SOLUTION: The mean preparation fee H&R Block charged retail customers last year was $183 (the Wall Street Journal, March 7, 2012). Use this price as the population mean and assume the popula

Question 1202706: The mean preparation fee H&R Block charged retail customers last year was $183 (the Wall Street Journal, March 7, 2012). Use this price as the population mean and assume the population standard deviation of preparation fees is $50.
a. What is the probability that the mean price for a sample of 30 H&R Block retail customers is within $8 of the population mean?
b. What is the probability that the mean price for a sample of 50 H&R
Block retail customers is within $8 of the population mean?
c. What is the probability that the mean price for a sample of 100 H&R Block retail customers is within $8 of the population mean?
d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a .95 probability that the sample mean is within
$8 of the population mean?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
answer to your questions:

a. What is the probability that the mean price for a sample of 30 H&R Block retail customers is within $8 of the population mean?

that probability is equal to 0.61916352

b. What is the probability that the mean price for a sample of 50 H&R
Block retail customers is within $8 of the population mean?

that probability is equal to 0.742100965

c. What is the probability that the mean price for a sample of 100 H&R Block retail customers is within $8 of the population mean?

that probability is equal to 0.890401417

d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a .95 probability that the sample mean is within
$8 of the population mean?

none of them.
to have at least .95 probability that the sample mean is within 8 dollars of the population mean, the sample size would have to be 151 or greater.

the excel spreadsheet shown below contains the calculations necessary for resolutuion of this problem.

graphical confirmation of these results are shown below:

to calculate the minimum sample size required so that the probability of getting a score within 8 dollars of the mean, you would do the following.

critical z-score at 95% two tail confidence interval is equal to plus or minus z = 1.959963985.
standard error is equal to standard deviation divided by square root of sample size.
when standard deviation is 50, this becomes:
standard error is equal to 50 / square root of sample size.

z-score formula is:
z = (x-m)/s
z is the z-score
(x-m) is the margin of error
s is the standared error, which is equal to 50/sqrt(n), where n is the sample size.
the formula becomes z = 8/s
when s = 50/sqrt(n), the frmula becomes z = 8/(50/sqrt(n))
this is simplified to z = 8/50 * sqrt(n)
solve for sqrt(n) to get sqrt(n) = z*50/8
when z = 1.959963985, the formula becomes:
sqrt(n) = 1.959963985 * 50 / 8 = 12.24977491
n is equal to sqrt(n)^2 = 150.0569853.
n has to be an integers, so round up to the next integer to get a sample size of at least 151 so that the probasbility of getting a margin of error of less than 8 is greater than 95%.

let me know if you have any questions.
theo

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