574. From a normal general population with a known standard deviation $\sigma=5.2$, a sample of size $n=100$ was extracted, and the sample mean $\bar{x}=27.56$ was found. It is required to test the null hypothesis $H_{0}: a=$ $=a_{0}=26$ against the alternative hypothesis $H_{1}: a \neq 26$ at a significance level of 0.05.
Solution. Let's find the observed value of the criterion:

$$
U_{\text {obs } 6 \pi}=\left(\bar{x}-a_{0}\right) \cdot \sqrt{n} / \sigma+(27.56-26) \cdot \sqrt{100} / 5.2=3 .
$$

According to the condition, the alternative hypothesis is $a \neq a_{0}$, so the critical region is two-sided.

Let's find the critical point from the equality

$$
\Phi\left(u_{\mathrm{cp}}\right)=(1-\alpha) / 2=(1-0.05) / 2=0.475
$$

From the Laplace function table (see Appendix 2), we find $u_{\text {cp }}=1.96$.

Since $U_{\text {obs }}>u_{\text {cp }}$-we reject the null hypothesis. In other words, the sample and hypothetical population means differ significantly.