Question

**Ages of students:** A simple random sample of 100
U.S. college students had a mean age of 22.68 years. Assume the
population standard deviation is *σ* = 4.74 years. Construct
a 99% confidence interval for the mean age of U.S. college
students.

The answer I posted my instructor says it is wrong. I came up with 21.45 to 23.91 a 99% confidence interval for the mean of the U.S college students

She said the z procedures are needed.

Thanks.

Answer #1

Solution :

Given that,

Sample size = n = 100

Z_{/2}
= 2.576

Margin of error = E = Z_{/2}*
(
/n)

= 2.756 * (4.74 / 100)

= 1.22

At 99% confidence interval estimate of the population mean is,

- E < < + E

22.68 - 1.22 < < 22.68 + 1.22

21.46 < < 23.90

A 99% confidence interval for the mean : **(21.46 ,
23.90)**

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