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Let’s say I 'm testing if students improve their mathematical knowledge by hitting them. I want an $\alpha=5\%$, because I will have problems with their parents. I ‘m joking, of course... It's just an example.

The mean of their degrees is $\mu_0$ and the standard deviation is $\sigma_0$. I took a sample of students, I hit them all and then I find that their sample mean is $\bar{x}$ with standard error $\sigma_{\bar{x}}$. I find in this case that $\bar{x}=\mu_0+1.8\sigma_{\bar{x}}$.

As I said before, I want to find if the sample belongs to a hypothetical population with mean $\mu$ different (specifically: bigger) than $\mu_0$.

Two questions:

  1. With two-tailed z-test I find that $\bar{x}\in(\mu_0-1.96\sigma_{\bar{x}},\mu_0+1.96\sigma_{\bar{x}})$, so I can’t accept $H_1:\mu\neq\mu_0$. But $\bar{x}>\mu_0+1.65\sigma_{\bar{x}}$, so with a one-tailed z-test I find that $H_1:\mu>\mu_0$. How can $\mu>\mu_0$, but not $\mu\neq\mu_0$?
  2. The $95\%$ confidence interval for the $\mu$ is $(\bar{x} -1.96\sigma_{\bar{x}},\bar{x}+1.96\sigma_{\bar{x}})$, so $\mu_0$ is in that interval. How can be $\mu>\mu_0$?

I believe that I shouldn’t use one-tailed hypothesis test in that case. I believe that I should use it if it wasn’t possible to have negative effects to the students. The probability of this test is assuming that no negative effects would happen.

Am I wrong? Should I hit myself?

Thanks in advance!

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  • $\begingroup$ The equation $\bar x=\mu_0+1.8\sigma_{\bar x}$ is wrong. $\bar x$ cannot be computed from knowing its mean and its standard deviation. $\endgroup$ Jul 26, 2021 at 11:54
  • $\begingroup$ @Lewian, it's not wrong. I calculate $\bar{x}$ and I find that $\bar{x}=\mu_0+1.8\sigma_{\bar{x}}$. $\endgroup$ Jul 26, 2021 at 12:08
  • $\begingroup$ Ah OK. Maybe say "I find in this case that..." to make that clearer. $\endgroup$ Jul 26, 2021 at 12:12
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    $\begingroup$ Hitting students might plausibly make them tend to perform worse. So (as usual) there is not justification for a one-tailed test that they perform better, especially if you are deciding the nature of the test after seeing the data and want it to be significant $\endgroup$
    – Henry
    Jul 26, 2021 at 12:15
  • $\begingroup$ @Henry , thanks for your answer. I believe the same with you, but I thing I missing something at interpreting conditional probability $\alpha$ for the case of one and two tailed test. $\endgroup$ Jul 26, 2021 at 18:42

1 Answer 1

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Generally a test will never tell us what the truth is. I also think it's bad wording and misleading (despite in pretty general use) to say that any hypothesis or alternative is "accepted" (normally this is used for the $H_0$ by the way, you use it for the alternative, but I think we should avoid it altogether).

Now regarding one-sided vs. two-sided testing: If we test $H_0:\ \mu=\mu_0$ against $H_1:\ \mu\neq\mu_0$, it means that data count as evidence against $H_0$ if $\bar x$ is either much smaller or much larger than $\mu_0$. If we test $H_0:\ \mu= \mu_0$ against $H_1:\ \mu>\mu_0$, it only counts as evidence against $H_0$ if $\bar x$ is much larger than $\mu_0$. This means that in the first case, evidence against $\mu=\mu_0$ is easier to find. But in order to make sure that only things that are unlikely really are interpreted as evidence against the $H_0$, we have to define an unlikely event under $H_0$. In the two-sided case, because evidence can be found on both sides of $\mu_0$, in order to make our "rejection event" unlikely, we need to make it a bit harder on each side to reject, in order to balance the fact that we have two sides for finding evidence.

If we are willing to assume that in fact $\mu>\mu_0$ (we don't even have to assume this, we could also say that this is the only deviation of interest to us; the $H_0$ of the one-sided test can also be given as $\mu\le\mu_0$) we only reject on one side ($\bar x$ too big), so in order to achieve the same rejection probability, we can reject a bit more easily than if we have to take into account two sides.

If indeed $\bar x$ is so that we reject the $H_0$ in a one sided test but not in a two-sided test, it doesn't mean that we can be sure that $\mu>\mu_0$ but not $\mu\neq\mu_0$ (which of course is a contradiction). Rather it means that we have found an $\bar x$ that is quite unlikely if we have committed ourselves to only look in one direction (and therefore counts as evidence against the $H_0$), but not quite so unlikely if we have allowed ourselves to look in two directions. To some extent this illustrates something that is often said, namely that we shouldn't only interpret tests in a binary "accept/reject" manner. For example, the p-value of the one-sided test may be 0.03 ("reject at 5% level"), and 0.06 for the two-sided test. This will not formally reject but is still quite small.

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  • $\begingroup$ thanks a lot! So you are saying that $\mu>\mu_0$ is not a real situation, it's only a prediction assuming that something happened at the left, as I said at my post. Am I wright? But in that case if we reject $H_0:\mu\leq\mu_0$ with $\alpha=5\%$, then we reject $\mu<\mu_0$ and $\mu=\mu_0$ simultaneously with $\alpha=5\%$. So again $\mu\neq\mu_0$. Am I wrong? $\endgroup$ Jul 26, 2021 at 18:40
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    $\begingroup$ "So you are saying that $\mu>\mu_0$ is not a real situation" - no, I'm not saying that. You are right that rejecting $\mu\le\mu_0$ implies rejecting $\mu=\mu_0$, but this does not imply $\mu\neq \mu_0$ (the test won't tell us what the truth is), it only implies that there is a certain amount of evidence that $\mu>\mu_0$, and $\mu\neq\mu_0$ by implication. (Don't forget that if you test one-sided, you commit yourself to not being interested in whether $\mu<\mu_0$, so $ \mu>\mu_0$ and $\mu\neq\mu_0$ are basically the same.) $\endgroup$ Jul 26, 2021 at 22:56
  • $\begingroup$ I think I understand... Thanks for your patience! $\endgroup$ Jul 27, 2021 at 6:05

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