0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $SSW$).

$\endgroup$
1
  • $\begingroup$ This is great! Thank you very much! $\endgroup$ – good_guy Dec 21 '16 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.