613. Based on a sample of size $n=100$, extracted from a two-dimensional normal population ($X, Y$), a correlation table 12 has been compiled.

Table 12

| $Y$ | $x$ |  |  |  |  |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | 10 | 15 | 20 | 25 | 30 | 35 | $n_{y}$ |
| 35 | 5 | 1 | - | - | - | - | 6 |
| 45 | - | 6 | 2 | - | - | - | 8 |
| 55 | - | - | 5 | 40 | 5 | - | 50 |
| 65 | - | - | 2 | 8 | 7 | - | 17 |
| 75 | - | - | - | 4 | 7 | 8 | 19 |
| $n_{x}$ | 5 | 7 | 9 | 52 | 19 | 8 | $n=100$ |

Required: a) find the sample correlation coefficient; b) at a significance level of 0.05, test the null hypothesis that the population correlation coefficient is zero against the alternative hypothesis $H_{1}: r_{\mathrm{r}} \neq 0$.
Solution. a) To simplify the calculations, we switch to conditional variants

$$
u_{i}=\left(x_{i}-C_{1}\right) / h_{1}, \quad v_{i}=\left(y_{i}-C_{2}\right) / h_{2}
$$

where $C_{1}$ and $C_{2}$ are false zeros (it is advantageous to take a variant located approximately in the middle of the variation series; in this case, we take $C_{1}=25, C_{2}=55$); $h_{1}=u_{i+1}-u_{i}$, i.e., the difference between two adjacent variants (step); $h_{2}=v_{i+1}-v_{i}$.

Practically, the correlation table in conditional variants is compiled as follows: in the first row, instead of the false zero $C_{1}=25$, we write zero; to the left of zero, we write sequentially $-1,-2,-3$, and to the right of zero, we write $1,2,3$. Similarly, in the first column, instead of the false zero $C_{2}=55$, we write zero; above zero, we write sequentially $-1,-2,-3$, and below zero, we write $1,2,3$. The frequencies are copied from the correlation table in the original variants. As a result, we obtain the correlation table 13.

Table 13

| $\boldsymbol{V}$ | $\boldsymbol{U}$ |  |  |  |  |  | $n_{\boldsymbol{n}}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | -3 | -2 | -1 | 0 | 1 | 2 |  |
| -2 | 5 | 1 | - | - | - | - | 6 |
| -1 | - | 6 | 2 | - | - | - | 8 |
| 0 | - | - | 5 | 40 | 5 | - | 50 |
| 1 | - | - | 2 | 8 | 7 | - | 17 |
| 2 | - | - | - | 4 | 7 | 8 | 19 |
| $n_{u}$ | 5 | 7 | 9 | 52 | 19 | $\mathbf{8}$ | $=100$ |

We will use the formula for calculating the sample correlation coefficient in conditional variants:

$$
r_{B}=\left(\sum n_{u v} u v-n \bar{u} \bar{v}\right) /\left(n \sigma_{u} \sigma_{v}\right)
$$

By calculating the quantities $\bar{u}, \bar{v}, \sigma_{u}, \sigma_{v}$ entering this formula using the method of products or direct calculation, we get: $\bar{u}=-0.03 ; \bar{v}=0.35 ; \sigma_{u}=1.153 ; \sigma_{v}=1.062$.

Using the calculation table (see problem 535, table 7), we find $\sum n_{u \gamma} u v=99$.

Therefore, the sample correlation coefficient is

$$
r_{\mathrm{B}}=\frac{\sum n_{u v} u v-n \overline{u v}}{n \sigma_{n} \sigma_{v}}=\frac{99-100 \cdot(-0.03) \cdot 0.35}{100 \cdot 1.153 \cdot 1.062}=0.817
$$

b) Let's test the null hypothesis that the population correlation coefficient is zero.

We calculate the observed value of the criterion:

$$
T_{\text {obs } n}=\frac{r_{\mathrm{B}} \sqrt{n-2}}{\sqrt{1-r_{\mathrm{B}}^{2}}}=\frac{0.817 \cdot \sqrt{100-2}}{\sqrt{1-0.817^{2}}}=14.03 .
$$

According to the condition, the alternative hypothesis is $r_{\mathrm{r}} \neq 0$, so the critical region is two-sided. From the table of critical points of the Student's distribution (see appendix 6), at the significance level $\alpha=0.05$, placed in the top row of the table, and the number of degrees of freedom $k=n-2=100-2=98$, we find the critical point of the two-sided critical region $t_{\text {cr }}(0.05 ; 98)=1.99$.

Since $T_{\text {obs }}>t_{\text {cr }}$, we reject the null hypothesis that the population correlation coefficient is zero. In other words, the correlation coefficient significantly differs from zero; therefore, $X$ and $Y$ are correlated.