Question

Question 16 of 18 (1 point) Attempt 1 of Unlimited 74 Section Exercise 12 (cal Ages of students: A simple random sample of 110 U.S. college students had a mean age of 23.02 years. Assume the population standard deviation is o = 4.77 years. Construct a 98% confidence interval for the mean age of U.S. college students. Round the answers to two decimal places. A 98% confidence interval for the mean age of U.S. college students is

Answer 1

As population standard deviation is known, we will use z distribution to find CI.

z value for 98% CI is 2.33 as P(-2.33<z<2.33)=0.98

So Margin of Error is $E=z*\frac{\sigma }{\sqrt{n}}=2.33*\frac{4.77}{\sqrt{110}}=1.06$

Hence CI is $CI=\overline{x}\pm E=23.02 \pm 1.06=(21.96,24.08)$

Hence answer here is $21.96<\mu<24.08$