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March 31, 2015

March 31, 2015

Posted by **Anonymous** on Wednesday, July 20, 2011 at 4:00pm.

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?

- statistics -
**BUSSTAT**, Wednesday, July 20, 2011 at 4:01pmPlease help and explain the best you can. I am so lost.

Thank you in advance

- statistics -
**MathGuru**, Thursday, July 21, 2011 at 7:16pmI'll help you with the first two:

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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