Question

The 90 million Americans of age 50 and over control 50 percent of all discretionary income. AARP estimates that the average annual expenditure on restaurants and carryout food was $1,864 for individuals in this age group. Suppose this estimate is based on a sample of 80 persons and that the sample standard deviation is $600. Round your answers to the nearest whole numbers. a. At 99% confidence, what is the margin of error? b. What is the 99% 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? million 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 $1,864 ? We would expect the median to be less than the mean

Answer 1

a)

sample mean 'x̄=	1864.000
sample size   n=	80.00
sample std deviation s=	600.000
std error 'sx=s/√n=	67.082

for 99% CI; and 79 df, value of t=		2.6400
margin of error E=t*std error    =		177.10
lower bound=sample mean-E =		1686.90
Upper bound=sample mean+E =		2041.10

a)

margin of error=177

b)

99% CI =1687 , 2041

c)

total amount estimate =90*1864 = 167760

d)