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

Answer)

As the population standard deviation is known here we can use standard normal z table to estimate the answers

A)

Critical value z from z table for 95% confidence level is 1.96

Margin of error (MOE) = Z*S.D/√N

MOE = 1.96*300/√60 = 75.9104735856

Interval is given by

(Mean - MOE, Mean + MOE)

(669.09, 820.91)

B)

Null hypothesis Ho : u = 838

Alternate hypothesis Ha : u not equal to 838

As the interval does not contain the null hypothesised value 838

We reject the null hypothesis Ho

So we do not have enough evidence to conclude that mean is 838

C)

As the confidence level increases width of the interval also increase

As then we have more confidence in our interval

And moreover width of the interval depends on the margin of error

Greater the error greater the width

And error is directly proportional to the confidence level.

D)

Critical value z from z table for 99% confidence level is 2.58

50 = 2.58*300/√n

N = 240