Suppose a 99% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$34,393, $47,207]. The population standard deviation used for the analysis is known to be $14,900. Required:
a. What is the point estimate of the mean salary for all college graduates in this town?
b. Determine the sample size used for the analysis.

a. $40,800 b. 36

Step-by-step explanation:

Given : a 99% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$34,393, $47,207].

a. Since Point estimate of of the mean = Average of upper limit and lower limit of the interval.

Therefore , the point estimate of the mean salary for all college graduates in this town =

= 40,800

hence, the point estimate of the mean salary for all college graduates in this town = $40,800

b.  Since  lower limit = Point estimate - margin of error, where Margin of error is the half of the difference between upper limit and lower limit.

Margin of error

Also, margin of error = , where z= critical z-value for confidence level and n is the sample size.

z-value for 99% confidence level  = 2.576

So,

The sample size used for the analysis =36

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step-by-step explanation: