Question

3. “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, 838 is not in the interval it means we can reject the null hypothesis

c)

the width of the confidenc einterval would be wider because as the confidenc elevel increases the width of 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