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A textbook of mine has online worked solutions, and the answer in the back of the book does not match the answer in the online solutions - they're not even close.

My book says that the power of a test is $1-P(\mbox{Type II Error})$, where $$P(\mbox{Type II Error}) = P(\mbox{Accept } H_0\,|\,H_0 \mbox{ is false})$$

The method that I use - and the online solutions use - is: $$P(\mbox{Accept } H_0\,|\,H_0 \mbox{ is false}) + P(\mbox{Reject } H_0\,|\,H_0 \mbox{ is false})=1$$ $$\implies P(\mbox{Type II Error}) = 1-P(\mbox{Reject } H_0\,|\,H_0 \mbox{ is false})$$ $$\implies \mbox{Power} = P(\mbox{Reject } H_0\,|\,H_0 \mbox{ is false})$$

The example in the book is as follows: Bird watchers want to test if the average birds per day has changed from $10$. If on day one, we have $4 \le X_1 \le 17$ then we conclude no change. If on day one $X_1 \le 3$ and on day two $X_2 \le 2$, then we conclude a decrease in numbers. If on day one $X_1 \ge 18$ and on day two $X_2 \ge 19$, then we conclude an increase in numbers.

I took the null hypothesis $H_0$ would be that $X_i \sim \mathrm{Po}(10)$. Assume instead that numbers have changed and that $X_i \sim \mathrm{Po}(5)$ instead. The question is to find the power of the test.

The probability $P(\mbox{Reject }H_0)=P(X_1\le 3)P(X_2 \le 2)+P(X_1\ge 18)P(X_2\ge 19)$. Assuming $X_i \sim \mbox{Po}(5)$ gives $$P(\mbox{Reject }H_0)=0.2650\cdot 0.1247+5.416\times 10^{-6} \cdot 1.402\times 10^{-6} = 0.03304$$

This is the answer in the online solutions. However, at the back of the book it says $0.3567$.

The question is a bit vague about $H_0$ and $H_1$. I think I'm treating $H_0: X_i \sim \mbox{Po}(10)$ and $H_1:X_i \sim \mbox{Po}(5)$. Perhaps I should have $H_1:X_i \sim \mbox{Po}(\lambda)$, where $\lambda \in \mathbb R_{\ge 0}$ and $\lambda \neq 10$?

Edit: Changed $\lambda \in \mathbb N$ to $\lambda \in \mathbb R_{\ge 0}$, and fixed a rounding error: $0.03303$ to $0.03304$.


Edit2: Added the following copy of the question:

enter image description here

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    $\begingroup$ Answers in the back are frequently wrong--they're rarely carefully reviewed and half the fun of looking at them is finding the errors, anyway. What may be of more interest to you, in terms of understanding power, would be to explore the apparent fixation on integers: why do you attempt to restrict the Poisson parameter $\lambda$ to whole numbers? $\endgroup$ – whuber May 20 '19 at 20:21
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    $\begingroup$ The probability calculation is right. As you'd expect the chance of falling in the RHS reject region is ignorably low. The product of the two probabilities correctly reflects the "and" statement. $\endgroup$ – AdamO May 20 '19 at 20:21
  • $\begingroup$ @whuber You're right, $\lambda$ should be a non-negative real number, not an integer. If edited my question to reflect this. Thank you! Besides that slip, did I calculate the "power" of the test correctly? $\endgroup$ – Fly by Night May 21 '19 at 15:53
  • $\begingroup$ The revision contains a couple of otherwise's towards the end that may clarify the situation. It seems that altogether there are 3 distinct ways in which one can fail to reject. $\endgroup$ – BruceET May 21 '19 at 18:59
  • $\begingroup$ @BruceET Go on... $\endgroup$ – Fly by Night May 21 '19 at 19:54
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A problematic test: The test procedure described seems seriously flawed: Power against alternative $H_a: \lambda = 5$ is only about $3.3\%.$ In R,

ppois(3, 5)*ppois(2, 5)
## 0.03303602 

Perhaps more precisely, as below, it's $3.304\%.$ [R code changed from original post. However, conclusions from some outcomes very unlikely under Poisson models, such as $P(X_1 \le 3, X_2 \ge 19),$ are not explicitly stated.]

ppois(3,5)*ppois(2,5) + (1-ppois(17,5))*(1-ppois(18,5))
## 0.03303602   

So by any computation, the power seems much too low for a test at a reasonable level of significance, which may require two days of bird-counting.

A better test from two days of counting: A more reasonable test would reject $H_0: \lambda = 20$ against $H_a: \lambda \ne 20,$ for $T \le 12$ or $T \ge 29,$ where $T$ is the total number of birds seen in two days. Then the significance level is $\alpha = 4.32\%.$

1 - sum(dpois(12:29,20))
## 0.04320504

The power against a rate of 5 birds a day or $\lambda = 10$ birds in two days is almost 70%.

1-sum(dpois(12:29,10))
## 0.6967764

Note: In my experience, bird watchers are an extraordinarily patient lot, so it might be feasible to seek a still-higher power based on a 3-day bird-counting expedition.

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  • $\begingroup$ This answer may further confuse the issue rather than illuminate it because (a) the question does not concern how good the test is and (b) although as a practical matter it's fine to ignore part of the power calculation, knowing in advance it will contribute only a small amount, it's deceptive not to include it. (You fail to exhibit two terms that are correctly included in the question itself.) $\endgroup$ – whuber May 20 '19 at 21:33
  • $\begingroup$ Perhaps too hastily, I decided to abandon the test proposed in the question based on @AdamO's Comment. Do you see any interpretation or simple modification by which the proposed test has reasonable significance level and power? At the very least, having one criterion for for 'failing to reject' and another for 'rejection' seems unorthodox? $\endgroup$ – BruceET May 20 '19 at 21:37
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    $\begingroup$ It depends on what you mean by "reasonable"! This one has low power because its level is so high (c. 99.87%). Perhaps it could be understood and analyzed as a quality control process--it's a kind of CUSUM test. That thought, at least, intrigued me when I first read the question. But following up on it would take us far afield... . $\endgroup$ – whuber May 20 '19 at 21:43
  • $\begingroup$ @BruceET Thanks for your reply. I didn't create this test, it's just a question in a book. Because of my lack of knowledge, I don't really understand much of your answer. Did I calulate the "power" of the test correctly as $0.03304$? $\endgroup$ – Fly by Night May 21 '19 at 15:56
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    $\begingroup$ What constitutes a Rejection is not totally specified, so there might be quibbles about the computation of power. By any interpretation, I think your computation is reasonable. If I were grading your paper, I'd mark it correct. (But if I were your instructor I hope I wouldn't have assigned the problem--at least not without some clarifications.) $\endgroup$ – BruceET May 21 '19 at 16:04

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