Determine the linear correlation coefficient.

In an area of the Great Plains, records were kept on the relationship between the rainfall (in inches) and the yield of wheat (bushels per acre). Calculate the linear correlation coefficient.

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
Rainfall (in inches), $x$ & 9.4 & 7.7 & 12.3 & 11.4 & 17.7 & 9.2 & 5.9 & 14.5 & 14.9 \\
\hline
Yield (bushels per acre), $y$ & 46.5 & 42.2 & 54.8 & 55.0 & 78.4 & 45.2 & 27.9 & 72.0 & 74.8 \\
\hline
\end{tabular}
\][/tex]

A. 0.981
B. 0.998
C. 0.899
D. 0.900



Answer :

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.