Question

Three points in the $x$–$y$ plane are $(-1,0.8)$, $(0,2.2)$, and $(1,2.8)$. The value of the slope of the best fit straight line in the least square sense is \underline{\hspace{1cm}} (Round off to 2 decimal places).

Hint:

In least squares fitting, always center data first by checking $\sum x$. If $\sum x=0$, slope reduces to $\frac{\sum xy}{\sum x^2}$. This simplifies calculations greatly.

Answer 1

Step 1: Recall formula for slope in least squares fit.
\[ m = \frac{N \sum xy - \sum x \sum y}{N \sum x^{2} - (\sum x)^{2}} \]

Step 2: Compute summations.
Points: $(-1,0.8), (0,2.2), (1,2.8)$ \[ \sum x = -1 + 0 + 1 = 0 \] \[ \sum y = 0.8 + 2.2 + 2.8 = 5.8 \] \[ \sum x^2 = (-1)^2 + 0^2 + 1^2 = 2 \] \[ \sum xy = (-1)(0.8) + (0)(2.2) + (1)(2.8) = -0.8 + 0 + 2.8 = 2.0 \] $N=3$.

Step 3: Apply formula.
\[ m = \frac{3(2.0) - (0)(5.8)}{3(2) - (0)^2} = \frac{6}{6} = 1.0 \] % Final Answer \[ \boxed{1.00} \]