Question

4. In a sample of 200 students who plan to apply to Ivy League universities, the average annual family income was $143,325 with a standard deviation of $25,418. Compute a 95% confidence interval for the mean income of families with a student who plans to apply to an Ivy League university. For each problem, a) attach StatCrunch output, b) write a sentence of interpretation, and c) state the interval in margin of error form.

Answer #1

for 95% confidence , t = 1.96 as n >> 30

(143325 - 1.96 * 25418/sqrt(200), 143325 + 1.96 * 25418/sqrt(200))

= ( 139802.2449 , 146847.7551)

b)

we are 95% confident that the true average annual income lies in this confidence interval

c)

margin of error = 1.96* 25418/sqrt(200) = 3522.7551

(143325 3522.7551)

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