Find the quadratic regression equation for the data points (10, 2.9), (20, 4.0) and (30, 4.9), and compute the coefficient of determination R squared. Show the derivation.

Step-by-step solution

1. 1

   Realise first that no least-squares machinery is needed. A quadratic has three free parameters $a$, $b$, $c$, and there are exactly three data points. Three points with distinct $x$-values determine a unique parabola, so the least-squares fit is the interpolating parabola: the minimum possible sum of squared residuals is $0$, and it is attained. This is why the whole problem reduces to solving a $3\times3$ linear system.

2. 2

   Substitute each point into $y=ax^2+bx+c$.

   $100a+10b+c=2.9\qquad(1)$

   $400a+20b+c=4.0\qquad(2)$

   $900a+30b+c=4.9\qquad(3)$

3. 3

   Eliminate $c$ by subtracting consecutive equations. Differencing removes $c$ immediately because its coefficient is the same in all three:

   $(2)-(1):\quad300a+10b=1.1\;\Longrightarrow\;30a+b=0.11\qquad(A)$

   $(3)-(2):\quad500a+10b=0.9\;\Longrightarrow\;50a+b=0.09\qquad(B)$

4. 4

   Difference once more to find $a$, then back-substitute.

   $(B)-(A):\quad20a=-0.02\;\Longrightarrow\;a=-0.001$

   $\text{from }(A):\;30(-0.001)+b=0.11\;\Longrightarrow\;b=0.14$

   $\text{from }(1):\;-0.1+1.4+c=2.9\;\Longrightarrow\;c=1.6$

   The regression equation is

   $y=-0.001x^2+0.14x+1.6$

   The double differencing is the second difference of the data ($0.9-1.1=-0.2$ per step of $10$), which is why $a$ came out negative: the increments are shrinking, so the curve bends downward.

5. 5

   Compute the fitted values and the residual sum of squares.

   $\hat y(10)=-0.1+1.4+1.6=2.9,\quad \hat y(20)=-0.4+2.8+1.6=4.0,\quad \hat y(30)=-0.9+4.2+1.6=4.9$

   Every fitted value equals its observed value, so

   $SS_{\text{res}}=\sum(y_i-\hat y_i)^2=0$

6. 6

   Finish with $R^2$ — and check the denominator is not zero. With $\bar y=\frac{2.9+4.0+4.9}{3}=3.9333$,

   $SS_{\text{tot}}=\sum(y_i-\bar y)^2=(-1.0333)^2+(0.0667)^2+(0.9667)^2=2.0067\neq0$

   so $R^2$ is well defined and

   $R^2=1-\frac{SS_{\text{res}}}{SS_{\text{tot}}}=1-\frac{0}{2.0067}=1$

7. 7

   Read the result with the right amount of scepticism. $R^2=1$ here is a structural certainty, not evidence that the quadratic model is good. Any three points, however noisy, give $R^2=1$ for a quadratic fit, exactly as any two points give $R^2=1$ for a straight line. To learn anything about model quality you need more data points than parameters — with $n=3$ and $3$ parameters there are zero residual degrees of freedom, and the adjusted $R^2$ is undefined.