In the mathematical statistics course today, we started talking about tests, and decision rules. To illustrate all the concepts introduced today, we considered the case where we have a sample with . And we want to test
In the course, we’ve seen that we could use a test based on the order statistics . The test would be
i.e. if we choose , and if , we choose .
From the definition of the first order risk,
we can easily get that
Thus, the power is then
To visualize it, use the following parameters
n=5 alpha=.1 theta0=1
C1=theta0*(1-alpha)^(1/n) theta=seq(0,2,by=.01) P1=(1-(theta0/theta)^n*(1-alpha))*(theta>C1) plot(theta,P1,type="l",lwd=2,col="blue",xlab="",ylab="Power")
Note that, so far, we did never consider the maximum of our sample. Assume that the maximum is , then we can compute the -value,
Here it is
Now, why not consider another test, based on the minimum (since we have the distribution of the minimum of a sample from a uniform distribution). The test is the same as before
but here, the threshold is
The power of the test is here
This test has the same significance level (by construction), but the power of the test is clearly lower than the one we got using the maximum of our sample, when
C2=theta0*(1-alpha^(1/n)) P2=(1-(theta0/theta)*(1-alpha^(1/n)))^n*(theta>C2) lines(theta,P2,type="l",lwd=2,col="red")
Why not consider a test based on ? The problem is that we need the distribution (more specifically the survival distribution) of . We can compute it, numerically. But that might be painful. An alternative is to consider some approximation, based on the central limit theorem, i.e.
Our test is based on , and to get the same significance as before, use
The power of the test is then
Here it is
mu=2*(theta0/2) s2=2^2*(theta0^2/12)/n C3=qnorm(1-alpha,mu,sqrt(s2)) (P=1-pnorm(C3,theta,sqrt(s2)))*(theta>C3) lines(theta,P)
Observe here that the test based on the maximum is not more powerful than the one based on the average (I just wonder if it could be due to the Gaussian approximation…).