Question

Suppose a 95% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$34,321, $41,279]. The population standard deviation used for the analysis is known to be $14,200.

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. (Round "z" value to 3 decimal places and final answer to the nearest whole number.)

Answer 1

Solution :

Given that,

Lower confidence interval = 34321

Upper confidence interval = 41279

(a)

  = (Lower confidence interval + Upper confidence interval ) / 2

= (34321 + 41279) / 2

= 75600 / 2 = 37800

= 37800

The point estimate of the mean salary for all college graduates in this town is 37800 .

Margin of error = E = Upper confidence interval -   = 41279 - 37800 = 3479

Margin of error = E = 3479

(b)

Population standard deviation = = 14200

Margin of error = E =

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size = n = (Z_/2* / E) ²

n = (1.96 * 14200 / 3479)²

n = 64

sample size = 64