Pooled variance
Definition taught in Confidence intervals, part of Statistical inference
Pooled variance
| Type | Definition |
| Taught in | Confidence intervals |
| Part | Statistical inference |
| Properties | 1 |
| Worked examples | 2 |
| Questions using it | 2 |
For two independent samples assumed to share a variance \sigma^2, the pooled sample variance is
s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}.
It is unbiased for \sigma^2, and (n_1+n_2-2)s_p^2/\sigma^2 is \chi^2_{n_1+n_2-2}.
Properties
Two-sample t pivot
Main article: Two-sample t pivot
For two independent normal samples with a common variance,
T=\frac{(\overline X_1-\overline X_2)-(\mu_1-\mu_2)}{s_p\sqrt{1/n_1+1/n_2}}
has a t distribution with n_1+n_2-2 degrees of freedom.
Examples
First- and third-year weights
Worked example from Estimation examples
15 first-year students have mean weight 62.0 kg and standard deviation 15 kg; 10 third-year students have mean 71.5 kg and standard deviation 12 kg.
Question. Give a 95% interval for the difference in mean weights without assuming equal variances.
Solution. The standard error is \sqrt{15^2/15+12^2/10}=5.422, and the effective degrees of freedom are 22.10, so t_{22.10;0.975}=2.073. The interval is
-9.5\pm2.073\times5.422=(-20.74,\ 1.74).
It contains 0, so the data do not show a difference in mean weight. The pooled interval, which assumes equal variances, is (-21.24,\ 2.24) on 23 degrees of freedom, a little wider here because the larger sample has the larger variance. With the raw data, t.test(x, y) gives the Welch interval, R’s default, and t.test(x, y, var.equal = TRUE) the pooled one.
Lost hours after a safety programme
Question page: Lost hours after a safety programme
Lost working hours averaged 91 per month over 50 months after a safety programme, with standard deviation 14.1, and 108 per month over the 50 months before, with standard deviation 14.3. Assume two independent normal samples with a common variance.
Compute the pooled standard deviation and a 95% confidence interval for \mu_1-\mu_2. Use t_{98;0.975}=1.984.
Someone reports the result as -17\pm2.84. What coverage does that interval have?
- The pooled variance is
s_p^2=\frac{49\times14.1^2+49\times14.3^2}{98}=\frac{14.1^2+14.3^2}{2}=201.65,\qquad s_p=14.20.
The standard error of \overline X_1-\overline X_2 is 14.20\sqrt{1/50+1/50}=14.20/5=2.84. The 95% interval is
-17\pm1.984\times2.84=-17\pm5.64=(-22.6,\ -11.4).
- -17\pm2.84 is the estimate plus or minus one standard error. With 98 degrees of freedom the coverage is \mathbb{P}(|T|<1) for T\sim t_{98}, about 0.68. It is a 68% interval, not a 95% one.
Used in
- Taught in Confidence intervals, where the derivation and examples are.
- Also used in Hypothesis tests
- Questions that use it (2) — Lost hours after a safety programme (answer) · Two groups (answer)
- All 6 questions on Confidence intervals
See also
Confidence interval · Pivot · One-sample t pivot · Two-sample t pivot · Variance-ratio F pivot · Wald interval