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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.

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In part (c) you need a continuity correction, just as in the earlier parts: $$ \Pr\left( X\le 39.5 \right) = \Pr\left( \frac{X-50}{\sqrt{50}} \le \frac{39.5-50}{\sqrt{50}} \right) \approx\Pr(Z\le -1.485) \approx 0.069. $$

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  • $\begingroup$ Is this still the case since we're asked for the p value for X = 39 rather than any $X \lt or \le 39$? Thanks. $\endgroup$ – user24013423 Dec 14 '14 at 21:01
  • $\begingroup$ A p-value is by definition the probability of getting data at least as strongly opposed to the null hypothesis as the data actually observed, given that the null hypothesis is true. So you need $\Pr(X\le 39)$, not $\Pr(X=39)$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 14 '14 at 23:11

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