4
$\begingroup$

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?

$\endgroup$

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.