Now if the distribution of the data is normal the sample mean has a
t-distribution with 5 degrees of freedom. But to use the t-distribution
we need the population standard deviation, which we will estimate with:
s = sqrt[1/(n-1) sum (x_1 - m)^2] = 12.11
Then the interval is:
(m - cv*s/sqrt(n), m + cv*s/sqrt(n))
where cv is the critical value for a 95% interval for a t-distribution with n-1
degrees of freedom.
The critical t-value for a 95% interval with 5 degrees of freedom is 2.57,
so the 95% interval would then be:
(33.33-2.57*12.11/sqrt(6), 33.33+2.57*12.11/sqrt(6))~= (20.62, 46.04)
We use the t-distribution because we have a sample size n=6, which is
a small sample so the use of the normal distribution is inappropriate here
as that requires large samples.
Note the use of the t-distribution is also inappropriate unless we have some
reason to believe the underlying distribution that the data is sampled from