563. The accuracy of an automatic machine is checked by the variance of the controlled size of the products, which should not exceed $\sigma_{0}^{2}=0.1$. A sample of 25 randomly selected products was taken, and the following measurement results were obtained:

controlled size

$$
\begin{array}{lllllllll}
\text { of the sample products } & x_{1} & 3.0 & 3.5 & 3.8 & 4.4 & 4.5 \\
\text { frequency } & n_{i} & 2 & 6 & 9 & 7 & 1
\end{array}
$$

It is required to check at a significance level of 0.05 whether the machine ensures the required accuracy.
Solution. Null hypothesis $H_{0}: \sigma^{2}=\sigma_{0}^{2}=0.1$. We will take as the alternative hypothesis $H_{1}: \sigma^{2}>0.1$.

Let's find the corrected sample variance. For simplification of the calculation, we will transition to conditional variants. Taking into account that the sample mean is approximately equal to 3.9, let $u_{i}=10 x_{i}-39$. The frequency distribution takes the form

$$
\begin{array}{llllll}
u_{i} & -9 & -4 & -1 & 5 & 6 \\
n_{i} & 2 & -6 & 9 & 7 & 1
\end{array}
$$

Let's find the auxiliary corrected variance of the conditional variants:

$$
s_{u}^{2}=\frac{\sum n_{i} u_{i}^{2}-\left[\sum n_{i} u_{i}\right]^{2} / n}{n-1} ;
$$

substituting the data from the problem, we get $s_{u}^{2}=19.75$.

Let's find the required corrected variance:

$$
s^{2} X=s_{u}^{2} / 10^{2}=19.75 / 100=0.2
$$

Let's find the observed value of the criterion:

$$
\chi_{\text {Obs }}^{2}=\frac{(n-1) s_{X}^{2}}{\sigma_{0}^{2}}=\frac{(25-1) \cdot 0.2}{0.1}=48
$$

The alternative hypothesis is of the form $\sigma^{2}>\sigma_{0}^{2}$, so the critical region is one-sided (Rule 1).

Let's find the critical point $\chi_{\text {crit }}^{2}(0.05 ; 24)=$ $=36.4$ from the table in Appendix 5. We have $\chi_{\text {obs }}^{2}>\chi_{\text {crit }}^{2}$, therefore, we reject the null hypothesis; the machine does not ensure the required accuracy and requires adjustment.