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

a)

sample mean, xbar = 745
sample standard deviation, σ = 300
sample size, n = 60

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 300/sqrt(60)
ME = 75.91

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (745 - 1.96 * 300/sqrt(60) , 745 + 1.96 * 300/sqrt(60))
CI = (669.09 , 820.91)

b)

No, it is not in the interval so, we reject the null hypothesis

c)

The width of the interval increases

d)

The following information is provided,
Significance Level, α = 0.01, Margin or Error, E = 50, σ = 300

The critical value for significance level, α = 0.01 is 2.58.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (2.58 * 300/50)^2
n = 239.63

Therefore, the sample size needed to satisfy the condition n >= 239.63 and it must be an integer number, we conclude that the minimum required sample size is n = 240
Ans : Sample size, n = 240