13.1.3. By the $\chi^{2}$ criterion, test the hypothesis about the normal distribution of the general population at the significance level $\alpha=0.05$.
Solution. We will test the hypothesis about the normal distribution of the general population using the $\chi^{2}$ (Pearson) criterion at $\alpha=0.05$.

The basis of the criterion is the comparison of frequencies $m_{i}$ and theoretical frequencies $m_{i}^{T}$, calculated under the assumption of a normal distribution of the general population. The Pearson criterion does not definitively confirm the correctness or incorrectness of the hypothesis, but only establishes its agreement or disagreement with the sample data at the given significance level. The criterion is chosen as the quantity

$$
\chi^{2}=\sum_{i=1}^{7} \frac{\left(m_{i}-m_{i}^{T}\right)^{2}}{m_{i}^{T}}
$$

Its value is compared with the critical value $\chi_{\text {kp }}^{2}$, determined from the corresponding table of values at the given significance level $\alpha=0.05$ and the number of degrees of freedom $k=p-1-r$, where $p$ is the number of intervals, $r$ is the number of parameters of the normal distribution law. In this case, $p=7 ; r=2 ; k=4$.

From the $\chi^{2}$ distribution table with $k=7-2-1=4$ degrees of freedom at the significance level $\alpha=0.05$, we find $\chi_{\mathrm{kp}}^{2}=9.49$.

If the inequality $\chi^{2}\chi_{\text {kp }}^{2}$ is satisfied as a result of the calculations, the hypothesis is rejected. We will apply the Pearson criterion to this sample. For this, we will create a calculation table, finding the theoretical frequencies $m_{i}^{T}$ for the normal distribution using the formula

$$
m_{i}^{T}=\frac{N h_{x}}{\sigma_{x}} \cdot \phi \frac{x_{i}-\bar{X}}{\sigma_{x}}
$$

where $\phi(z)=\frac{1}{\sqrt{2 \pi}} \cdot e^{-x^{2} / 2}$.

| $x_{i}$ | $z_{i}=\frac{x_{i}-\bar{X}}{\sigma_{x}}$ | $\phi\left(z_{i}\right)$ | $m_{i}^{T}$ | $m_{i}$ | $m_{i}-m_{i}^{T}$ | $\frac{\left(m_{i}-m_{i}^{T}\right)^{2}}{m_{i}^{T}}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0.2 | $-1.97$ | 0.06 | 4.42 | 5 | 0.58 | 0.076 |
| 1.4 | $-1.27$ | 0.20 | 13.73 | 13 | $-0.73$ | 0.038 |
| 2.6 | $-0.57$ | 0.37 | 26.15 | 25 | $-1.15$ | 0.051 |
| 3.8 | 0.13 | 0.44 | 30.58 | 25 | $-5.58$ | 1.017 |
| 5 | 0.82 | 0.31 | 21.94 | 19 | $-2.94$ | 0.393 |
| 6.2 | 1.52 | 0.14 | 9.66 | 10 | 0.34 | 0.012 |
| 7.4 | 2.22 | 0.04 | 2.61 | 3 | 0.39 | 0.059 |
| $\Sigma$ |  |  |  |  |  | 1.65 |

By summing the numbers in the last column of the table, we get $\chi^{2}=1.65$.

Since $\chi^{2}<\chi_{\text {kp }}^{2}$, the hypothesis about the normal distribution of the general population is accepted.

Thus, at the significance level $a=0.05$, it can be considered that the general population is distributed according to the normal law with parameters $a=\bar{X}=3.58 ; \sigma_{x}=1.72$.

Answer. $\bar{X}=3.584 ; D_{x}=2.949 ; \sigma_{x}=1.72 ;$ the hypothesis about the normal distribution of the general population is accepted.

## 13.2. Construction of the Regression Line Equation

A two-dimensional sample of the results of joint measurements of features $x$ and $y$ with a volume of $N=100$ measurements is given by the correlation table:

| $i$ | $j$ | 1 | 2 | 3 | 4 | 5 | $m_{x_{i}}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | $y_{j}$ | 0.5 | 1.3 | 2.1 | 2.9 | 3.7 |  |
| 1 | 0.2 | 2 | 3 |  |  |  | 5 |
| 2 | 1.4 | 3 | 8 | 2 |  |  | 13 |
| 3 | 2.6 |  | 9 | 16 |  |  | 25 |
| 4 | 3.8 |  |  | 15 | 10 |  | 25 |
| 5 | 5 |  |  | 9 | 10 |  | 19 |
| 6 | 6.2 |  |  | 3 | 6 | 1 | 10 |
| 7 | 7.4 |  |  |  | 1 | 2 | 3 |
| $m_{y_{j}}$ |  | 5 | 20 | 45 | 27 | 3 | $N=100$ |