 Solution

Four groups have $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ pairwise comparisons. Testing each at test size $\alpha =0.05$ without adjustment generally increases the familywise error rate, the chance of at least one type I error. If the six rejection events were independent, that chance would be $1-(1-0.05{)}^6\simeq 0.265$; shared samples make the actual comparisons dependent, so this expression is illustrative rather than exact.

A single omnibus ANOVA tests equality of all four means with one stated test size, avoiding that six-test inflation for the global question. If the aim is to identify differing pairs, follow with suitably adjusted comparisons, such as Bonferroni correction or a simultaneous pairwise method. A significant omnibus test does not automatically license six unadjusted claims. The benefit concerns control of type I error, rather than a guarantee that the model's independence and common-variance assumptions hold.