Friday in the course of statistics, we started the section on confidence interval, and like always, I got a bit confused with the degrees of freedom of the Student (should it be 
or 
?) and which empirical variance (should we consider the one where we divide by or the one with ?).
And each time I start to get confused, the student obviously see it, and start to ask tricky questions… So let us make it clear now. The correct formula is the following: let
then
is a confidence interval for the mean of a Gaussian i.i.d. sample.
But the important thing is neither the n-1 that appear as degrees of freedom nor the that appear in the estimation of the standard error. Like always in mathematical result, the most important part of that result is not mentioned here: observations have to be i.i.d. and to be normally distributed. And not “almost” normally distributed….
Consider the following case: we have =20 observations that are almost normally distributed. Hence, I consider a student t distribution
An Anderson Darling normality test accepts a normal distribution in 2 cases out of 3.
With a true normal distribution if would be 95% of the cases, so in some sense, I can pretend that I generate almost normal samples.
For those samples, we can look at bounds of the 90% confidence interval for the mean, with three different formulas,
i.e. the correct one, or the one where I considered degrees of freedom instead of ,
and the one were we condired a Gaussian quantile instead of a Student t one,
for(s in 1:10000){ X=rt(n,df=3) m[s]=mean(X) sd=sqrt(var(X)) IC1[s]=m[s]-qt(.95,df=n-1)*sd/sqrt(n) IC2[s]=m[s]-qt(.95,df=n)*sd/sqrt(n) IC3[s]=m[s]-qnorm(.95)*sd/sqrt(n) }
One the graph below are plotted the distributions of the values obtained as lower bound of the 90% confidence interval,
(the curves with and degrees of freedom in quantiles are the same, here).
The dotted vertical line is the true lower bound of the 90%-confidence interval, given the true distribution (which was not a Gaussian one).
If I get back to the standard procedure in any statistical textbook, since the sample is almost Gaussian, the lower bound of the confidence interval should be (since we have a Student t distribution)
mean(IC1) [1] -0.605381
instead of
mean(IC3) [1] -0.5759391
(obtained with a Gaussian distribution instead of a Student one). Actually, both of them are quite different from the correct one which was
quantile(m,.05) 5% -0.623578
As I mentioned in a previous post (here), an important issue is that if we do not know a parameter and substitute an estimator, there is usually a cost (which means usually that the confidence interval should be larger). And this is what we observe here. From a teacher’s point of view, it is an important issue that should be mentioned in statistical courses….
But another important point is also that confidence interval is valid only if the underlying distribution is Gaussian. And not almost Gaussian, but really a Gaussian one. So since with =20 observations everything might look Gaussian, I was wondering what should be done in practice… Because in some sense, using a Student quantile based confidence interval on some almost Gaussian sample is as wrong as using a Gaussian quantile based confidence interval on some Gaussian sample…
i.i.d. with distribution 
. Here we note
is a random variable. The idea is to assume that 
. Thus, we need to compute the distribution of 
which is here extremely simple (due to properties of the Gaussian distribution), i.e.
as an estimator of given our sample data (and thus, we also have a confidence interval since we know the distribution of 
. Note, first, that \theta has support 
. So we need a distribution on that support. Why not a beta distribution ? E.g.
proportional to
, and 
being a parameter that will change, from 0 to 4.
or 
, depending whether 
or 
is the smallest parameter.
and 
