To determine the linear correlation coefficient, we follow these detailed steps:
1. Organize the Data:
We have two sets of data. The rainfall in inches (let's call it [tex]\( x \)[/tex]) and the yield of wheat in bushels per acre (let's call it [tex]\( y \)[/tex]).
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline
\text{Rainfall }(x) & 9.4 & 7.7 & 12.3 & 11.4 & 17.7 & 9.2 & 5.9 & 14.5 & 14.9 \\
\hline
\text{Yield }(y) & 46.5 & 42.2 & 54.8 & 55.0 & 78.4 & 45.2 & 27.9 & 72.0 & 74.8 \\
\hline
\end{array}
\][/tex]
2. Calculate the Mean:
Find the mean (average) of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
[tex]\[
\bar{x} = \frac{\sum x_i}{n} = \frac{9.4 + 7.7 + 12.3 + 11.4 + 17.7 + 9.2 + 5.9 + 14.5 + 14.9}{9} = \frac{102.0}{9} \approx 11.33
\][/tex]
[tex]\[
\bar{y} = \frac{\sum y_i}{n} = \frac{46.5 + 42.2 + 54.8 + 55.0 + 78.4 + 45.2 + 27.9 + 72.0 + 74.8}{9} = \frac{496.8}{9} \approx 55.20
\][/tex]
3. Calculate the Covariance:
Covariance is a measure of the joint variability of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
[tex]\[
\text{cov}(x, y) = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{n - 1}
\][/tex]
4. Calculate the Standard Deviations:
The standard deviation for [tex]\( x \)[/tex] and for [tex]\( y \)[/tex] are given by:
[tex]\[
s_x = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
\][/tex]
[tex]\[
s_y = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n - 1}}
\][/tex]
5. Calculate the Correlation Coefficient:
The correlation coefficient [tex]\( r \)[/tex] is given by:
[tex]\[
r = \frac{\text{cov}(x, y)}{s_x s_y}
\][/tex]
By computing these steps, we find that the linear correlation coefficient is approximately:
[tex]\[
r \approx 0.9808
\][/tex]
Interpretation: The correlation coefficient of approximately 0.981 indicates a very strong positive linear relationship between the amount of rainfall and the yield of wheat in this dataset. The closer the value is to 1, the stronger the linear relationship.
