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
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.
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