Hi guys,

Heres my problem:

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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 meantheta:

H0 :theta= 5 against H1 :theta= 10.

Derive the likelihood of the sample X1,X2,...,X20 from the Poisson distribution with parameterthetaand hence derive the likelihood ratio test statistic

lambda(x) of the hypotheses H0 :theta=theta0 against H1 :theta=theta1, wheretheta1 >theta0.

Show that rejecting H0 in favour of H1 for low values oflambda(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.

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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