# statistics

The 92 million Americans of age 50 and over control 50% of all discretionary income (AARP Bulletin, March 2008). AARP estimated that the average annual expenditure on restaurants and carry out food was \$1873 for individuals in this age group. Suppose this estimate is based on a sample of 80 persons and that the sample standard deviation is \$550.

a. At 95% confidence, what is the margin of error?
b. What is the 95% confidence interval for the population mean amount spent on restaurants and carryout food?
c. What is your estimate of the total amount spent by Americans of age 50 and over on restaurants and carryout food?
d. If the amount spent on restaurants and carryout food is skewed to the right, would you expect the median amount spent to be greater or less than \$1873?

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posted by BUSSTAT

For a):
The margin of error = 1.96(sd/√n)
sd = 550
n = 80
1.96 represents 95% using a z-table

For b):
CI95 = mean + or - margin of error
mean = 1873

Substitute the values into the formula and calculate.

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