Imagining that the mean IQ of the population is 100, sd is 15. If the calculated IQ of the academic is 125, a two way t-test fails to reject the hypothesis that the professor is significantly smarter than the average person in the 5% significant, 100+1.96*15 ~ 130 is the rejection margin. Now a one way t-test however rejects the hypothesis with the margin being at 124.67.

My question is knowing that the academic cannot be less gifted than the lower 2.5% percentile or to that effect, even the average person, is it logical to use the one way test, in this circumstance?

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    $\begingroup$ A t-test is not applicable: the SD is presumed known. All you are doing is finding the percentile corresponding to the professor's IQ. $\endgroup$ – whuber Sep 1 '16 at 16:24
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    $\begingroup$ I note crucial objections to thinking about this as a test at all. That said, it's more common to talk of a 5% significance level, not a 95% significance level, for what you are trying. $\endgroup$ – Nick Cox Sep 1 '16 at 16:54
  • $\begingroup$ Student's t-test, obviously. Erm... I'll get my coat. $\endgroup$ – Mick Sep 2 '16 at 1:07

You don't need any significance tests. You already have the IQ of the professor in question, namely, 125, which is not equal to 100. The situation that you seem to have this situation confused with is when you have a sample of people from some population and you want to make an inference about the mean IQ of the population. Saying that the sample mean is "significantly" different from 100 means that you've decided that the population mean isn't 100. But there's no population to make inferences about here, since the mean IQ of the population you're comparing the professor to is already known.

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    $\begingroup$ I see your point. So I should simply report the percentile here. Saying that the mentioned person has a higher IQ than 95% of the population? $\endgroup$ – Theoden Sep 1 '16 at 17:12
  • $\begingroup$ @Theoden Yep, that makes sense. $\endgroup$ – Kodiologist Sep 1 '16 at 18:12
  • $\begingroup$ Just one more thing, wouldn't the original question make sense, if we assume that the IQ test itself has a sd of let's say 15? We will have an estimated mean and a standard deviation in that case. $\endgroup$ – Theoden Sep 1 '16 at 18:16
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    $\begingroup$ @Theoden Do you mean to say that you would be presuming that the IQ is measured with normally distributed measurement error, and the standard error of measurement is 15? You can then talk about estimating the professor's true IQ, yes, since you're no longer assuming his IQ is exactly what was measured (125). But significance testing would be an odd approach to that problem. $\endgroup$ – Kodiologist Sep 1 '16 at 18:37
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    $\begingroup$ One (somewhat silly) way the OP's question could make sense is if you consider the professor's IQ to be a sample of size 1 from the population of professors, and you were interested in testing the hypothesis that the population mean for professors were different than the general population. But then you would not be doing a t-test. $\endgroup$ – Cliff AB Sep 2 '16 at 5:15

If the professor scored significantly below average, than that would be worth reporting. Therefore you are interested in both ends of the tails and need to perform a two tailed test.

Usually you compare your p with some alpha. That alpha is agreed upon to be 0.05 just as testing is agreed to be two tailed.

These are two reasons for two tails.

  • $\begingroup$ (+1) Although I think no significance test is called for here, your treatment of one- versus two-tailed tests is on the money, and that's one of the issues OP brought up. $\endgroup$ – Kodiologist Sep 1 '16 at 17:04
  • $\begingroup$ @Kodiologist I agree, that the test is probably useless. It tests, whether any one professor is unlikely to be a random sample from the total population. Also, I would not call it a t-test but rather a z-test. The question of why we teach one tailed tests in order to use two tailed tests in science almost exclusively remain difficult to answer - I tried to cut a long story short. $\endgroup$ – Bernhard Sep 1 '16 at 21:21

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