Hi currently studying for a final exam and I just want to confirm my approach/answers to this problem are correct:
Suppose that $X \sim \mathrm{Poisson}$. We wish to test $H_0: \lambda = 50$ vs $H_1: \lambda < 50$, based on observing a single $X$. We employ the test: reject $H_0$ if $X \le 36$, accept $H_0$ if $X \ge 37$.
(a) Use a normal approximation with continuity correction to approximate the type 1 error of this test.
(b) Again, using a normal approximation with continuity correction, approximate the power of this test at $\lambda = 30$.
(c) Give the approximate $p$ value if we observe $X = 39$.
My answers/attempts:
(a) $\mu = 50, \sigma = \sqrt{50}$
$\displaystyle P(X \le 36) = P(X \lt 36.5) = P(Z \lt ({36.5 - 50 \over \sqrt{50}}) = -1.91))$, which gives a probability of $0.0281$.
(b) $\displaystyle 1 - P(X \lt 36.5) = 1 - P(Z \lt ( {36.5 - 30 \over \sqrt{30}}) = 1.19) = 1 - 0.8830 = 0.117$.
(c) (the one I'm most unsure of): ${39 - 50 \over \sqrt{50}} = -1.56$. $p = 0.0594$. I know in cases where $\mu = \mu_0$ we generally need to employ $2P(Z\ge |t|)$ etc. I'm generally confused about exactly the way in which $p$ is determined for $X = 39$.
Any help would be immensely appreciated.
In part (c) you need a continuity correction, just as in the earlier parts: $$ \Pr\left( X\le 39.5 \right) = \Pr\left( \frac{X-50}{\sqrt{50}} \le \frac{39.5-50}{\sqrt{50}} \right) \approx\Pr(Z\le -1.485) \approx 0.069. $$