Question

4. The department of HR at the University of Michigan wants to estimate the amount of an annual healthcare premium needed for an Assistant Professor as part of a new recruiting program. In a sample of 50 Assistant Professors, they found that the average yearly premium needed is $16,000 with a standard deviation of $3,500. (a) What is the population mean? What is the best estimate of the population mean? (b) Develop an 90% confidence interval for the population mean. (c) Assuming a 99% level of confidence, how large of a sample is required with an acceptable error of $400?

Answer 1

a) Here we dont know population mean we use sample mean and confidence interval to estimante range of population mean, best estimate of population mean is 16000.

b) x?= 16000

sigma= 3500

n= 50

alpha=0.1 then Z(alpha/2)= 1.645

Margin of error E=Z(alpha/2)*sigma/sqrt(n)

=1.645*3500/sqrt(50)

=814.233

90% Confidence interval for population mean =(x?-E,x?+E)

lower bound= 15185.76654

upper bound= 16814.23346

c)

sigma= 3500

MOE= 400

Zalpha/2= 2.576

we know the relation,

E=Z(alpha/2)*sigma/sqrt(n)

400=2.576*3500/sqrt(n)

n= 508.0516

n= 509

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