# Thread: Likelihood Ratio Test Statistic

1. ## Likelihood Ratio Test Statistic

Hi guys,

Heres my problem:

**********
Data X1,X2,....,Xn, n = 20, are available of numbers of flaws per 10 metres
of bales of a manufactured material. Assuming that the Poisson model is valid
for these data, the probability mass function of X is:

fx(x; theta) = [(theta^x).(exp(-theta))] / x! where x = 0,1,2... ; theta > 0 and with mean E(X) = theta

The average number of flaws per 10 metres was known to be 5 prior to installation of new machinery. It is suspected that this average may have increased.
We now wish to test the following hypotheses about the value of the mean theta:
H0 : theta = 5 against H1 : theta = 10.
Derive the likelihood of the sample X1,X2,...,X20 from the Poisson distribution with parameter theta and hence derive the likelihood ratio test statistic
lambda(x) of the hypotheses H0 : theta = theta0 against H1 : theta = theta1, where theta1 > theta0.
Show that rejecting H0 in favour of H1 for low values of lambda(x) corresponds
exactly to a rejection region of the form {x:t(x) > c}, where T(X) = Sigma(Xi) (i=1 to n), the sample total, and state what optimal property this test possesses.
************

Ive got the likelihood L(theta;x) and think i have the likelihood ratio as:

lambda(x) = [(theta0^SigmaXi)(exp(-n.theta0))] / [(theta1^SigmaXi).(exp(-n.theta1))]...?

Is this first part right? if not i'm completely lost and if so i'm not entirely sure how to go about doing the rest of the question.

Please help and apologies for the long winded question...it was too hard to explain in any other way.

Much appreciated

2. I'll give it a try, but since I myself am a student, I cannot guarantee my reply is the correct one. It may bump up the thread however to the attention of experts.

If I am testing $H_0:\theta_0=5$ against $H_1: \theta_1=10$, I will use the likelihood ratio that you already stated:

$r=\frac{\theta_0^{\Sigma{X_i}}e^{-n\theta_0}}{\theta_1^{\Sigma{X_i}}e^{-n\theta_1}}$

This shows to me how likely the Ho compared to H1. To be considered a test, this ratio should be compared to some number (k) that is determined by the confidence level of the test (alpha):

$r=\frac{\theta_0^{\Sigma{X_i}}e^{-n\theta_0}}{\theta_1^{\Sigma{X_i}}e^{-n\theta_1}}\geq{k_{\alpha}}$

Then I'll just substitute theta's and n that I am given and find the expression for the test statistic vs k:

$r=(\frac{5}{10})^{\Sigma_{X_i}}e^{-20(5-10)}\geq{k_{\alpha}}$

$r=(\frac{1}{2})^{\Sigma_{X_i}}e^{100}\geq{k_{\alph a}}$

After some manipulations (in particular, taking log, which allows to keep the same sign of the inequality, and simplifying algebraically)

$\Sigma_{X_i}\geq\frac{100-lnk}{ln2}$ that will be your $T(x)\geq{c}$

The last question 'optimal property that this test possesses' I am not sure what exactly this is supposed to mean. I think it should be clear from the focus of your study guide, or other similar questions in the textbook.

3. The test is, for all values of c, most powerful of its size (you can't, however, get whatever size you want by varying c).

4. Originally Posted by theodds
The test is, for all values of c, most powerful of its size (you can't, however, get whatever size you want by varying c).
Ah! That's the property then ))) Is that because c's value does not depend on $H_1$? (only on the $\Sigma_{X_i}$).

Is the rest of my answer OK?

5. LRT's are most powerful of their size by the NP Lemma for testing simple-simple hypotheses (but not all most powerful tests are LRT's, and it doesn't extend beyond simple-simple as far as I know).

The test is also UMP for testing $\theta \le \theta_0$ against $\theta > \theta_0$ by Karlin-Rubin (this part is related to the fact that the alternative hypothesis has no effect on the choice of c).

Your solution looks okay, but as I stated you can't find a $k_\alpha$ for arbitrary $\alpha$ in this problem. This highlights the fact that LRT's are most powerful of their size in this context, but that not all most powerful tests are LRT's (a size $\alpha$ NP test always exists).