Question

Calculate coefficient of of correlation from the following data and interpret the result.

$\begin{array}{cccccccc}NumberofYearsof& 0& 2& 4& 6& 8& 10& 12\\ SchoolingofFarmers& & & & & & & \\ AnnualYieldperArea& 4& 4& 6& 10& 10& 8& 7\\ \left(i{n}^{\prime }000\right)& & & & & & & \end{array}$

Answer 1

Let X and Y denote number of years of schooling of farmers and annual yielding per acre respectively.

Calculation of Coefficient of Correlation

$\begin{array}{ccccccc}X& x=(X-\overline{X})& X^2& Y& y=(Y-\overline{Y})& y^2& xy\\ & \overline{X}=6& & & \overline{Y}=7& & \\ 0& -6& 36& 4& -3& 9& 18\\ 2& -4& 16& 4& -3& 9& 12\\ 4& -2& 4& 6& -1& 1& 2\\ 6& 0& 0& 10& 3& 9& 0\\ 8& 2& 4& 10& 3& 9& 6\\ 10& 4& 16& 8& 1& 1& 4\\ 12& 6& 36& 7& 0& 0& 0\\ \sum X=42& & \sum X^2=112& \sum y=49& & \sum y^2=38& \sum xy=42\end{array}$
Here, N= 7, $\sum X=42,\sum X^2=112,\sum y=49,\sum y^2=38$ and $\sum xy=42$
Now, $\overline{X}=\frac{\sum X}{N}=\frac{42}{7}=6$ and $\overline{Y}=\frac{\sum Y}{N}=\frac{49}{7}=7$
$\therefore r=\frac{\sum xy}{\sqrt{\sum x^2\times \sum y^2}}=\frac{42}{\sqrt{112\times 38}}$
$=\frac{42}{4256}=\frac{42}{65.24}=0.644$
It shows a moderate degree of positive correlation between number of years of schooling of farmers and annual yield per acre.