Question

In a simple random sample of 120 Kansas seniors you found the mean out-of-pocket health care expenses for 2008 to be $4, 897. Assume σ=2,125.
i) How many people would you have to sample to estimate the mean out-of-pocket health care expenses in 2008 for all Kansas seniors to within $200 with 99% confidence? (5 points)

ii) Calculate and interpret a 98% confidence interval for the mean out-of-pocket health care expenses in 2008 for all Kansas seniors. (5 points)
iii) What is the margin of error for the interval calculated in part ii? (2 points)

Answer 1

a)

The following information is provided,
Significance Level, α = 0.01, Margin or Error, E = 200, σ = 2125

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 * 2125/200)^2
n = 751.45

Therefore, the sample size needed to satisfy the condition n >= 751.45 and it must be an integer number, we conclude that the minimum required sample size is n = 752
Ans : Sample size, n = 752 or 751

b)

sample mean, xbar = 4897
sample standard deviation, σ = 2125
sample size, n = 120

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, Zc = Z(α/2) = 2.33

ME = zc * σ/sqrt(n)
ME = 2.33 * 2125/sqrt(120)
ME = 451.99

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (4897 - 2.33 * 2125/sqrt(120) , 4897 + 2.33 * 2125/sqrt(120))
CI = (4445.01 , 5348.99)

c)

ME = zc * σ/sqrt(n)
ME = 2.33 * 2125/sqrt(120)
ME = 451.99