3. At a significance level of 0.05, test the null hypothesis of homogeneity of two samples of sizes: $n_{1}=40$ and $n_{2}=50$ under the alternative hypothesis $H_{1}: F_{1}(x) \neq F_{2}(x)$, if it is known that in the combined variational series, composed of the variants of both samples, the sum of the ordinal numbers of the variants of the first sample $W_{\text {obs }}=1800$.
Solution. According to the condition, the competing hypothesis has $F_{1}(x)=F_{2}(x)$, so the critical region is two-sided.

Let's find $\boldsymbol{z}_{\text {kp }}$ using the equality

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

Using the Laplace function table (see Appendix 2), we find $z_{\text {kp }}=1.96$. Substituting $n_{1}=40, n_{2}=50, z_{\text {kp }}=1.96, Q=0.05 / 2=0.025$ into the formula $w_{\text {lower.crit }}\left(Q ; n_{1}, n_{2}\right)=\left[\frac{\left(n_{1}+n_{2}+1\right) n_{1}-1}{2}-z_{\mathrm{kp}} \sqrt{\frac{n_{1} n_{2}\left(n_{1}+n_{2}+1\right)}{12}}\right]$, we get $w_{\text {lower.crit }}=1578$.

Let's find the upper critical point: $w_{\text {upper.crit }}=\left(n_{1}+n_{2}+1\right) n_{1}$ $-w_{\text {lower.crit }}=(40+50+1) \cdot 40-1578=2062$. Since $w_{\text {lower.crit }}<W_{\text {obs. }}<$ $<w_{\text {upper.crit }}(1578<1800<2062)$, there are no grounds to reject the null hypothesis of sample homogeneity.