627. At a significance level of 0.01, test the null hypothesis $H_{0}: F_{1}(x)=F_{2}(x)$ about the homogeneity of two samples, the sizes of which are $n_{1}=6, n_{2}=7$ (the first row contains the variants of the first sample; the second row contains the variants of the second sample):

$$
\begin{array}{rrrrrrrr}
x_{i} & 3 & 4 & 6 & 10 & 13 & 17 & \\
y_{i} & 1 & 2 & 5 & 7 & 16 & 20 & 22
\end{array}
$$

Accept as the alternative hypothesis $H_{1}$ : $F_{1}(x) \neq F_{2}(x)$.
Solution. The competing hypothesis has $F_{1}(x) \neq F_{3}(x)$, so the critical region is two-sided. Arrange the variants of both samples in the form of a single variational series and number them:

| ordinal |  |  |  |  |  |  |  |  |  |  |  |  |  |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| number | 1 | 2 | 4 | 5 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |  |  |
| variant | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 10 | 13 | 16 | 17 | 20 | 22 |

Find the observed value of the Wilcoxon criterion - the sum of the ordinal numbers (they are in italics) of the variants of the first sample:

$$
W_{\text {obs }}=3+4+6+8+9+11=41
$$

Find the lower critical point of the critical region from the table, considering that $Q=0.01 / 2=0.005, n_{1}=6, n_{2}=7$: $\boldsymbol{\omega}_{\text {lower.crit }}(0.005 ; 6,7)=24$. Find the upper critical point: $w_{\text {upper.crit }}=\left(n_{1}+n_{2}+1\right) n_{1}-\omega_{\text {lower.crit }}(6+7+1) \cdot 6-24=60$. Since $w_{\text {lower.crit }}<W_{\text {obs. }}<w_{\text {upper.crit }}(24<41<60)$-there is no basis to reject the null hypothesis of sample homogeneity.