# Multiple means hypothesis F test

Suppose that I have m samples. Size of each sample is $n_1, n_2,... n_m$.Elements in each sample are independent N($\mu_i$,$\sigma^2$), $i=\overline{1,m}$. If I want to test hypothesis $\mu_1=\mu_2= ... =\mu_m$ I could use F test - $S_2/S_1$, where $S_1$ is variance, $S_2$ is variance under hypothesis. I am struggling to test the hypothesis $\mu_a$= $c\mu_{a-1}$. I guess it is the same hypothesis as $\mu_m=c\mu_{m-1}= c^2\mu_{m-2}... =c^{m-1}\mu_1$. Can you show me how F test is applied in this case?

## 1 Answer

It's a bit tricky. First, the easy part, the denominator: Denote $\bar{x}_i=\frac1{n_i}\sum_{j=1}^{n_i}x_{ij}$. Then $SSW=\sum_{i=1}^m\sum_{j=1}^{n_i}(x_{ij}-\bar x_i)^2$. The ratio $SSW/\sigma^2$ is distributed by $\chi^2(n_1+\dots+n_m-m)$ both under the null and the alternative hypotheses. Also, note that $SSW$ and all $\bar x_i$ are independent (this follows from the Fisher's lemma).

Now the tricky part, the numerator. If the null hypothesis holds, then $\bar x_i\sim\mathcal N(c^{i-1}\mu_1,\sigma^2/n_i)$, or if we denote $y_i=\sqrt{n_i}\bar x_i$, then $y_i=\mu_1\sqrt{n_i}c^{i-1}+\epsilon_i$, where $\epsilon_i\sim\mathcal N(0,\sigma^2)$ and independent between each other. This means we can apply the linear regression technique, get the OLS estimator $$\hat\mu_1=\frac{\sum_{i=1}^m\sqrt{n_i}c^{i-1}y_i}{\sum_{i=1}^mn_i(c^{i-1})^2}=\frac{\sum_{i=1}^mn_ic^{i-1}\bar x_i}{\sum_{i=1}^mn_i(c^{i-1})^2},$$ and get the following distribution of the sum of residuals squared: $SSR/\sigma^2\sim\chi^2(m-1)$, where $$SSR=\sum_{i=1}^m\left(y_i-\hat\mu_1\sqrt{n_i}c^{i-1}\right)^2=\sum_{i=1}^mn_i\left(\bar x_i-\hat\mu_1c^{i-1}\right)^2.$$ The final $F$-statistic is constructed as a ratio of normalized $SSR$ and $SSW$ (to eliminate unknown $\sigma^2$): $$F=\frac{SSR/(m-1)}{SSW/(n_1+\dots+n_m-m)}$$ and have the Fisher's $F(m-1,n_1+\dots+n_m-m)$ distribution under the null (since the numerator is constructed from $\{\bar x_i\}$ only, which don't depend on $SSW$).

• This is great! Thank you very much! – good_guy Dec 21 '16 at 18:02