Paired differences

Reduce to one column, then proceed as before

With paired data the analysis begins by computing one difference per pair, $d_i=x_{i,\text{after}}-x_{i,\text{before}}$. The $n$ differences are a single sample from a population with mean ${\mu }_D$, and everything that follows is Module 10’s one-sample t on that sample:

$$t=\frac{\bar{d}-{\delta }_0}{s_d/\sqrt{n}},\mathrm{df}=n-1(\text{pairs}-1),\bar{d}\pm t_{\alpha /2;n-1}\frac{s_d}{\sqrt{n}}$$

Here $s_d$ is the sample SD of the differences, computed from the column of $d$’s with $n-1$ in the denominator. It is not $\sqrt{s_1^2+s_2^2}$ and not anything else built from the two original columns — pairing works precisely because the differences vary far less than the raw values do.

Worked example — did the promotion lift weekly sales?

Eight Maple Ledger client stores report weekly sales (\$000) the week before and the week after a promotion:

store    1   2   3   4   5   6   7   8
before  42  51  38  47  55  60  44  49
after   45  53  41  46  58  66  47  50
d        3   2   3  -1   3   6   3   1

Did mean weekly sales change? Test at $\alpha =0.01$, bracket the p-value, and give a 95 % interval for the mean change.

Step 1. Paired (the same eight stores). $H_0:{\mu }_D=0$, $H_a:{\mu }_D\ne 0$, where ${\mu }_D$ is the mean change in weekly sales (after − before).

Step 2. $\sum d=20$, so $\bar{d}=20/8=2.5$. $\sum d^2=9+4+9+1+9+36+9+1=78$, and the shortcut formula gives

$$s_d^2=\frac{\sum d^2-n{\bar{d}}^2}{n-1}=\frac{78-8\times 6.25}{7}=\frac{28}{7}=4,s_d=2$$

$$\mathrm{SE}=\frac{2}{\sqrt{8}}=0.7071,t=\frac{2.5}{0.7071}=3.536,\mathrm{df}=7$$

Step 3. Two-sided at 1 %: column 0.005, row 7: $t_{0.005;7}=3.499$. Bracketing: 3.536 lies beyond the last column (3.499 at 0.005), so one-sided $p<0.005$ and two-sided $p<0.01$.

Step 4. $3.536>3.499$: reject $H_0$ at the 1 % level. There is evidence that mean weekly sales changed after the promotion — and since $\bar{d}>0$, that they rose.

The 95 % interval. $t_{0.025;7}=2.365$:

$$2.5\pm 2.365\times 0.7071=2.5\pm 1.67=(0.83,4.17)$$

With 95 % confidence the promotion raised mean weekly sales by between \$830 and \$4,170 per store. Entirely positive, as a two-sided test that rejects even at 1 % implies.

Why the columns’ SDs are useless here

The before column has SD 7.1 and the after column 8.1 — the stores differ in size. Treated as independent samples (the wrong design) the SE would be about 3.8 and $t$ about 0.66: nothing. The pairing removes the store-size variability, leaving differences with SD 2 and a decisive $t$. Note one shortcut that does hold: $\bar{d}={\bar{x}}_{\text{after}}-{\bar{x}}_{\text{before}}=50.75-48.25=2.5$, by linearity. There is no such shortcut for $s_d$; it must come from the differences.

The df counts pairs

Eight stores measured twice are 16 numbers but 8 pairs and 7 df. Writing $\mathrm{df}=14$ (as if independent) or 15 (as if 16 observations) reads the wrong row of the table — and changes the critical value.