Answer
The linear model is reasonable, as the graph of the regression line is a very good fit for the data points.
Work Step by Step
Plotting the data points and graphing the regression line, we would say that the regression line seems to be a good fit.
Calculating the correlation coefficient, we add another column to the table,
$\left[\begin{array}{llllll}
& S & I & SI & S^{2} & I^{2}\\
& & & & & \\
\hline & 10 & 200 & 2000 & 100 & 40,000\\
& 15 & 500 & 7500 & 225 & 250,000\\
& 20 & 600 & 12,000 & 400 & 360,000\\
& 25 & 900 & 22,500 & 625 & 810,000\\
\hline & & & & & \\
\sum & 70 & 2200 & 44,000 & 1350 & 1,460,000\\
& & & & &
\end{array}\right]$
and calculate
$r=\displaystyle \frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{n(\sum x^{2})-(\sum x)^{2}}\cdot\sqrt{n(\sum y^{2})-(\sum y)^{2}}}\approx 0.9838699101$,
which is very close to $1$, indicating a very good fit.
