Answer: There is evidence that students from college \(P\) take longer on average. A \(90\%\) confidence interval for \(\mu_P-\mu_Q\) is \(\boxed{[0.09,\ 2.71]}\) seconds.

(i) The sample means are

\[\bar x=\frac{2620}{50}=52.4,\qquad \bar y=\frac{3060}{60}=51.0.\]

So the observed difference is

\[\bar x-\bar y=1.4.\]

Use the hypotheses

\[H_0:\mu_P=\mu_Q,\qquad H_1:\mu_P>\mu_Q.\]

Estimate the two population variances separately:

\[s_P^2=\frac{138200-2620^2/50}{49}=18.61\ldots,\]

and

\[s_Q^2=\frac{157000-3060^2/60}{59}=15.93\ldots.\]

The estimated variance of \(\bar X-\bar Y\) is

\[\frac{s_P^2}{50}+\frac{s_Q^2}{60}=\frac{18.61\ldots}{50}+\frac{15.93\ldots}{60}=0.6378\ldots.\]

Therefore the standard error is

\[\sqrt{0.6378\ldots}=0.7986\ldots.\]

The test statistic is

\[z=\frac{1.4}{0.7986\ldots}=1.75\quad\text{approximately}.\]

For a one-tailed test at the \(10\%\) significance level, the critical value is

\[z_{0.90}=1.282.\]

Since

\[1.75>1.282,\]

we reject \(H_0\). There is evidence that students from college \(P\) take longer, on average, to run 400 metres.

(ii) For a \(90\%\) confidence interval, use \(z_{0.95}=1.645\).

The interval for \(\mu_P-\mu_Q\) is

\[(\bar x-\bar y)\pm1.645(0.7986\ldots).\]

So

\[1.4\pm1.31.\]

Hence the confidence interval is

\[\boxed{[0.09,\ 2.71]}\]

seconds, to 2 decimal places.