# How to test the hypothesis that two populations' means are different, given two sets of samples?

I am able to draw any number of samples from two different populations. I would like to test the hypothesis that the population mean of the first population is smaller than the population mean of the second population. The null hypothesis is that the first mean is greater than or equal to the other one.

I have no other information about the populations, other than that they are probably normally distributed, but the samples can be assumed independent.

How can I test this hypothesis to a given confidence level?

(this might be a textbook question but I don't have any statistics textbooks to refer to - feel free to tag as homework if you feel it's appropriate, although it's more of a hobby thing)

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If you know or are willing to assume the variances are the same then this may lead to a t-test; if not then the Behrens-Fisher problem – Henry Nov 24 '11 at 12:50
@Henry thanks for those links. I can probably assume the variances are the same. I'll check these out. – romkyns Nov 24 '11 at 12:56

Using Henry's links, here's what I ended up doing. I chose the Welch's t test as the approximate solution to Behrens-Fisher problem. I then applied the specified formulae to obtain $t$ and $\nu$.

I then computed the Student's t-distribution value for the $|t|$ and $\nu$. The two-tailed variant gives the confidence that the means are the same, while the one-tailed variant gives confidence that $\bar X_1 > \bar X_2$ (or the opposite, depending on the sign of $t$).

Here's a complete example:

$$\bar X_1 = 0.2527, \bar X_2 = 0.2329$$ $$s_1^2 = 0.001121, s_2^2 = 0.001366$$ $$N_1 = 59, N_2 = 62$$ $$\implies t = 3.09, \nu = 119$$

The confidence that the means differ is 99.8% here, and the confidence that $\bar X_1 > \bar X_2$ is 99.9%.

A different example:

$$\bar X_1 = 0.2527, \bar X_2 = 0.2525$$ $$s_1^2 = 0.001121, s_2^2 = 0.001031$$ $$N_1 = 59, N_2 = 62$$ $$\implies t = 0.041, \nu = 118$$

The confidence that the means are the same is 96.7%, and the confidence that $\bar X_1 > \bar X_2$ is 48.4%.

P.S. this is not necessarily correct, use with care.

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No one mentioned that the populations have to have normal distributions. In the non-normal case you use the Wilcoxon rank sum test which is nonparametric.

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