Comparing two population means

What you’ll learn

• Find the standard error of a difference of independent means.
• Test a stated difference using known variances.
• Recognise the large-sample extension and the independence requirement.

Understand the idea

For two independent samples, the difference of sample means estimates the difference of population means. Its variance is the sum of the two sample-mean variances. Subtract any hypothesised difference from the observed difference before standardising.

With independent Normal populations and known variances, the resulting statistic is exactly standard Normal under the null. Large samples allow the population variances to be replaced by their separate sample estimates and can extend the method to non-Normal populations. Keep each variance paired with its own sample size.

Measurements before and after treatment on the same people are paired, so the two sample means are not independent. The independent-sample standard error is then unjustified. The design matters as much as the arithmetic; conclusions should address the population means and the chosen direction of the alternative.

Key formulae

$$\operatorname{Var}(\bar X-\bar Y)=\frac{\sigma_X^2}{n_X}+\frac{\sigma_Y^2}{n_Y}$$

$$Z=\frac{(\bar X-\bar Y)-\delta_0}{\sqrt{\sigma_X^2/n_X+\sigma_Y^2/n_Y}}$$

$$\widehat{\operatorname{SE}}(\bar X-\bar Y)=\sqrt{\frac{s_X^2}{n_X}+\frac{s_Y^2}{n_Y}}$$

Worked example

Independent Normal samples have means 42 and 39, sizes 36 and 64, and known population standard deviations 6 and 8 respectively. Test equality of the population means against a two-sided alternative at 5%, using 1.96.

1. State $H_0:\mu_X-\mu_Y=0$ and $H_1:\mu_X-\mu_Y\ne0$.
2. The standard error is $\sqrt{6^2/36+8^2/64}=\sqrt2$.
3. The statistic is $(42-39)/\sqrt2=2.121$.
4. Because $2.121>1.96$, reject equality of the means at 5%.

Answer: There is evidence at 5% that the population means differ.