Answer: The least possible value is \(T=90\).

Under the null hypothesis that the median is still \(m_0\), the Wilcoxon signed-rank statistic has, for \(n=28\), mean and variance

\(\mu=\frac{n(n+1)}{4}=\frac{28\times29}{4}=203\),

\(\sigma^2=\frac{n(n+1)(2n+1)}{24}=\frac{28\times29\times57}{24}=1928.5\).

Since the researcher believes the median has changed, the test is two-tailed. At a \(1\%\) significance level, the lower critical standard normal value is approximately \(-2.576\).

To find the least value of \(T\) which would not lead to rejection, use the lower-tail boundary with a continuity correction:

\(\frac{T+0.5-203}{\sqrt{1928.5}}\gt -2.576\).

Solving,

\(T+0.5\gt 203-2.576\sqrt{1928.5}\),

\(T\gt 89.4\) approximately.

Since \(T\) is an integer, the least possible value is therefore \(T=90\).