Most Powerful Tests

[Sinistral]

I am getting:


how to say that c=3 when no alpha is given.

[Sinistral]

i am not able to find a critical region at all :( . all I am getting is constants on both the side of the inequality.
plz help

[Sinistral]

again the same problem:

[Amit Goyal]

Given
H0: U(0, 1)
HA: U(0, 2)

(a) For a sample of size n = 1, the test procedure for which alpha = 0 and beta is minimized is the following:
Reject the null if X ≥ 1, accept otherwise.
alpha = Pr(X ≥ 1|U(0,1)) = 0
beta = Pr(X<1|U(0,2)) = 0.5

(b) For a sample of size n, the test procedure for which alpha = 0 and beta is minimized is the following:
Reject the null if max{X1, X2,...,Xn} ≥ 1, accept otherwise.
alpha = Pr(max{X1, X2,...,Xn} ≥ 1|U(0,1)) = 0
beta = Pr(max{X1, X2,...,Xn} < 1|U(0,2)) = 0.5^n

[Sinistral]

In part (b) how did you arrive at a test: Reject the null if max{X1, X2,...,Xn} ≥ 1
because even if  max{X1, X2,...,Xn} ≥ 1 there will be some Xi's which will be < 1. so there will be some non-zero type I error.
for alpha to be zero shouldn't critical region be min{X1, X2,...,Xn} ≥ 1 ???

I am getting super confused. how do u decide a test? by intuition, inspection? I was trying Neyman Pearson theorem but didnt succeed.

[Amit Goyal]

max{X1, X2, ..., Xn} ≥ 1  is equivalent to saying that at least one of n observations is ≥ 1. If the null distribution is U(0, 1) then its impossible to observe something ≥ 1. In other words, max{X1, X2, ..., Xn} ≥ 1 can never happen if null is true. Hence, alpha is 0.
You can use intuition or Neyman Pearson or inspection, all methods work. Think.

[Sinistral]

ok.
I think now I understood. thanx a ton.
But I seriously need to come at these critical regions on my own.

I'll again try it using neyman pearson tonight.