# Auxiliary random experiment

This question is from Robert Hogg's Introduction to Mathematical Statistic 6th version exercise 8.5.2 page 461. The question is:

Let $X_1, X_2,...,X_{10}$ be a random sample of size 10 from a Poisson distribution with parameter $\theta$. Let $L(\theta)$ be the joint pdf of $X_1, X_2,...,X_{10}$. The problem is to test $H_0:\theta=\frac{1}{2}$ against $H_1:\theta=1$

(a) Show that $L(\frac{1}{2})/L(1) \le k$ is equivalent to $y=\sum_1^n x_i\ge c$

(b)In order to make $\alpha=0.05$， show that $H_0$ is rejected if y>9 and, if y=9, reject $H_0$ with probability $\frac{1}{2}$(using some auxiliary random experiment)

The solution says (a) and (b) are straightforward, I think (a) is simple, but I cannot see straightforward at (b)

My some thoughts on (b): since $X_1,...,X_{10}$ are random sample for a Poisson so $y=\sum_1^{10} x_i\sim Poisson (5)$ with $H_0$ and $y=\sum_1^{10} x_i\sim Poisson (10)$ under $H_1$

Under $H_0:$

$y>9, \Rightarrow p=1-ppois(9,5)=0.03182806$

$y=9, \Rightarrow p=dpois(9,5)= 0.03626558$

They are all less than $\alpha=0.05$ what is the meaning of "reject $H_0$ with probability $\frac{1}{2}$"?

Thank you very much.

• You need to finish the calculation! What is  (1-ppois(9,5)) + dpois(9,5)/2 equal to? (Incidentally, the probability "1/2" in (b) appears to be an approximation to a number close to $0.5010796$.)
– whuber
Jun 24 '16 at 13:49
• Thank you very much whuber. Now I get the solution. I will write down the straightforward solution which is not straightforward for me at all. Jun 25 '16 at 14:03

$(1-ppois(9,5)) + dpois(9,5)*p=0.05\\\Rightarrow p=0.5010796$