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When studying two independent samples means, we are told we are looking at the "difference of two means". This means we take the mean from population 1 ($\bar y_1$) and subtract from it the mean from population 2 ($\bar y_2$). So, our "difference of two means" is ($\bar y_1$ - $\bar y_2$).

When studying paired samples means, we are told we are looking at the "mean difference", $\bar d$. This is calculated by taking the difference between each pair, and then taking the mean of all those differences.

My question is: Do we get the same ($\bar y_1$ - $\bar y_2$) versus its $\bar d$ if we calculated them from two columns of data, and the first time considered it two independent samples, and the second time considered it paired data? I have played around with two columns of data, and it seems that the values are the same! In that case, can it be said that the different names are used for just non-quantitative reasons?

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    $\begingroup$ Think about it this way: how would you compute $\bar d$ with unpaired data? $\endgroup$
    – shadowtalker
    Commented Apr 29, 2015 at 14:10
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    $\begingroup$ @ssdecontrol Especially if the sample sizes are different. $\endgroup$
    – Alexis
    Commented Apr 29, 2015 at 19:08
  • $\begingroup$ If you do the calculations, the difference of means (278.75-277.75 = 1) is exactly the same as the mean of differences (1+1+2+0)/4 = 1 $\endgroup$
    – Dave C
    Commented Jun 1, 2021 at 10:07

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(I'm assuming you mean "sample" and not "population" in your first paragraph.)

The equivalence is easy to show mathematically. Start with two samples of equal size, $\{x_1,\dots,x_n\}$ and $\{y_1,\dots,y_n\}$. Then define $$\begin{align} \bar x &= \frac{1}{n} \sum_{i=1}^n x_i \\ \bar y &= \frac{1}{n} \sum_{i=1}^n y_i \\ \bar d &= \frac{1}{n} \sum_{i=1}^n x_i - y_i \end{align}$$

Then you have: $$\begin{align} \bar x - \bar y &= \left( \frac{1}{n} \sum_{i=1}^n x_i \right) - \left( \frac{1}{n} \sum_{i=1}^n y_i \right) \\ &= \frac{1}{n} \left( \sum_{i=1}^n x_i - \sum_{i=1}^n y_i \right) \\ &= \frac{1}{n} \left( \left( x_1 + \dots + x_n \right) - \left( y_1 + \dots + y_n \right) \right) \\ &= \frac{1}{n} \left( x_1 + \dots + x_n - y_1 - \dots - y_n \right) \\ &= \frac{1}{n} \left( x_1 - y_1 + \dots + x_n - y_n \right) \\ &= \frac{1}{n} \left( \left( x_1 - y_1 \right) + \dots + \left( x_n - y_n \right) \right) \\ &= \frac{1}{n} \sum_{i = 1}^n x_i - y_i \\ &= \bar d. \end{align}$$

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    $\begingroup$ But two confidence intervals calculated for "the difference of the means" and "the mean difference" will be different, right? This can be seen by looking at $A = [1, 2, 3, 4, 5, ...]$ and $B = [..., 5, 4, 3, 2, 1]$. A paired "mean difference" will be different for $A - A$ (which is all zero) versus $A - B$ (which is not all zero); the difference of the means is unaffected by the order of the elements. $\endgroup$
    – bers
    Commented Dec 15, 2015 at 19:37
  • $\begingroup$ Can't edit my previous post any longer. The 3rd sentence should begin "A sequence of paired 'mean differences' ..." $\endgroup$
    – bers
    Commented Dec 16, 2015 at 13:07
  • $\begingroup$ @bers what does $A-A$ have to do with it? $\endgroup$
    – shadowtalker
    Commented Dec 16, 2015 at 13:23
  • $\begingroup$ Assume $C=A$. Then $A-C$ and $A-B$ are two different sequences. The confidence interval for the mean paired difference will certainly be different in both cases. But the difference of the means, and so it's confidence interval, will be indentical both for $A-C$ and $A-B$. Or am I wrong? $\endgroup$
    – bers
    Commented Dec 16, 2015 at 13:27
  • $\begingroup$ @bers I think you're confused, but I'm confused as to what you're confused about. $\endgroup$
    – shadowtalker
    Commented Dec 16, 2015 at 13:40
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the distribution of the mean difference should be tighter then the distribution of the difference of means. See this with an easy example: mean in sample 1: 1 10 100 1000 mean in sample 2: 2 11 102 1000 difference of means is 1 1 2 0 (unlike samples itself) has small std.

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  • $\begingroup$ This is only the case because your two example vectors are ridiculously correlated and var(X - Y) = var(X) + var(Y) - 2cov(X, Y). $\endgroup$
    – einar
    Commented Oct 20, 2020 at 9:24
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d bar is for paired data that is correlated, and will have a different test statistic compared to what is used for a two sample t-test of independent sample means. in my opinion saying that mean difference represents d bar and difference between sample means represents the different between y bar sub 1 and y bar sub 2 is more accurate and precise, being consistent with the mathematics of calculating the statistics and how their statistical tests are conducted.

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