Goal: interpret “t test formula weighted” in two-sample inference
The phrase t test formula weighted most naturally refers to the pooled two-sample t test (independent samples, equal but unknown variances). The “weighted” part is the pooled variance: each sample variance contributes in proportion to its degrees of freedom, \((n_1-1)\) and \((n_2-1)\).
Symbols and assumptions
| Symbol | Meaning | Notes (equal-variance case) |
|---|---|---|
| \(n_1, n_2\) | Sample sizes for groups 1 and 2 | Independent samples |
| \(\bar{x}_1, \bar{x}_2\) | Sample means | Estimate \(\mu_1, \mu_2\) |
| \(s_1^2, s_2^2\) | Sample variances | Estimate the same population variance \(\sigma^2\) when the equal-variance assumption holds |
| \(\Delta_0\) | Null hypothesized difference \(\mu_1-\mu_2\) | Often \(\Delta_0=0\) |
| \(s_p^2\) | Pooled variance (weighted average of \(s_1^2\) and \(s_2^2\)) | Weights are \((n_1-1)\) and \((n_2-1)\) |
Equal-variance (homoscedasticity) assumption: both groups come from populations with the same variance \(\sigma^2\), while means may differ. If variances are clearly unequal, the Welch t test is typically preferred.
Step 1: The weighted pooled variance (where “weighted” enters)
\[ s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}. \]
The weights are the degrees of freedom: group 1 contributes \((n_1-1)\) copies of variability information and group 2 contributes \((n_2-1)\). Equivalently, writing explicit weights:
\[ s_p^2=w_1 s_1^2+w_2 s_2^2,\qquad w_1=\frac{n_1-1}{n_1+n_2-2},\quad w_2=\frac{n_2-1}{n_1+n_2-2},\quad w_1+w_2=1. \]
Visualization: degrees-of-freedom weights in the pooled variance
Step 2: The pooled two-sample t statistic (equal variances)
After \(s_p^2\) is computed, the pooled standard deviation is \(s_p=\sqrt{s_p^2}\). The test statistic is:
\[ t=\frac{(\bar{x}_1-\bar{x}_2)-\Delta_0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}},\qquad \text{df}=n_1+n_2-2. \]
Step 3: Worked numerical example (showing the weighted formula clearly)
Two independent groups are compared (e.g., reaction-time means under two conditions). Assume equal unknown variances.
| Quantity | Group 1 | Group 2 |
|---|---|---|
| Sample size | \(n_1=12\) | \(n_2=10\) |
| Mean | \(\bar{x}_1=52\) | \(\bar{x}_2=46\) |
| Standard deviation | \(s_1=8 \Rightarrow s_1^2=64\) | \(s_2=7 \Rightarrow s_2^2=49\) |
Compute the weighted pooled variance:
\[ s_p^2=\frac{(12-1)\cdot 64+(10-1)\cdot 49}{12+10-2} =\frac{11\cdot 64+9\cdot 49}{20} =\frac{704+441}{20} =57.25. \]
\[ s_p=\sqrt{57.25}\approx 7.566. \]
Compute the standard error and the t statistic (take \(\Delta_0=0\)):
\[ \text{SE}=s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} =7.566\sqrt{\frac{1}{12}+\frac{1}{10}} =7.566\sqrt{0.08333+0.10} =7.566\sqrt{0.18333} \approx 3.240. \]
\[ t=\frac{52-46}{3.240}\approx 1.852, \qquad \text{df}=12+10-2=20. \]
Step 4: Decision rule (how the formula is used)
For a two-sided test at \(\alpha=0.05\) with \(\text{df}=20\), the usual critical value is about \(t_{0.975,20}\approx 2.086\). Since \(|t|\approx 1.852<2.086\), the observed difference is not statistically significant at the 5% level under the pooled-variance model.
Summary: the “t test formula weighted” in the pooled two-sample setting is the weighting inside \(s_p^2\), where each variance \(s_i^2\) is weighted by its degrees of freedom before forming the standard error and the t statistic.