The 92 million Americans of age 50 and over control 50% of all discretionary income. AARP estimates that the average annual expenditure on restaurants and carryout food was $1,873 for individuals in the age group. Suppose this estimate is based on a sample of 80 persons and that the sample standard deviation is $550.At 95% confidence, what is the margin of error?What is the 95% confidence interval for the population mean amount spent on restaurants and carryout food?What is your estimate of the total amount spent by Americans of age 50 and over on restaurants and carryout food?If the amount spent on restaurants and carryout food is skewed to the right, would you expect the median amount spent to be greater than or less than$1,873?

Name: The 92 million Americans of age 50 and over control 50% of all discret
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Description: The 92 million Americans of age 50 and over control 50% of all discretionary income. AARP estimates that the average annual expenditure on restaurants and carryout food was $1,873 for individuals in the age group. Suppose this estimate is based on a

Given Mean=$1873 Standard deviation=$550 n=80

Answered: The 92 million Americans of age 50 and…

MATLAB: An Introduction with Applications

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Author:Amos Gilat

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The 92 million Americans of age 50 and over control 50% of all discretionary income. AARP estimates that the average annual expenditure on restaurants and carryout food was $1,873 for individuals in the age group. Suppose this estimate is based on a sample of 80 persons and that the sample standard deviation is $550.
1. At 95% confidence, what is the margin of error?
2. What is the 95% confidence interval for the population mean amount spent on restaurants and carryout food?
3. What is your estimate of the total amount spent by Americans of age 50 and over on restaurants and carryout food?
4. If the amount spent on restaurants and carryout food is skewed to the right, would you expect the median amount spent to be greater than or less than$1,873?

Definition Definition Middle value of a data set. The median divides a data set into two halves, and it also called the 50th percentile. The median is much less affected by outliers and skewed data than the mean. If the number of elements in a dataset is odd, then the middlemost element of the data arranged in ascending or descending order is the median. If the number of elements in the dataset is even, the average of the two central elements of the arranged data is the median of the set. For example, if a dataset has five items—12, 13, 21, 27, 31—the median is the third item in ascending order, or 21. If a dataset has six items—12, 13, 21, 27, 31, 33—the median is the average of the third (21) and fourth (27) items. It is calculated as follows: (21 + 27) / 2 = 24.

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