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Usually I am very familiar with how to do hypothesis tests, but I have never seen an alternative hypothesis where $\mu$ is equal to a particular value. How would one proceed in this situation? This is an example I came across:

"Assuming normality with variance $σ^2 = 9$, test the null hypothesis $\mu = 60.0$ against the alternative hypothesis $\mu = 57.0$ using a sample size of $20$ with $\bar x = 58.05$ and choosing $\alpha = 0.05$."

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    $\begingroup$ (Explaining my edit) ... It would be unusual to call the hypothesized value under the alternative $\mu_0$ (you'd call the value under $H_0$ "$\mu_0$" -- i.e. $60$ is the thing you'd label $\mu_0$). Under $H_1$ you'd call it $\mu_1$ (i.e. $57$ would be referred to as $\mu_1$). To do otherwise is just asking to be misunderstood. $\endgroup$ – Glen_b -Reinstate Monica Jun 23 '16 at 19:29
  • $\begingroup$ This case is important from a theoretical point of view, because it's possible to construct an optimal (in a certain sense) test. Check out the Neyman-Pearson lemma: en.wikipedia.org/wiki/Neyman%E2%80%93Pearson_lemma $\endgroup$ – Zen Jun 23 '16 at 19:37
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It's standard to look at point null vs point alternative when first introducing the Neyman Pearson lemma, so if you've seen that you'll probably have seen this simple-alternative case done already.

Very little is different in the simple-null-simple-alternative case from the simple-null case, you are just in a situation where those (60 and 57) are the only two possible values for $\mu$.

Clearly unusually small values of $\bar{X}$ would lead you to consider $H_0$ to be untenable, but large values (larger than 60) would not lead to you conclude the mean is $57$ instead of $60$, so you only reject on one side.

So all that's left to do is give a test statistic whose distribution under the null hypothesis can be calculated, in order to determine a rejection region for that statistic that corresponds to small values being taken by $\bar{X}$.

You already know a test statistic which will have a known distribution under $H_0$ (... and if you use the Neyman-Pearson lemma, you can argue it will be the most powerful test in this circumstance).

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  • $\begingroup$ Interesting! So I'm assuming this is not a test you would ever do in the real world? $\endgroup$ – Joe Jun 23 '16 at 19:21
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    $\begingroup$ Ever? No, such situations can occasionally arise, but they're relatively rare. $\endgroup$ – Glen_b -Reinstate Monica Jun 23 '16 at 19:25
  • $\begingroup$ Funnily enough that was quite common in my statistics tests. $\endgroup$ – Firebug Jun 23 '16 at 19:33
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I'm not as expert at statistics as some people here so please bear with me; but I'd like to throw in my two cents worth. As I understand the example problem you quote, it is basically asking if a sample mean of 57 is more likely, or has a lower (more statistically significant) p-value, than a sample mean of 60, if the true mean is 58.05. So your H0 is that mu=60 and H1 is that mu=57. Or, "is 57 truly lower than 60, given the sample size (20), true mean (58.05) and the alpha level (.05) stated?" and at what p-level. So (again as I understand it) you would test the hypothesis in the usual way and use a one-sided p-value to test if 57 or lower is significantly different from 60, given the above parameters. (my intuition tells me it's not different from 60, because 60 is farther away from the mean of 58.05 than 57 is and the distribution is normal, i.e., symmetrical, and both are within one standard deviation (3) from the mean).

Your actual question however seems to be asking what is the probability of your H1 mu being exactly equal to 57; is that correct? To answer that may require a different approach, perhaps a probability density function.

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