Question

The 94 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 $1,871 for Individuals in this age group. Suppose this estimate is based on a sample of 80 persons and that the sample standard deviation is $500. Round your answers to the nearest whole numbers 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? $ 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 $1,871 ? Select your answer -

Answer 1

a)

Given
X bar	1871
S	590
n	80

α=	0.05
df	79	n-1

95% CI

CI
tc	1.99045	T.INV.2T(alpha,df)
ME	131.2981	tc*(S/SQRT(n))
Upper	2002.298	X bar + tc*(S/SQRT(n))
Lower	1739.702	X bar - tc*(S/SQRT(n))

a)

Margin of error (ME) = 131 (rounded)

b)

95% CI = (1740, 2002)

c)

Estimate of the total amount

Xbar * n = 1871*80 = $149680

d)

the amount spent on restaurants and carryout food is skewed to the right, the median amount spent to be less than $1871