Let $x_1,\dots,x_m$ be iid Poisson$(\lambda_1)$ and $y_1,\dots,y_m$ be iid Poisson$(\lambda_2)$ and independent of $x_1,\dots,x_m$. $S_1=\displaystyle\sum_{i=1}^mx_i$ and $S_2=\displaystyle\sum_{i=1}^ny_i$ are independent Poissons with parameters $m\lambda_1$ and $n\lambda_2$.
a) Show that the conditional distribution of $S_1$, given that $S_1+S_2=S$ is binomial with parameters $(s,p)$ where $$p=\frac{m\lambda_1}{m\lambda_1+n\lambda_2}$$b) Test the null hypothesis $H_0:\theta=1$ versus $H_1:\theta<1$, where $\theta=\frac{\lambda_1}{\lambda_2}$. What is $p$ under $H_0$? Should you reject the null hypothesis for large or small values of $S_1$?
c) Suppose that $m=n=25$ and $S_1=6$ and $S_2=18$. Compute the one-sided $p$-value for testing $H_0:\theta=1$ versus $H_1:\theta<1$ using the conditional binomial test suggested above.
I was able to figure out the first part, but need help figuring out parts b and c. Any help will be appreciated.
First of all
Second:
you literally wrote $y_1,...,y_m$ are iid poisson with parameter $\lambda_2$; this means that the two random sample have the same size $m$ but then you show the sum of $y_i$ going to $n$, that is a different size from $X-$ sample
(clarify if the two random samples have the same size or not)
(b) it is evident that $(p|\mathcal{H}_0)=\frac{m}{m+n}$
(c) under the given data verify that $\mathcal{H}_0:\theta=1$ against the alternative $\theta<1$ means to verify that, in a Binomial
$$B(24;p)$$
the parameter $p=0.5$ against the alternative that $p<0.5$
Using the conditional distribution, as suggested, this is the drawing
As you can see, for $S_1\leq 6$ the pvalue is less than 2% (it's about $1.133\%$) thus, i.e. you can reject the null hypothesis with any significance level $\alpha > p_{value}$
In other words, you reject the null hypothesis with a significance level $\alpha=5\%$ but you cannot reject it with a significance level $\alpha=1\%$
This also answer to the second (b) question