A class survey in a large class for firstyear college studen

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 440 students was x = 15 hours. Suppose that we know that the study time follows a Normal distribution with standard deviation 8 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.

Solution

Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=15
Standard deviation( sd )=8
Sample Size(n)=440
Confidence Interval = [ 15 ± Z a/2 ( 8/ Sqrt ( 440) ) ]
= [ 15 - 2.58 * (0.381) , 15 + 2.58 * (0.381) ]
= [ 14.016,15.984 ]