Question

A class survey in a large class for first-year college students asked, "About how many hours do you study in a typical week?". The mean response of the 451 students was x¯¯¯x¯ = 15 hours. Suppose that we know that the study time follows a Normal distribution with standard deviation 9 hours in the population of all first-year students at this university.

What is the 99% confidence interval (±±0.001) for the population mean?

Confidence interval is from_____ to ____ hours.

Answer 1

Sample size = n = 451

Sample mean = $\bar{x}$ = 15

Population standard deviation = $\sigma$ = 9

a) We have to construct 99% confidenc interval for the population mean.

Here population standard deviation is known so we have to use one sample z-confidence interval.

z confidence interval

$(\bar{x}-E , \bar{x}+E)$

Here E is a margin of error

$E = \frac{Zc*\sigma}{\sqrt{n}}$

Zc = 2.58 ( Using z table)

$E = \frac{2.58*9}{\sqrt{451}}=1.093387$

So confidence interval is ( 15 - 1.093387 , 15 + 1.093387) = > (13.9066 , 16.0934)