# Finding the degrees of freedom for a confidence interval for the difference of two means

This is more a question to make sure I have a good understanding.

1. Assuming that the variance of each sample is similar i.e., homogeneity of variance and $n_1$ and $n_2$ are the same. $$\text{df} = (n_1-1) + (n_2-1)$$ also the $$\text{MSE} = \frac{s_1^2 + s_2^2}{2}$$

2. Assuming that the variance of each sample is similar ie Homogeneity of Variance but $n_1\neq n_2$ $$s_{x_1-x_2} = \sqrt{\frac{2\left(s_1^2+s_2^2\right)/ \left(n_1+n_2-2\right)}{1/n_1+1/n_2}}$$ $$\text{df}= (n_1-1) + (n_2-1)$$

3. Also, if you can assume that the two samples have the same variance then you can pool them $$s_p = \sqrt{\frac{\left(n_1-1\right)s_1^2 + \left(n_2-1\right)s_2^2}{n_1+n_2-2}}$$ with $\text{df} = n_1+n_2-2$ but have also read not to use this method

4. I have also read that to be more exact to use $$\text{df} = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1}\left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1}\left(\frac{s_2^2}{n_2}\right)^2}$$ Works well when both $n_1$ and $n_2$ are both larger than 5.

5. I have just read about Welch-Satterthwaite estimate,used when $\sigma_1 \neq \sigma_2$ $$\text{df} = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{s_1^4}{n_1^2\left(n_1-1\right)}+\frac{s_2^4}{n_2^2\left(n_2-1\right)}}$$

To be honest, I don't know which one to use. I think using 4 is good, but not sure which test it goes with. Also, why is pooling bad (and if it's so bad why put it in the book)? I am studying for the AP statistics exam.

• Man, AP tests are just getting harder and harder... or maybe students are getting smarter. – Crashworks Dec 4 '13 at 20:47