1817. The distribution of features $X$ and $Y$ is presented in the following correlation table:

|  |  | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | $m_{x}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| -2 |  |  |  | 1 | 2 | 1 |  |  |  | 4 |  |
| -1 |  |  | 1 | 3 |  | 3 | 1 |  |  | 8 |  |
| 0 |  | 2 | 4 |  |  |  | 4 | 2 |  | 12 |  |
| 1 | 1 | 5 |  |  |  |  |  | 5 | 1 | 12 |  |
| 2 | 3 |  |  |  |  |  |  |  |  |  |  |

Find the correlation ratios $\eta_{y / x}$ and $\eta_{x / y}$ and compare them with the corresponding linear correlation coefficient.
Solution. The calculation of correlation relationships will be carried out according to the following scheme (table on p. 410):

$$
\begin{aligned}
& \bar{X}=\frac{\Sigma_{x} m_{x} x}{\Sigma_{x} m_{x}}=\frac{8}{42}=0.190 ; \bar{X}^{2}=\frac{\Sigma_{x} m_{x} x^{2}}{\sum_{x} m_{x}}= \\
& =\frac{60}{42}=1.424 ; \sigma_{X}^{2}=\overline{X^{2}}-(\bar{X})^{2}=1.388 ;
\end{aligned}
$$

$$
\begin{gathered}
\bar{V}=\frac{\Sigma m_{y} v}{\sum_{y} m_{y}}=\frac{0}{42}=0 ; \overline{V^{2}}=\frac{\Sigma m_{y} v^{2}}{\Sigma m_{y}}=\frac{302}{42}=7.190 ; \sigma_{V}^{2}=7.190 ; \\
\overline{\left(\overline{V_{x}}\right)^{2}}=\frac{\sum_{x} \frac{1}{m_{x}}\left(\sum_{y}^{\Sigma} m_{x y} v\right)^{2}}{\Sigma_{x} m_{x}}=\frac{0}{42}=0 ; \sigma^{2}\left(\overline{V_{x}}\right)=\overline{\left(\bar{V}_{x}\right)^{2}}- \\
-(\bar{V})^{2}=0-0=0 ; \\
\overline{\left.(\bar{X})_{y}\right)^{2}}=\frac{\left.\sum_{y} \frac{1}{\bar{m}_{y}}\left(\sum_{x} m_{x y}\right)^{x}\right)^{2}}{\Sigma_{y}^{\Sigma m_{y}}}=\frac{52.542}{42}=1.251 ; \sigma^{2}\left(\overline{X_{v}}\right)=\overline{\left(\overline{\bar{X}_{v}}\right)^{2}}- \\
-(\bar{X})^{2}=1.215 .
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-410.jpg?height=1747&width=1020&top_left_y=57&top_left_x=112)

From this, considering property 5, \(\eta_{g / x}=\eta_{o / x}=\sqrt{\frac{\sigma^{2}\left(\bar{V}_{\mathrm{x}}\right)}{\sigma_{V}^{2}}}=\sqrt{\frac{0}{7.190}}=0 ; \eta_{x / g}=\eta_{x / D}=\sqrt{\frac{\sigma^{2}\left(\bar{X}_{v}\right)}{\sigma_{X}^{2}}}=\sqrt{\frac{1.215}{1.388}}=\sqrt{0.875}=\) \(=0.935\). Obviously, \(\rho_{y / x}=\rho_{x / y}=0\) and \(r(X, Y)=0\).