Question

In the survey "How America Pays for College" the average annual amount the n=1643 families paid for college was $20977. (a) If the population standard deviation is $7500, give the 90% confidence interval (±0.01) for μ, the average amount a family pays for a college undergraduate The interval is from $ to $ . (b) Would the margin of error for 99% confidence be larger or smaller? A 99% confidence interval would have a larger margin of error A 99% confidence interval would have a same margin of error A 99% confidence interval would have a smaller margin of error

Answer 1

a)

90% confidence interval for is

- Z/2 * / sqrt(n) < < + Z/2 * / sqrt(n)

20977 - 1.645 * 7500 / sqrt(1643) < < 20977 + 1.645 * 7500 / sqrt(1643)

20672.63 < < 21281.37

90% CI is (20672.63, 21281.37)

b)

The 99% confidence interval considers more accuracy than 90% confidence interval, so margin of error will

also be larger for 99% confidence than 90% confidence.

A 99% confidence interval would have a larger margin of error.