Probability and Statistics > T-Distributions > Satterthwaite approximation

Before you read this article, you may want to read this one first:

Pooled standard error (how to find it).

Watch the video or read the article below:

## What is the Satterthwaite Approximation?

The Satterthwaite approximation is a way to account for two different sample variances. Basically, there are two ways to account for two sample variances:

- Use the pooled standard error formula: S
_{p}√ (1/n_{1}+ 1/n_{2}) - Use Satterthwaite’s: S
_{e}= √ (s_{1}^{2}/n_{1}+ s_{2}^{2}/n_{2})

The two formulas are essentially equivalent — they will both give exactly the same answer. However, there are significant differences when the variances for the two samples are different. The pooled SE can only be used when your variances are equal — which almost never happens in real life. When your variances are not equal, use the Satterthwaite approximation. In fact, the Satterthwaite is **always** correct, so you may want to consider always using it over the pooled SE.

## Satterthwaite Approximation: Steps

**Sample problem:** Use the Satterthwaite approximation for the following sample data:

Sample 1: s=20, n=50.

Sample 2: s=15, n=40.

Step 1: **Insert your data into the formula. **Note that:

n_{1} is the sample size from the first sample; n_{2} is the sample size from the second sample;

s_{1} is the standard deviation from the first sample; s_{1}^{2}is the variance.

s_{2} is the standard deviation from the first sample; s_{2}^{2}is the variance.

S_{e} = √ (20^{2}/50 + 15^{2}/40)

Step 2: **Solve**:

S_{e} = √ (8 + 5.625) ≈ 3.691

The Satterthwaite approximation is roughly equal to 3.691.

*That’s it!*

I believe there is a mistake in your formula.

The S1 and S2 are the standard deviations of samples S1 and S2 and not the variances.

Hello, Emilio,

Thanks for catching that! I have updated the formula.