560. A sample of size $n=21$ has been drawn from a normal population, and the corrected sample variance $s^{2}=16.2$ has been calculated. It is required to test the null hypothesis $H_{0}$ : $\sigma^{2}=\sigma_{0}^{2}=15$ at a significance level of 0.01, using the alternative hypothesis $H_{1}: \sigma_{0}^{2}>15$.
Solution. Let's find the observed value of the criterion:

$$
\chi_{\text {obs }}^{2}=\frac{(n-1) s^{2}}{\sigma_{0}^{2}}=\frac{(21-1) \cdot 16.2}{15}=21.6 .
$$

According to the condition, the alternative hypothesis is $\sigma^{2}>15$, so the critical region is right-sided (Rule 1). From the table in Appendix 5, at a significance level of 0.01 and degrees of freedom $k=n-1=21-1=20$, we find the critical point $\chi_{\text {crit }}^{2}(0.01 ; 20)=37.6$.

Since $\chi_{\text {obs }}^{2}<\chi_{\text {crit }}^{2}$, there is no reason to reject the null hypothesis that the population variance is equal to the hypothetical value $\sigma_{0}^{2}=15$. In other words, the difference between the corrected variance $(16.2)$ and the hypothetical population variance (15) is not significant.