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A bag contains $N$ balls numbered $1,2,3,…,N$ to test the hypothesis $H_0 : N=10 $ vs $H_1:N=20.$ Draw $2$ balls from the bag without replacement, let $X$ denote the larger of the $2$ numbers in the drawn balls and reject $H_0$ if $X>9$. What is the power of this test?
My attempt.
When $H_1$ is true, $N=20$. And now we need to calculate the probability that $H_0$ is rejected which is $\frac{\binom{20}{2}-\binom{9}{2}}{\binom{20}{2}}$. Am I correct?

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By definition, the power of a test for a particular alternative hypothesis $H_1$ is the chance of rejecting the null when $H_1$ governs the data: that is, when $N=20$. Rejection occurs when the larger of the two balls drawn exceeds $9$.

The chance of rejection is most simply computed by finding the complementary chance that both balls have values of $9$ or less, and subtracting this chance from $1$. That's because there are $\binom{9}{2}$ two-ball subsets of the numbers $\{1,2,\ldots, 9\}$ that are selected from the $\binom{N}{2}=\binom{20}{2}$ equiprobable two-ball subsets when $N=20$. Consequently, the power is

$$1 - \binom{9}{2}/\binom{20}{2} = \frac{\binom{20}{2} - \binom{9}{2}}{\binom{20}{2}},$$

as stated in the question.

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Let $A$ and $B$ be the numbers on the balls. \begin{align*} P(X > 9) &= 1-P(X \le 9) \\ &= 1-P(A \le 9 \text{ and } B \le 9) \end{align*} Can you take it from here?

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