565. A batch of products is accepted if the variance of the controlled size does not significantly exceed 0.2. The corrected sample variance, found from a sample of size \( n=121 \), turned out to be \( s_{x}^{2}=0.3 \). Can the batch be accepted at a significance level of 0.01?
Solution. Null hypothesis $H_{0}: \sigma^{2}=\sigma_{0}^{2}=0.2$. Alternative hypothesis $H_{1}: \sigma^{2}>0.2$.

Let's find the observed value of the criterion:

$$
\chi_{\text {Obs }}^{2}=(n-1) \cdot s_{X}^{2} / \sigma_{0}^{2}=120 \cdot 0.3 / 0.2=180 .
$$

The alternative hypothesis is $\sigma^{2}>0.2$, hence the critical region is right-tailed. Since the table in Appendix 5 does not contain the number of degrees of freedom $k=120$, we will find the critical point approximately using the Wilson-Hilferty equality:

$$
\chi_{\mathrm{kp}}^{3}(\alpha ; k)=k\left[1-2 /(9 k)+z_{\alpha} \sqrt{2 /(9 k)}\right]^{3}
$$

First, let's find (considering that $\alpha=0.01$) $z_{\alpha}=$ $=z_{0.01}$ from the equality

$$
\Phi\left(z_{0.01}\right)=(1-2 \alpha) / 2=(1-2 \cdot 0.01) / 2=0.49
$$

Using the table of the Laplace function (see Appendix 2) and linear interpolation, we find: $z_{0.01}=2.326$. Substituting $k=120, z_{\alpha}=$ $=2.326$ into the Wilson-Hilferty formula, we get $\chi_{\text {kp }}^{2}(0.01 ; 120)=$ $=158.85$. (This approximation is quite good: in more complete tables, the value 158.95 is given). Since $\chi_{\text {obs }}^{2}>\chi_{\text {kp }}^{2}$, we reject the null hypothesis. The batch cannot be accepted.