Solved – How to test if one value is significantly different from two others

small-samplestatistical significancet-test

I'm analysing classroom interaction data. I'm using a simple quantitative index counting the number of interactions of a certain kind. I want to know if it significantly changed between years 1 and 2, where I used method A to teach, and year 3, where I used method B.

I thus have $x_1$ and $x_2$ from the first two years to compare against a single value $x_3$ for the third year.

It is correct to use a $t$-test to check if $x_3$ is significantly different from $x_1$ and $x_2$ by checking if, following a $t$ distribution with 1 degree of freedom, the probability of getting a value at least as extreme as $x_3$ is lower than some $p$-value?

That is, whether this holds:

$$P_{t(df=1)}\left(X \geq \frac{x_3 – \bar{x}}{s}\right) \leq 0.05,$$

where $\bar{x} = \frac{1}{n}\sum_{i=1}^n x_n$ and $s=\sqrt{\frac{1}{n-1}\sum_{i=1}^n (x_i-\bar{x})^2}$, so in this case where $n=2$ we would have $\bar{x} = \frac{x_1 + x_2}{2}$ and $s=\sqrt{(x_1 – \bar{x})^2 + (x_2 – \bar{x})²}$.

Is this correct, or shall I rather use $\frac{x_3 – \bar{x}}{s/\sqrt{2}}$ instead of $\frac{x_3 – \bar{x}}{s}$; i.e., divide by the standard error of the mean instead of the sample standard deviation?

(I am aware that this makes the assumption that the population follows a normal distribution.)

Best Answer

If you just want to look for improvement from years 1 and 2 to year 3, then you could treat years 1 and 2 as a single sample and do a t-test. That sounds like what you want to do.

If it's helpful, you can check out this page I made for my intro stats class for comparing means from two independent samples.

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