Answer
The linear model is reasonable, as the graph of the regression line is a 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,
$\left[\begin{array}{llllll}
& S & E & SE & S^{2} & E^{2}\\
& & & & & \\
\hline & 10 & 2 & 20 & 100 & 4\\
& 15 & 2.5 & 37.5 & 225 & 6.25\\
& 20 & 3 & 60 & 400 & 9\\
& 25 & 4.5 & 112.5 & 625 & 20.25\\
\hline & & & & & \\
\sum & 70 & 12 & 230 & 1350 & 39.5\\
& & & & &
\end{array}\right]$
and calculate
$r=\displaystyle \frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{n(\sum x^{2})-(\sum x)^{2}}\cdot\sqrt{n(\sum y^{2})-(\sum y)^{2}}}\approx 0.956182887468$,
which is very close to $1$, indicating a good fit.
