Hi guys in this problem we have X one X two and up to X. M be a sample from an exponential distribution with parameter lambda. So it can be shown that to Lemaitre times some nation over I from one to end for X. I has a tai chi squared distribution with two N degrees of freedom. So we use this fact to drive a test statistics and critical region for testing such that each note the null hypothesis lambda equals lambda. Note against the alternative hypothesis such as that llamada is less than La Madonna. Okay, so it's given that statistic statistics is uh to limit the times is a submission of X. I. Which has a chi squared distribution with two and degrees of freedom. So if this is true, F H note is true. Obviously we are going to reject each note for extreme values for statistics. Okay, for the one sided alternative, we must think carefully about which deal of the distribution to use. Should we be rejecting? It's not for every large or very small values of the test. So under the alternative hypothesis, um we have two llamada times as a submission for X. I from R equals one to end has a chi square distribution. So this sequins lama donald over llamada times two lambda. Times submission of X I from R equals one to end. And lambda over llamada note over a kilometer. There's more than one. Okay, so we expect to get large values of the test statistics under each one. Okay, so we therefore reject each note in favor of each one. Such that llama to less than la Madonna. Statistics statistics we see it's it's appropriate other percentage point of the chi square distribution. So the critical value is chosen to obtain the specialized type one error probability. So we'll let X square of alpha over to an M denote the upper 100 alpha percent a percentage point of excess square distribution. Okay so that assuming that each note is true. Therefore our test is to reject each note. Okay so the critical region for the test such that X note of alpha and to end and infinity. Okay then we consider the right tail. Test that test it's note lambda equals lambda and all against each one. Such a debt llamada moors and llama donut. Okay again statistics statistics is um to la Mcdonald times assassination over X I from equals one to end. Right. Okay. Which has a chi squared distribution with two n degrees of freedom of each note is true. So applying the same process as in the first part we conclude that appropriate to reject its nose for some very small values of statistics statistics then the critical value is now that lower down 100 alpha percentage point. So it's X squared of one minus alpha over two. Yeah. Yeah it's one over alpha and to em okay it's not all for over two of the chi square distribution was two and degrees of freedom. So that probability of to lambda and the submission from equals one to end for X. I. Um This is less than X squared for one minus alpha and two and it's alpha. Okay, assuming that it's not, it is true. Okay. So we reject it's not uh this value and the critical region for the test from zero two X squared of one minus alpha and to end. Okay, okay. Them for the last part of the problem we consider the two tail test. That's the test. Each note lambda equals lambda node against each one. Such as that llama. That does not equal la Madonna. Again, that is the statistics to lambda note times the submission for X Y. Which has a chi square distribution with two and degrees of freedom. Okay, so if each one is true in this case is a critical values are ah x squared of alpha over two and um and next one squared X. Described of one minus alpha over two and two. And so the upper and lower limits are 100 alpha over two percentage points. Of the chi square distribution was two and degrees of freedom. Okay, so let's find the probability of X squared of one minus alpha over two and two and it's less than or equal to land an old times submission from I equals one to N for X I and less than or equal X squared off also over two and two. And so it's one minus. Okay, now we reject it, know what if the test, the statistics falls outside this interview, that is. We reject each not in favor of each one if, either to lambda times as a submission ah for X. I less than X squared of one minus alpha over two and two and over the second part to limit the note times and submission for X I from I equals one to enter more than X squared of also over to and two. And okay, that's it.