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 minutes do you study on a typical weeknight?” The mean response of the 463 students was  = 118 minutes with a standard deviation of 65 minutes in the population of all first-year students at this university. Create a 99% confidence interval for the mean study time of all first-year students. Assume all necessary assumptions and conditions apply given the large sample size.        Interpret your interval in context.    

Solution

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)=118
Standard deviation( sd )=65
Sample Size(n)=463
Confidence Interval = [ 118 ± Z a/2 ( 65/ Sqrt ( 463) ) ]
= [ 118 - 2.58 * (3.021) , 118 + 2.58 * (3.021) ]
= [ 110.206,125.794 ]
Interpretations:
1) We are 99% sure that the interval [ 110.206,125.794 ] contains the
true population mean

2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contain the true
population mean