Question

Fit a straight-line trend by the method of least squares for the following data:

\[ \begin{array}{lcccccc} \text{Year:} & 2011 & 2012 & 2013 & 2014 & 2015 & 2016 \\ \text{Production (in tons):} & 210 & 225 & 275 & 220 & 240 & 235 \end{array} \]

Hint:

To fit a straight-line trend: code $x$ with midpoint as 0, compute $a = \bar{y}$, and $b = \dfrac{\sum xy}{\sum x^2}$.

Answer 1

Let \( x \) be the coded variable such that 2013 = 0:
So years become: \( x = -2, -1, 0, 1, 2, 3 \)
Let \( y \) = production values.

Create a table:
\[ \begin{array}{c|c|c|c} \text{Year} & x & y & xy \\ \hline 2011 & -2 & 210 & -420 \\ 2012 & -1 & 225 & -225 \\ 2013 & 0 & 275 & 0 \\ 2014 & 1 & 220 & 220 \\ 2015 & 2 & 240 & 480 \\ 2016 & 3 & 235 & 705 \\ \end{array} \]
Step 1: Compute sums:
\[ \sum x = 3, \quad \sum y = 1405, \quad \sum xy = 760, \quad \sum x^2 = 19 \]
Number of terms: \( n = 6 \)

Step 2: Apply least squares formula:
The equation of the trend line is: \[ y = a + bx \]
Calculate slope and intercept:
\[ b = \frac{\sum xy}{\sum x^2} = \frac{760}{19} = 40 \\ a = \frac{\sum y}{n} = \frac{1405}{6} \approx 234.17 \]
So, the trend equation is:
\[ y = 234.17 + 40x \]