This is the problem (t for theta)://<![CDATA[ aax_getad_mpb({ "slot_uuid":"f485bc30-20f5-4c34-b261-5f2d6f6142cb" }); //]]>

Test H0: t <= 1 vs. H1: t >1 using the generalized likelihood ratio test where you have a random sample from X {X1, X2, ... , X50} and the sum of all Xi = 35. Use alpha = .05 (the probability of a type I error)Code (Text):

X ~ Expo(t) = t * e ^ (-t * x), x>0, t >0

0 otherwise

My professor actually did much of this in class and I've asked him and he said it's not supposed to be a hard problem. I don't know. In class, this is the test he arrived at for the GLRT:

reject H0 if

((t0 ^ n)* e^(-t0 * (sum over Xi))) / ((n/(sum over Xi))^n * e^(-n)) <= k

n is the number of observations (50), t0 is... hmm, I think t0 is 1 in this case. k is the number that makes the test agree with alpha = .05.

So now I need to simplify that expression so that I can find the probability that the resulting distribution is less than k. I can mess with k to make the left side simpler but I think I shouldn't do anything with (sum over Xi)--I think k should not depend on the values of the Xi's. I can only simplify it this far:

n * ln (sum over Xi) - t0 * (sum over Xi) <= k1

Do I need to use transformations to figure out how that messy thing on the left is distributed or am I doing something wrong? He said it shouldn't be that hard of a problem so I'm hesitant to use transformations.

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# Generalized likelihood ratio test

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