Question

Americans have become increasingly concerned about the rising cost of Medicare. In 1990, the average annual Medicare spending per enrollee was $3267; in 2003, the average annual Medicare spending per enrollee was $6883 (Money, Fall 2003). Suppose you hired a consulting firm to take a sample of fifty 2003 Medicare enrollees to further investigate the nature of expenditures. Assume the population standard deviation for 2003 was $2100.

When calculating values for z, round to two decimal places.

1. Calculate the standard error of the mean amount of Medicare spending for a sample of fifty 2003 enrollees (to 2 decimals).
   
   
2. What is the probability that the sample mean will be within +/- $300 of the population mean (to 4 decimals)?
   
   
3. What is the probability that the sample mean will be greater than $7500 (to 4 decimals)?
   
   
   If the consulting firm tells you the sample mean for the Medicare enrollees it interviewed was $7500, would you question whether the firm followed correct simple random sampling procedures?
   - Select your answer -Yes, because the probability of attaining that sample mean is very highYes, because the probability of attaining that sample mean is very lowNo, because the probability of attaining that sample mean is very highNo, because the probability of attaining that sample mean is very low

Answer 1

(A) standard error = (standard deviation/sqrt(n))

where standard deviation = 2100 and sample size n = 50

so, SE = 2100/sqrt(50)

= 296.98

(B) Using normalcdf

setting the following values

lower = 6883-300 = 6583

upper = 6883+300 = 7183

mean = 6883

sigma = 296.98

P(6583<X<7183) = normalcdf(6583,7183,6883,296.98)

= 0.6876

(C)

Using normalcdf

setting the following values

lower = 7500

upper = 99999

mean = 6883

sigma = 296.98

P(X>7500) = normalcdf(7500,99999,6883,296.98)

= 0.0189

We can see that the probability is less than 0.05, so we can  question whether the firm followed correct simple random sampling procedures because the probability is unusual

Yes, because the probability of attaining that sample mean is very low