Answer: The regression line is \(\boxed{x=1.88y-0.721}\), the correlation coefficient is \(\boxed{0.931}\), and there is evidence of positive correlation.

(i) The regression line of \(y\) on \(x\) passes through \((\bar{x},\bar{y})\). We are given \(\bar{y}=6.24\), so

\[6.24=1.1664+0.4604\bar{x}.\]

Hence

\[\bar{x}=\frac{6.24-1.1664}{0.4604}=11.02.\]

Since \(n=10\),

\[\sum x=10\bar{x}=110.2,\qquad \sum y=10\bar{y}=62.4.\]

Now calculate

\[S_{xx}=\sum x^2-\frac{(\sum x)^2}{n}=1419.98-\frac{110.2^2}{10}=205.576,\]

and

\[S_{yy}=\sum y^2-\frac{(\sum y)^2}{n}=439.68-\frac{62.4^2}{10}=50.304.\]

The slope of the regression line of \(y\) on \(x\) is

\[\frac{S_{xy}}{S_{xx}}=0.4604.\]

So

\[S_{xy}=0.4604S_{xx}=0.4604(205.576)=94.647.\]

The slope of the regression line of \(x\) on \(y\) is

\[b=\frac{S_{xy}}{S_{yy}}=\frac{94.647}{50.304}=1.8815\ldots.\]

Therefore the line is

\[x-11.02=1.8815(y-6.24).\]

So

\[x=1.88y-0.721\]

to 3 significant figures.

(ii) The product moment correlation coefficient is

\[r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{94.647}{\sqrt{205.576\cdot50.304}}.\]

Hence

\[r=\boxed{0.931}.\]

(iii) Test

\[H_0:\rho=0,\qquad H_1:\rho>0.\]

For \(n=10\) at the \(5\%\) significance level in a one-tailed test, the critical value is

\[r=0.549.\]

Since

\[0.931>0.549,\]

we reject \(H_0\). There is evidence of positive correlation between the two variables.