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Chapter 8. Estimation: One or Two Populations

Unit 8.4.4. Independent samples. Case III (Satterthwaite's formula)

Assumptions. There are two samples of sizes $latex n_X,\, n_Y$, respectively, drawn from normal populations with unknown population means. Population variances $latex \sigma_X^2,\, \sigma_Y^2$ are unknown and unequal.

This is another instance of a complex and long preamble.

Preamble. Equations (8.8) continue to be true. Denote $latex D = \bar X - \bar Y$. Because of Eq. (8.8), as usual, $latex \sigma_D^2$ is estimated by $latex s_D^2 = \frac{\sigma_X^2}{n_X} + \frac{\sigma_Y^2}{n_Y}$ and instead of $latex z_0 = \frac{D - \mu_D}{\sigma_D} = \frac{\bar X - \bar Y - (\mu_X - \mu_Y)}{\sigma_D}$ we use

$latex t = \frac{\bar X - \bar Y - (\mu_X - \mu_Y)}{s_D}. \qquad (\textrm{Eq. 8.13})$

The idea employed in Case II now sounds like this: find the number of degrees of freedom $latex v$ in such a way that

$latex \chi_v^2 = \frac{v s_D^2}{\sigma_D^2}. \qquad (\textrm{Eq. 8.14})$

Suppose Eq. (8.14) is true. Taking variance of both sides and using homogeneity and Property 3 of the chi-square distribution from Sampling distribution of sample variance we get $latex 2v = V(\chi_v^2) = \left( \frac{v}{\sigma_D^2} \right)^2 V(s_D^2)$ or, rearranging,

$latex v = \frac{2 \sigma_D^4}{V(s_D^2)}. \qquad (\textrm{Eq. 8.15})$

To find $latex V(s_D^2)$ we use the result from Exercise 7.2:

$latex V(s_D^2) = \frac{1}{n_X^2} V(s_X^2) + \frac{1}{n_Y^2} V(s_Y^2) = \frac{2 \sigma_X^4}{n_X^2(n_X - 1)} + \frac{2 \sigma_Y^4}{n_Y^2(n_Y - 1)}.$

Combining this with Eq. (8.15) yields

$latex v = \frac{\sigma_D^4}{\frac{1}{n_X - 1} \left( \frac{\sigma_X^2}{n_X} \right)^2 + \frac{1}{n_Y - 1} \left( \frac{\sigma_Y^2}{n_Y} \right)^2}. \qquad (\textrm{Eq. 8.16})$

Definition 8.8 (Satterthwaite's formula). Replacing in Eq. (8.16) population variances by their sample counterparts we obtain Satterthwaite's formula for the number of degrees of freedom:

$latex v = \frac{\left( \frac{s_X^2}{n_X} + \frac{s_Y^2}{n_Y} \right)^2}{\frac{1}{n_X - 1} \left( \frac{s_X^2}{n_X} \right)^2 + \frac{1}{n_Y - 1} \left( \frac{s_Y^2}{n_Y} \right)^2}.$

Exercise. Find the confidence interval for $latex \mu_X - \mu_Y$.

Solution.

Step 1. Calculate $latex \bar X, \bar Y, s_X^2, s_Y^2, v$.

Step 2. Choose $latex \alpha$ and find the critical value of the $latex t$ statistic $latex t_{v, \alpha/2}$.

Step 3. The confidence interval is $latex P(\bar X - \bar Y - ME < \mu_X - \mu_Y < \bar X - \bar Y + ME) = 1 - \alpha$, where $latex ME = t_{v, \alpha/2} \sqrt{\frac{s_X^2}{n_X} + \frac{s_Y^2}{n_Y}}.$

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