Question

In: Statistics and Probability

“A sample of 60 individuals in the Midwest showed the sample average per person annual expenditure...

“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.

Solutions

Expert Solution

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 ,


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