The 92 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 $1,873 for individuals in this age group. Suppose this estimate is based on a sample of 70 persons and that the sample standard deviation is $550.

a.) At 95% confidence, what is the margin of error in dollars? (Round your answer to the nearest dollar.) $_______

b)What is the 95% confidence interval for the population mean amount spent in dollars on restaurants and carryout food? (Round your answers to the nearest dollar.)$_________ to $__________

c) What is your estimate of the total amount spent in millions of dollars by Americans of age 50 and over on restaurants and carryout food? (Round your answer to the nearest million dollars.)$____________ million

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 $1,873?

a. We would expect the median to be greater than the mean of $1,873. The few individuals that spend much less than the average cause the mean to be smaller than the median.

b. We would expect the median to be less than the mean of $1,873. The few individuals that spend much less than the average cause the mean to be larger than the median.

c. We would expect the median to be less than the mean of $1,873. The few individuals that spend much more than the average cause the mean to be larger than the median.

d. We would expect the median to be greater than the mean of $1,873. The few individuals that spend much more than the average cause the mean to be smaller than the median.