Question

“A sample of 60 individuals in the Midwest showed the sample average per person annual expenditure for prescription drugs of $745. Suppose the population standard deviation expenditure for prescription drugs is $300.” a. Construct a 95% confidence interval estimate for the population mean annual expenditure for prescription drugs in the Midwest. b. Is $838 in the interval from part a? What would this imply about the population mean being $838? c. Without calculating: what would happen to the width of the confidence interval estimate if the confidence level changed to 99%? Explain. d. Determine the minimum sample size to ensure that a 99% confidence interval has a margin of error of $50. Assume the population standard deviation is $300.

Answer 1

Since , the population standard deviation is known.

Therefore , use normal distribution.

a) Now , ; From Z-table

Therefore , the 95% confidnece interval estimate for the population mean annual expenditure for the prescription drugs in the Midwest is ,

b) Here , the value $838 is not in the interval ( $669.0895 , $820.9105)

c) Now , the 99% confidence interval is ($645.0770,844.9230$)

The width of the 95% confidence interval is ,

Width=Upper confidence limit-Lower confidence limit=820.9105-669.0895=$151.0155

The width of the 99% confidence interval is ,

Width=Upper confidence limit-Lower confidence limit=844.9230-645.0770=$199.8460

Therefore , As the confidence level increases , the width of the confidence interval is also increases.

d) Since , Margin of error=E=$50

Population standard deviation=$300

Significance level=0.01

; From Z-table

Therefore , the required sample size is ,