Question

  

Suppose a 99% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$39,986, $48,414]. The population standard deviation used for the analysis is known to be $14,700.

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

Given that , 99% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$39,986, $48,414]

i.e. Confidence level = C = 0.99

Population SD =

Lower limit of C.I  .............(Equation 1 )

Upper limit of C.I     ...........(Equation 2)

Where E = Margin of error

A ) we want to calculate, point estimate of the mean salary for all college graduates in this town

Note that , sample mean is point estimate of population mean

sample mean is calculate as

Add equation 1and equation 2

we get ,

we get sample mean

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

B) sample size ( n ) can be calculate as

Where , Zc = Critical value for C = 2.58 ..........( From Z table)

E = Margin of error

Margin of error ( E ) can be calculated as

Substract equation 2 - equation 1

we get

Therefor ,

we get ,

   ......( ANSWER)

Required sample size = n = 81 ......( ANSWER)