Welch t with the minimum-df rule

The statistic with s in place of σ

With the population SDs unknown and not assumed equal — the default case on the paper — the two sample SDs go into L11.2’s SE, and the statistic is a t:

$$t=\frac{({\bar{x}}_1-{\bar{x}}_2)-{\delta }_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$$

This is the Welch (unpooled) two-sample t. Its exact distribution under $H_0$ is not a t with any whole-number df, so a convention is needed. This course’s convention, by hand: $\mathrm{df}=\min (n_1-1,n_2-1)$. It is the most conservative choice — the smallest df, the heaviest tails, the hardest test to reject — and it is what the marker expects to see written on the paper.

Worked example — is the new billing system paid faster?

Kitchener Ridge Logistics runs two billing systems side by side. Twelve invoices under the current system: ${\bar{x}}_1=31.2$ days, $s_1=9.5$. Ten under the new system: ${\bar{x}}_2=24.1$, $s_2=6.8$. Is the new system paid faster? $\alpha =0.05$.

Step 1. $H_0:{\mu }_1-{\mu }_2=0$, $H_a:{\mu }_1-{\mu }_2>0$ (current minus new; faster means the current mean is larger). Upper-tailed.

Step 2.

$$\frac{s_1^2}{n_1}=\frac{90.25}{12}=7.521,\frac{s_2^2}{n_2}=\frac{46.24}{10}=4.624$$

$$\mathrm{SE}=\sqrt{7.521+4.624}=\sqrt{12.145}=3.485,t=\frac{31.2-24.1}{3.485}=\frac{7.1}{3.485}=2.037$$

Step 3. $\mathrm{df}=\min (11,9)=9$. Row 9: $t_{0.05;9}=1.833$. Bracketing $p$: 2.037 sits between 1.833 (0.05) and 2.262 (0.025), so $0.025<p<0.05$.

Step 4. $2.037>1.833$: reject $H_0$ at the 5 % level. There is evidence that invoices under the new billing system are paid faster on average. Assumptions to state: two independent random samples, and — with $n$ of 12 and 10 — roughly normal days-to-pay in each population.

What R does instead

Software does not use the min rule. R’s t.test uses the Welch–Satterthwaite approximation,

$$\mathrm{df}=\frac{{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}^2}{\frac{(s_1^2/n_1{)}^2}{n_1-1}+\frac{(s_2^2/n_2{)}^2}{n_2-1}}=\frac{12.145^2}{\frac{7.521^2}{11}+\frac{4.624^2}{9}}=19.62$$

and prints df = 19.62, p-value = 0.02767. The statistic is identical — $t=2.037$ — but with 19.6 df the t-curve has lighter tails than with 9, so the same $t$ sits further out: the critical value falls from 1.833 to 1.726 and the exact $p$ is 0.0277, at the low end of the hand bracket. More df → smaller p. Here both routes reject; they can disagree when $t$ lands between 1.726 and 1.833, and then the paper wants the hand answer with the min rule written out.

You are not asked to compute the Satterthwaite df on the paper. You are asked to write $\mathrm{df}=\min (n_1-1,n_2-1)=9$, and — when an R printout is shown — to say why its df is larger and its p smaller.

The two-sided version

Had the question been do the systems differ?, compare $|t|=2.037$ with $t_{0.025;9}=2.262$: do not reject at 5 %; bracket $0.05<p<0.10$. The same data, a two-sided question, a different decision — Module 10’s lesson again.