Show that $$ \sum\left(Y_{i}-\mu\right)^{2} / \sigma^{2}=(n-1) S^{2} / \sigma^{2}+\left[(\bar{Y}-\mu)^{2} n / \sigma^{2}\right] $$

can be changed into a form $$ \frac{1}{\sigma^{2}} \widehat{S}_{1}=\frac{1}{\sigma^{2}} \sum_{j=1}^{2} \sum_{k=1}^{20}\left(Y_{j k}-\mu_{j}\right)^{2}-\frac{20}{\sigma^{2}} \sum_{k=1}^{20}\left(\bar{Y}_{j}-\mu_{j}\right)^{2} $$ and deduce that $$ \frac{1}{\sigma^{2}} \widehat{S}_{1} \sim \chi^{2}(38) $$ Similarly show that $$ \frac{1}{\sigma^{2}} \widehat{S}_{0}=\frac{1}{\sigma^{2}} \sum_{j=1}^{2} \sum_{k=1}^{20}\left(Y_{j k}-\mu\right)^{2}-\frac{40}{\sigma^{2}} \sum_{j=1}^{2}(\bar{Y}-\mu)^{2} $$ and if $\mathrm{H}_{0}$ is true then $$ \frac{1}{\sigma^{2}} \widehat{S}_{0} \sim \chi^{2}(39) $$

Provided that $$ \begin{array}{l} \hat{S}_{0}=\sum \sum\left(Y_{j k}-\bar{Y}\right)^{2}, \text { where } \bar{Y}=\sum_{j=1}^{2} \sum_{k=1}^{K} Y_{j k} / 40 \\ \hat{S}_{1}=\sum \sum\left(Y_{j k}-\bar{Y}_{j}\right)^{2}, \text { where } \bar{Y}_{j}=\sum_{k=1}^{K} Y_{j k} / 20 \end{array} $$ for $j=1,2$

Intuitively I can explain why $\hat{S_0}/\sigma^2$ would look the way it is as we can simply consider $$ \sum\left(Y_{i}-\mu\right)^{2} / \sigma^{2}=(n-1) S^{2} / \sigma^{2}+\left[(\bar{Y}-\mu)^{2} n / \sigma^{2}\right] $$ is the expression for one gender, group, so for this to be true for both gender group, we have to sum across two genders. However, I am unable to show why $ \hat{S}_1$ looks the way it is. Also, Suppose the expression is true, why Chi-square distribution is with a degree of freedom 38 and 39?

  • $\begingroup$ There are two hypotheses: $H_0:\mu_1=\mu_2=\mu$ and $H_1:\mu_1\ne\mu_2$. We have $\hat{S}_0$ when $H_0$ holds (only one overall mean, 39 degrees of freedom), $\hat{S}_1$ when $H_1$ holds (two distinct group means, 38 d.o.f.) $\endgroup$
    – Sergio
    Commented Sep 14, 2020 at 10:44
  • $\begingroup$ @Sergio Yes, but I am struggling to get this conclusion from the formula which is part of my question $\endgroup$
    – JoZ
    Commented Sep 15, 2020 at 14:23

1 Answer 1


As to $\hat{S}_1$, I think that there is a small typo in book's equations.

Let's consider the first group, $j=1$. We have (see previous question/answer): $$\sum_{k=1}^{20} (y_{1k}-\bar{y}_1)^2=\sum_{k=1}^{20}(y_{1k}-\mu_1)^2-20(\bar{y}_1-\mu_j)^2$$ To get $\hat{S}_1=\sum_{j=1}^2\sum_{k=1}^{20}(y_{jk}-\bar{y}_j)^2$, we have to sum two equations putting $j=1,2$: $$\sum_{j=1}^2\sum_{k=1}^{20} (y_{jk}-\bar{y}_j)^2=\sum_{j=1}^2\sum_{k=1}^{20}(y_{jk}-\mu_j)^2-\sum_{j=1}^2 20(\bar{y}_j-\mu_j)^2$$ Not $\sum_{k=1}^2$ in the last term. If we divide by $\sigma^2$, then on the right side we have:

  • the sum of $40$ squared standard normal variables, independent because $y$ is i.i.d., $\sim\chi^2_{40}$;
  • minus the sum of two squared standard normal variables, independent because $y$ is i.i.d. and $\mu_1\ne\mu_2$ by $H_1$, $\sim\chi^2_2$;

so the left side is $\sim\chi^2_{38}$.


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