What's wrong with this approach to computing power? I've just been given an old set of lecture notes to look at. 
One of them contains the following question: if we are undertaking a paired $t$-test with a mean of differences $d=1$ and  standard deviation $s=2$, what is the minimum number of subjects that would be required to "ensure that the difference observed is significant". Let's leave aside for now that particularly monstrous phrasing. 
The "model answer" is as follows: the test statistic we wish to discuss is $\frac{d}{s/\sqrt{n}}$. Let's find a value of $n$ such that this test statistic is greater than the corresponding critical $t$-value. It transpires that this is about 19:
deeff<-18; c(qt(p=1-0.05/2, df=deeff), 1/(2/sqrt(deeff)))
[1] 2.100922 2.121320
deeff<-19; c(qt(p=1-0.05/2, df=deeff), 1/(2/sqrt(deeff))) 
[1] 2.093024 2.179449

Now, clearly if we do a power test "properly", using the above as a definition of Cohen's d, we get a very different estimate of n: 
library(pwr)
pwr.t.test(d=0.5, type="paired", power=0.8)
>   Paired t test power calculation 

              n = 33.36713
              d = 0.5
      sig.level = 0.05
          power = 0.8
    alternative = two.sided

NOTE: n is number of *pairs*

Now, clearly, this is the result of a huge conceptual misunderstanding. My question is this, what's the most concise way to explain why the original answer is wrong? 
 A: I assume that the d = 1 and s = 2 values are descriptions of the population from which the sample is drawn. If that's the case, then any sample you draw may overestimate or underestimate the values for d and s. However, you want to ensure (horrible phrasing aside) that the difference is significant even if your sample underestimates d or overestimates s. This is why your second power calculation results in a larger sample. N = 19 may produce a significant effect if d = 1 and s = 2 in the sample you draw, but N = 33-34 will result in a significant effect for 80% of the samples you could draw, given the population values provided.
A: I calculate the power as the following:
The critical value when the null is true: cv <- qt(p=1-0.05/2, df=33, ncp=0, lower.tail=T)
The probability of exceeding the critical value when the alternative is true:
pow <- pt(q=cv, df=33, ncp=1/(2/sqrt(33)), lower.tail=F)
which is 80% as the power calculator mentions. I think the first calculation is wrong. Whether that's the original or not doesn't matter I guess.
The reason why the original is wrong is that the expected test statistic exceeds the critical value when the power is usually only about 50% because the non-central T-distribution is mostly symmetric. Running those numbers actually verifies this.
