,000. Ta-da!

b. Now, to determine the sample size used for the analysis, we need to take a closer look at the confidence interval. The formula for the margin of error in a confidence interval is given by:

Margin of Error = (z-score) * (standard deviation / square root of sample size)

Since we're dealing with a 95% confidence interval, the z-score corresponding to this level of confidence is approximately 1.96. We also know that the standard deviation used for the analysis is $12,000. By rearranging the formula, we can solve for the sample size:

Sample Size = (z-score)^2 * (standard deviation)^2 / (margin of error)^2

Substituting in the given values, we have:

Sample Size = (1.96)^2 * ($12,000)^2 / (($43,920 - $36,080) / 2)^2

Solving this equation... *more calculating sounds*... the sample size comes out to be approximately 116.

So, the sample size used for the analysis is around 116 college graduates. Hope that clarifies things!

a. The point estimator of the mean salary of all college graduates in this town is the midpoint of the confidence interval. So, the point estimator is calculated as the average of the upper and lower bound of the confidence interval:

Point Estimator = (Lower Bound + Upper Bound) / 2
= ($36,080 + $43,920) / 2
= $40,000

Therefore, the point estimator of the mean salary is $40,000.

b. To determine the sample size used for the analysis, we can make use of the formula for the margin of error in a confidence interval:

Margin of Error = Z * (Standard Deviation / √n)

Here, Z represents the z-score corresponding to the desired confidence level (in this case, 95%), and n represents the sample size.

For a 95% confidence level, the z-score is 1.96. Given that the standard deviation of the population is $12,000 and the margin of error is the half-width of the confidence interval, we can substitute these values into the formula to solve for the sample size:

Half-Width = Z * (Standard Deviation / √n)
(Upper Bound - Lower Bound) / 2 = 1.96 * ($12,000 / √n)
($43,920 - $36,080) / 2 = 1.96 * ($12,000 / √n)
$3,920 = 1.96 * ($12,000 / √n)

Now we can solve for n:

√n = (1.96 * $12,000) / $3,920
√n = 6
n = 36

Therefore, the sample size used for the analysis is 36.

a. To find the point estimator of the mean salary, we use the midpoint of the confidence interval, which is the average of the upper and lower bounds. In this case, the confidence interval is [$36,080, $43,920]. So, the point estimator is:

Point estimator of the mean salary = (Lower bound + Upper bound) / 2

Point estimator of the mean salary = ($36,080 + $43,920) / 2

Point estimator of the mean salary = $39,000

Therefore, the point estimator of the mean salary of all college graduates in this town is $39,000.

b. The sample size used for the analysis can be determined using the formula for the confidence interval:

Sample size = (Z-value * population standard deviation) / (margin of error)

Given that the confidence level is 95%, the corresponding Z-value is approximately 1.96. The population standard deviation is $12,000.

Using the formula:

Sample size = (1.96 * $12,000) / (Half the width of the confidence interval)

The half the width of the confidence interval can be calculated by taking the difference between the upper and lower bounds and dividing it by 2:

Half the width of the confidence interval = (Upper bound - Lower bound) / 2

Half the width of the confidence interval = ($43,920 - $36,080) / 2

Half the width of the confidence interval = $3,920

Substituting the values into the formula:

Sample size = (1.96 * $12,000) / $3,920

Sample size ≈ 5,997

Therefore, the sample size used for the analysis is approximately 5,997.