In: Math
Suppose a 90% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$45,783, $57,017]. The population standard deviation used for the analysis is known to be $13,700. [You may find it useful to reference the z table.]
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.)
Solution :
Given that,
A ) Lower confidence interval = $45,783
Upper confidence interval = $57,017
= (Lower confidence interval + Upper confidence interval ) / 2
= (45,783 + 57,017) / 2
= 102800 / 2
= 51400
= 51400
Margin of error = ME = Upper confidence interval -
= 57,017 - 51400
= 5617
standard deviation = = $13,700
B ) margin of error = E = $ 5617
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
Z/2 = Z0.05 = 1.645
Sample size = n = ((Z/2 * ) / E)2
= ((1.645 * 13700) / 5617)2
= 16.088
Sample size = 16