00:01
Hello students, here in the given problem, here first part to show that the a for testing that is h0 is theta is equal to theta 0 and h1 is theta is not equal to theta 0 is a is equal to yn upon theta 0 when yn is less than equal to theta 0 and a is equal to 0 if yn is greater than theta 0.
00:18
So here we can consider the likelihood function.
00:21
So here likelihood function under h0 is which is here l of theta 0 which is equal to 1 upon theta 0 raise to n and under the h1 l of theta hat which is equal to 1 upon theta hat raise to n where the theta hat is equal to maximum of y1, y2 up to yn.
01:00
So here the likelihood ratio the likelihood ratio consider a is equal to l of theta hat divided by l of theta 0.
01:12
So here put the values the final answer theta 0 divided by theta raise to n.
01:22
Therefore the likelihood function a is equal to theta 0 upon theta raise to n define a as a given a is equal to yn upon y theta if yn is less than equal to theta and a is equal to 0 if yn is greater than theta 0.
01:39
So a is related to lambda since a is equal to theta hat divided by theta 0 and the rejection region is a is less than equal to a where a is constant.
02:12
This is the answer of the a part.
02:15
Now b part when here when h0 is true then minus 2 log of a has chi squared 2n degrees of freedom here to show that the minus 2 log a has a 2 chi squared 2n degrees of freedom here so this is because this is because the theta hat which is equal to yn follows a uniform distribution on 0 and theta 0 and you can show that here minus 2 log a has chi squared distribution with 2 degrees of freedom here minus 2 log a has show that chi squared 2 degrees of freedom...