Answer: The regression line is \(\boxed{y=0.782x+4.93}\), and there is not enough evidence to support the coach's belief.

(i) For the regression line of \(y\) on \(x\), use

\[y-\bar{y}=b(x-\bar{x}),\]

where

\[b=\frac{S_{xy}}{S_{xx}}.\]

The means are

\[\bar{x}=\frac{191}{8}=23.875,\qquad \bar{y}=\frac{188.8}{8}=23.6.\]

Now calculate

\[S_{xy}=4510.99-\frac{191(188.8)}{8}=3.39,\]

and

\[S_{xx}=4564.46-\frac{191^2}{8}=4.335.\]

So

\[b=\frac{3.39}{4.335}=0.782.\]

Thus

\[y-23.6=0.782(x-23.875).\]

Expanding,

\[\boxed{y=0.782x+4.93}.\]

(ii) Let

\[d=x-y.\]

The coach believes that the mean decrease is more than \(0.2\) seconds, so use

\[H_0:\mu_d=0.2,\qquad H_1:\mu_d>0.2.\]

The differences are

\[0.3,\ 0.2,\ 0,\ 0.6,\ 0.3,\ 0.5,\ 0.2,\ 0.1.\]

Their sum is \(2.2\), so

\[\bar{d}=\frac{2.2}{8}=0.275.\]

Also,

\[\sum d^2=0.88.\]

The unbiased estimate of the variance is

\[s^2=\frac{0.88-2.2^2/8}{7}=0.0393.\]

So

\[s=\sqrt{0.0393}.\]

The test statistic is

\[t=\frac{\bar{d}-0.2}{s/\sqrt8}.\]

Substitute the values:

\[t=\frac{0.275-0.2}{\sqrt{0.0393}/\sqrt8}=1.07.\]

There are \(8-1=7\) degrees of freedom. For a one-tailed \(10\%\) test, the critical value is

\[t_{7,0.90}=1.415.\]

Since

\[1.07<1.415,\]

we do not reject \(H_0\). Therefore there is not enough evidence, at the \(10\%\) level, to support the coach's belief.