# Minimum sample size determination for difference of two normally distributed populations

This is homework, but I am rather stuck and the TA for the class has yet to make his solutions. Any help with moving me in the proper direction would be most appreciated.

I am having some difficulty determining the smallest sample size needed to detect with a probability of 90% a one-mark difference between two means $\mu_1$ and $\mu_2$, when $\alpha = 0.05$.

Secondly, I am asked the value of $\beta$ for the test performed if the difference between $\mu_1$ and $\mu_2$ is 2.

The means come from two independent random variables drawn from normal distributions (e.g., $X_1 \sim \mathcal N(\mu_1, \sigma_1^2)$ and $X_2 \sim \mathcal N(\mu_2, \sigma_2^2)$, with $n_1 \neq n_2$.

I know for two-sided alternative hypothesis with significance level $\alpha$ and $n_1 = n_2$ in order to detect a difference between the means with power $1-\beta$ is: $$n \approx \frac{(Z_{\alpha / 2} + Z_{\beta})^2(\sigma_1^2 + \sigma_2^2)}{(\delta - \delta_0)^2}$$

For $n_1 \neq n_2$ does one standardize with the sample size for each distribution? i.e., $(\sigma_1^2 + \sigma_2^2)$ turns into $(\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2})$ or does one need to worry about that?

• You might find it helpful to review the derivation of that formula for $n$. – whuber Nov 8 '12 at 21:09
• Thank you. It took a little while to find a derivation, but it did help. – James Nov 12 '12 at 14:20