The latest available data showed health expenditures were $8086 per person in th

Question

The latest available data showed health expenditures were $8086 per person in the United States or 17.6% of Gross Domestic Product (Centers for Medicare & Medicaid Services website, April 1, 2012). Use $8086 as the population mean and suppose a survey research firm will take a sample of 100 people to investigate the nature of their health expenditures. Assume the population standard deviation is $2500. Show the standard deviation of the mean amount of health care expenditures for a sample of 100 people. Round your answer to nearest whole value. What is the probability the sample mean will be within Plus are mines $200 of the population mean? Round your answer to four decimal places. What is the probability the sample mean will be greater than $9000? Round your answer to four decimal places. If the survey research firm reports a sample mean greater than $9000, would you question whether the firm followed correct sampling procedures? Why or why not?

Explanation / Answer

a)

By central limit theorem, the standard error of the mean is

sigma(X) = sigma/sqrt(n) = 2500/sqrt(100) = $250. [ANSWER]

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B)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound = 8086-200=   7886      
x2 = upper bound = 8086+200=   8286      
u = mean =    8086      
n = sample size =    100      
s = standard deviation =    2500      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -0.8      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0.8      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.211855399      
P(z < z2) =    0.788144601      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.576289203   [ANSWER]  
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c)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    9000      
u = mean =    8086      
n = sample size =    100      
s = standard deviation =    2500      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    3.656      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   3.656   ) =    0.000128091 [ANSWER]

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d)

YES, because this is a very small probability.

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