Answer: \(\boxed{p=6.5}\), \(\boxed{k=5}\), \(\boxed{r=0.659}\), and there is no evidence of positive correlation at the \(5\%\) level.
(i) First calculate the sums needed for the regression line.
\[\sum x=4+5+7+8+10+14=48,\]
and
\[\sum x^2=4^2+5^2+7^2+8^2+10^2+14^2=450.\]
For the \(y\)-values,
\[\sum y=5+8+p+7+p+9=29+2p.\]
Also
\[\sum xy=4(5)+5(8)+7p+8(7)+10p+14(9)=242+17p.\]
Now
\[S_{xx}=\sum x^2-\frac{(\sum x)^2}{n}=450-\frac{48^2}{6}=66.\]
And
\[S_{xy}=\sum xy-\frac{(\sum x)(\sum y)}{n}.\]
So
\[S_{xy}=242+17p-\frac{48(29+2p)}{6}.\]
Since \(48/6=8\),
\[S_{xy}=242+17p-8(29+2p)=242+17p-232-16p=10+p.\]
The gradient of the regression line of \(y\) on \(x\) is
\[\frac{S_{xy}}{S_{xx}}.\]
We are given that this gradient is \(0.25\), so
\[0.25=\frac{10+p}{66}.\]
Therefore
\[10+p=16.5,\qquad p=\boxed{6.5}.\]
Now find \(k\). With \(p=6.5\),
\[\bar y=\frac{29+2(6.5)}{6}=\frac{42}{6}=7,\]
and
\[\bar x=\frac{48}{6}=8.\]
The regression line passes through \((\bar x,\bar y)\), so
\[7=0.25(8)+k.\]
Thus
\[7=2+k,\qquad k=\boxed{5}.\]
(ii) With \(p=6.5\),
\[\sum y=42.\]
Also
\[\sum y^2=5^2+8^2+6.5^2+7^2+6.5^2+9^2=303.5.\]
So
\[S_{yy}=303.5-\frac{42^2}{6}=9.5.\]
From part (i),
\[S_{xy}=10+p=16.5,\qquad S_{xx}=66.\]
The product moment correlation coefficient is
\[r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}.\]
Therefore
\[r=\frac{16.5}{\sqrt{66(9.5)}}=0.659\quad\text{to 3 significant figures}.\]
(iii) Test
\[H_0:\rho=0,\qquad H_1:\rho>0.\]
For \(n=6\), the one-tailed \(5\%\) critical value for the product moment correlation coefficient is
\[0.729.\]
Since
\[0.659<0.729,\]
we do not reject \(H_0\). Therefore there is no evidence of positive correlation between the variables at the \(5\%\) significance level.
