# Is it possible to use a two sample $t$ test here?

I want to do statistical analysis to compare the results of the different specimen sizes (which I am comparing) with each other. Seeing as I have at least 12 specimens for each specimen size, I thought it was possible to use the data sets to compare the means and identify the differences between these values. I then thought that I could do this using a two sample $t$ test to try and determine the difference in the population means.

My question is, is this possible? How can I justify the data is Normally distributed to allow this

EDIT: As well as this, is it correct wording to say:

I used a hypothesis test to test the differences in the two sample populations and tested the validity of the claim null hypothesis versus the alternate hypothesis. I determined a level of significance ($\alpha$ value) tested the null hypothesis against a probability value ($p$ value). If my $p$ value > $\alpha$ value, then I can accept my null hypothesis. Once I had accepted (or declined) the null for each data set, I can then carry out my $t$ test to compare the independent sample means.

Is $p$ value correct there, or should it be $t$ value?

• As described, and more generally under the classical null hypothesis testing framework, you don't accept the null. You can only reject it ($p\le\alpha$) or say that data are not incompatible with it ($p>\alpha$). – chl Apr 16 '13 at 7:15
• @chl Is $p$ the right letter there or would I need to use $t$? – Kaish Apr 16 '13 at 16:36
• $t$ usually stands for your test statistic, while $p$ is your p-value which reflects the probability of observing a result as least as extreme under the null. If p is low (i.e., below a given threshold, say $\alpha=0.05$, defined before seeing the data), then when conclude that such an extreme result is unlikely to be due to random fluctuations alone. – chl Apr 16 '13 at 21:24

It may be reasonable to do a two-sample t unless specimen sizes tend to be particularly skew or heavy-tailed. Alternatively you could use a permutation test ($\binom{24}{12}=$ 2.7 million is quite doable, though you can also sample them), or you could go to a Wilcoxon-Mann-Whitney if you're prepared to deal with the fact that (without some additional assumptions) it's not directly a test for means.

You cannot make this claim in this circumstance.

The sample size of 12 is insufficient to satisfy the Central Limit Theorem and so the samples will only be normally distributed if the population is normally distributed. To satisfy the CLT you need at least $n=30$.

If you have evidence that the population is normally distributed then include that evidence and make the assumption. If it isn't, take more samples.

• "To satisfy the CLT you need at least n=30." What is the basis for this statement? I can show cases where the distribution is non-normal yet $n$ much less that 30 gives good answers, and I can show cases where $n$ much greater than 30 is not sufficient. You can almost never have evidence that a population is normally distributed; indeed I doubt that's ever actually true. – Glen_b Apr 16 '13 at 3:46
• @Glen_b I got it from the statistics course at the University of New South Wales where it is given as a "rule of thumb". – Dale M Apr 16 '13 at 4:58
• (Oh, hey. I taught stats there for a while, during my PhD, long ago.) It's a commonly-stated "rule" in a lot of the more cookbooky texts, but you shouldn't believe everything you're told. The actual support for it always seems to be absent. – Glen_b Apr 16 '13 at 6:35
• @Glen_b Especially in a statistics course for an MBA in particular, I'm guessing! – Dale M Apr 16 '13 at 6:37
• That would fit the sort of context you see it. (Hmm. I taught some stats/prob at the AGSM while I was there, too.) To give you an example, consider samples from a Poisson with mean 0.001. If you sum 100 of those, you get a Poisson with mean 0.1 ... and a t-test computed on those isn't really very close to t-distributed. – Glen_b Apr 16 '13 at 6:54