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Ok, this might be an obvious question but it's had me stumped.
Let's say we're doing a test to see the improvement on test scores before and after some process.
We could state the hypothesis as:

H0: μ = 0
HA: μ > 0

assuming that there will be an increase. How would you state the conclusion if it turned out that there was actually a drop in the scores? This would be outside the parameters of the null hypothesis and the alternative hypothesis, simply saying that you reject the null hypothesis does not accurately answer the question.
Could the hypothesis be stated as

H0: μ ≤ 0

and if so why is there no option to do this in Minitab.

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Note that when you do a paired t-test, you are testing if the mean difference between pairs is significantly different from 0.

You describe a one-tailed t-test which gives you the option to test if there is a difference in one direction (e.g. 'greater than' in the case you describe).

'Less than' would also be a valid option in a one-tailed t-test. This would be denoted by the sign: <. I expect that should be possible in Minitab.

When you carry out a paired t-test, you are comparing your observed mean difference with what you would expect if the population mean difference was the value specified in your H0 (usually 0).

If your H0 is that the population mean difference is 0, with the paired t-test you consider a distribution of possible sample mean differences centred on 0. The t-test tells you whether you are at the extreme end of that distribution (with the one-tailed test for 'greater than', this is whether they are at the extreme upper end). If your H0 was that the population mean difference was -1, that distribution of possible sample mean differences would now be centred around -1, and your critical value would fall at a different value of sample mean difference.

So your H0 for a t-test needs to specify that your population mean difference = a single value, so that you have a single critical value to compare your sample mean difference to. Using 'less than or equal to 0' would give you a whole range of possible values, so you would no longer have a single critical value to test against - so no, what you propose is not possible (or at least, not as a single t-test).

If you do not have any strong reason to believe that the test score will only have improved, i.e. it is possible that it could in fact have got worse, then in my opinion it would be more correct to use a two-tailed t-test, that will test the alternative hypothesis that 'the mean difference between pairs is not equal to 0'.

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