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The conditions that I have learned are as follows:

  1. If the sample size less than 15 a t-test is permissible if the sample is roughly symmetric, single peak, and has no outliers.

  2. If the sample size at least 15 a t-test can be used omitting presence of outliers or strong skewness.

  3. With a larger sample the t-test can be use even if skewed distribution if the sample is greater than 30, but less than 10% of the population.

Why can't you use a the t-test when the sample size is larger than 10% of the population size? What happens then? Do you use the z-test?

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As in the answers above, I find rule 3 to be somewhat bewildering. And I am not certain the z-test is the reason behind it. I suspect it's based on some sort of argument that there is an unusual covariance between your sample and the rest of the populations. I.e. if we already draw the samples that are far from the rest of the distribution due to the skewness, they won't be in the rest of the population anymore. –  Erik Dec 10 '12 at 15:43

2 Answers 2

up vote 2 down vote accepted

First, you have to understand why there are two tests, for a same quantity. Let's say you have a sample $x_1, \dots, x_n$, drawn from an unknown distribution and you want to test if the mean of the distribution is zero or not.

So you compute the sample mean $\overline x = {1\over n} \sum_{i=1}^n x_i$. And you compute the sample variance $s^2 = {1\over n-1} \sum_{i=1}^n (x_i-\overline x)^2$. And finally, you reduce $\overline x$ by the standard error $s = \sqrt{s^2}$, considering ${\overline x \over s/\sqrt n}$.

There are two cases :

  • the underlying distribution is normal ; then ${\overline x \over s/\sqrt n}$ is distributed like a $t$ distribution (if the mean is zero), and you use a $t$ test. This is an exact procedure.

  • you don’t know whether the underlying distribution is normal or not. If $n$ is big enough, the central limit theorem tells you that ${\overline x \over s/\sqrt n}$ is approximately distributed like a standard normal distribution (if the mean is zero), and you use a $z$ test. This is an approximate procedure.

What you were stating are just guidelines to help you decide if the assumptions required for $t$ test are satisfied.

I don’t get rule 3. For me, it is just false. If the distribution is skewed, it is not normal, and you have no reason to think that the $t$ test will perform better than the $z$ test.

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When the distribution is skewed, the sampling distribution of the mean may still be (and often is) sufficiently close to Normal for the t test to apply. –  whuber Dec 10 '12 at 16:47
    
The sampling distribution of the mean will be close to normal, so we can use a $z$ test. There’s no reason for the sampling distribution of the reduced mean $\overline x / s$ to be closer to the $t$ distribution than to the normal distribution. Or is there any? –  Elvis Dec 10 '12 at 17:12
    
This is hard to answer theoretically, because it depends on circumstance. But overall if the underlying distribution has one or more heavy tails (compared to the Normal), I would expect the t distribution to be a better approximation to the distribution of $\bar{x}/s$ than the z distribution. –  whuber Dec 10 '12 at 17:21
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@whuber, I think you’re right. I tried with a variety of distributions, and in many cases a $t$ distribution is better than anormal distribution. Note however that with skewed distributions like an Exp(λ=0.1) both are terrible. –  Elvis Dec 11 '12 at 9:52
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I agree: I cannot conceive of any good justification for rule 3 and I can think of many exceptions. The (unpaired) t-test is notoriously sensitive to skewness and that "10% of population" threshold appears to be without foundation. –  whuber Dec 11 '12 at 16:13

You can actually use the t-test if you like -- it's just more conservative. As your sample size grows larger, the Central Limit Theorem says that the distribution of your mean approaches a normal distribution, regardless of the underlying population distribution. Therefore, you can use the Z-test, since that compares your statistic with a normal distribution.

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ok, but which formula do I use to find the degrees of freedom? –  MaoYiyi Dec 10 '12 at 5:14
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There are no degrees of freedom for a Z-test. If you are conducting a one sample t-test, you use your sample size minus one for the degrees of freedom: n-1. –  StatsStudent Dec 10 '12 at 5:39
    
also, why can't the sample be larger than 10% of the population? –  MaoYiyi Dec 10 '12 at 8:29
    
I was meaning with the t-test? there are several to choose from, which one is the best? –  MaoYiyi Dec 10 '12 at 8:30
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@whuber OK, how about "if the assumptions are met and you apply both a z-test and a t-test, sometimes the z-test will reject the null when the t-test fails to do so". In practice, of course, with large n, the 2 distributions are virtually identical. –  Peter Flom Dec 10 '12 at 22:14

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