Answer: (a) Reject \(H_0\). There is sufficient evidence at the \(5\%\) significance level to suggest that there is a difference in the average marks of students from schools \(A\) and \(B\).

(b) One required assumption is that the underlying distributions of marks for the two schools have the same shape.

(a) Let \(W\) be the sum of the ranks for the \(123\) students from school \(A\). We test

\(H_0:\) the population median mark for school \(A\) is equal to the population median mark for school \(B\).

\(H_1:\) the population median mark for school \(A\) is not equal to the population median mark for school \(B\).

Here \(n_A=123\), \(n_B=147\), and \(N=270\). Under \(H_0\), the rank sum for school \(A\) has

\(\operatorname{E}(W)=\frac{123(270+1)}{2}=16666.5\),

and

\(\operatorname{Var}(W)=\frac{123\times147\times(270+1)}{12}=408329.25\).

The observed rank sum is \(W=15355\). Since this is below the mean, using a continuity correction gives

\(z=\frac{15355+0.5-16666.5}{\sqrt{408329.25}}=-2.05\text{ approximately}.\)

This is a two-tailed test at the \(5\%\) level, so the critical values are \(\pm 1.96\). Since

\(-2.05\lt -1.96\),

the result is significant and we reject \(H_0\).

Therefore, there is sufficient evidence at the \(5\%\) significance level to suggest that the average marks of students from schools \(A\) and \(B\) are different.

(b) One assumption required is that the underlying distributions of marks for the two schools have the same shape, so that a comparison of location is meaningful.