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Let $(X_1, X_2, X_3)$ be a sample taken from $\mathcal{N}(\mu, \sigma^2 = 12.2^2)$ (mean unknown, variance known). We want to test the following hypotheses: $$ H_0 : \mu = 0 \\ H_a : \mu < 0 $$ It is determined that $H_0$ will be rejected if and only if $X_1 + X_2 + X_3 < -20\sqrt{3}$. If the true mean is $\mu = -20$, what is the statistical power of our test?

I proceed in the following manner:

The power of a statistical test is given by $\mathbb{P}(\text{accept }H_a \mid H_a \text{ is true})$. So, I go on to determine when exactly do we accept the alternative hypothesis in terms of $\bar{X}$ when $H_a$ is true, i.e. $\mu = -20$. We have:

$$ \begin{align*} &\mathbb{P}(X_1 + X_2 + X_3 < -20\sqrt{3} \mid \mu = -20) = \mathbb{P}\left(\bar{X} < - 20 / \sqrt{3} \mid \mu = -20\right) = \\ = \;&\mathbb{P}\left(\frac{\bar{X} - \mu}{\sigma / \sqrt{n}} < \frac{-20 / \sqrt{3} - \mu}{\sigma / \sqrt{n}} \mid \mu = -20\right) = \mathbb{P}\left(\underbrace{\frac{\bar{X} - (-20)}{12.2 / \sqrt{3}}}_{\mathcal{N}(0, 1)} < \frac{-20 / \sqrt{3} - (-20)}{12.2 / \sqrt{3}}\right) = \\ = \;&\mathbb{P}\left(\bar{X}^{*} < \frac{-20 / \sqrt{3} + 20}{12.2 / \sqrt{3}}\right) \approx \Phi(1.2) \approx 0.88493 \end{align*} $$

Are my approach and reasoning correct? I have doubts because this turns out to be surprisingly "powerful."

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There is a slight mistake of plus-minus sign from the second to third step. Naturally, the power will be reduced.

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  • $\begingroup$ Thanks a lot! Could you please check now? $\endgroup$
    – d125q
    Commented Dec 28, 2014 at 13:27
  • $\begingroup$ Seems all-right! $\endgroup$ Commented Dec 28, 2014 at 13:29

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