567. From two independent samples, the sizes of which are $n=40$ and $m=50$, extracted from normal populations, the sample means are found to be $\bar{x}=130$ and $\bar{y}=140$. The population variances are known: $D(X)=80$, $D(Y)=100$. It is required to test the null hypothesis $H_{0}: M(X)=M(Y)$ against the alternative hypothesis $H_{1}: M(X) \neq M(Y)$ at a significance level of 0.01.
Solution. Let's find the observed value of the criterion:

$$
z_{\text {obs }}=\frac{\bar{x}-\bar{y}}{\sqrt{D(X) / n+D(Y) / m}}=\frac{130-140}{\sqrt{80 / 40+100 / 50}}=-5
$$

According to the condition, the alternative hypothesis is $M(X) \neq M(Y)$, so the critical region is two-sided.

Let's find the right critical point from the equality

$$
\Phi\left(z_{\mathrm{cr}}\right)=(1-\alpha) / 2=(1-0.01) / 2=0.495
$$

From the Laplace function table (see Appendix 2), we find

$$
z_{\mathrm{cr}}=2.58
$$

Since $\left|Z_{\text {obs }}\right|>z_{\text {cr }}$, according to rule 1, we reject the null hypothesis. In other words, the sample means differ significantly.