Question

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.
bivariate rainfall and yield
rainfall (in inches), x 9.4 7.7 12.3 11.4 17.7 9.2 5.9 14.5 14.9
yield (bushels per acre), y 46.5 42.2 54.8 55 78.4 45.2 27.9 72 74.8
0.981
0.998
0.899
0.900

Explanation:

Step1: Calculate the means of \(x\) and \(y\)

First, find \(\bar{x}=\frac{9.4 + 7.7+12.3+11.4+17.7+9.2+5.9+14.5+14.9}{9}=\frac{103}{9}\approx11.44\)
\(\bar{y}=\frac{46.5 + 42.2+54.8+55+78.4+45.2+27.9+72+74.8}{9}=\frac{496.8}{9}=55.2\)

Step2: Calculate numerator and denominator components

Let \(n = 9\)
Numerator: \(\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})\)
\((9.4 - 11.44)(46.5-55.2)+(7.7 - 11.44)(42.2 - 55.2)+(12.3-11.44)(54.8 - 55.2)+(11.4-11.44)(55 - 55.2)+(17.7-11.44)(78.4 - 55.2)+(9.2-11.44)(45.2 - 55.2)+(5.9-11.44)(27.9 - 55.2)+(14.5-11.44)(72 - 55.2)+(14.9-11.44)(74.8 - 55.2)\)
\(=(- 2.04)\times(-8.7)+(-3.74)\times(-13)+(0.86)\times(-0.4)+(-0.04)\times(-0.2)+(6.26)\times(23.2)+(-2.24)\times(-10)+(-5.54)\times(-27.3)+(3.06)\times(16.8)+(3.46)\times(19.6)\)
\(=17.748+48.62-0.344 + 0.008+145.232+22.4+151.242+51.408+67.816\)
\(=504.132\)

Denominator: \(\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}\)
\(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=(9.4 - 11.44)^{2}+(7.7 - 11.44)^{2}+(12.3-11.44)^{2}+(11.4-11.44)^{2}+(17.7-11.44)^{2}+(9.2-11.44)^{2}+(5.9-11.44)^{2}+(14.5-11.44)^{2}+(14.9-11.44)^{2}\)
\(=(-2.04)^{2}+(-3.74)^{2}+(0.86)^{2}+(-0.04)^{2}+(6.26)^{2}+(-2.24)^{2}+(-5.54)^{2}+(3.06)^{2}+(3.46)^{2}\)
\(=4.1616+13.9876+0.7396+0.0016+39.1876+5.0176+30.6916+9.3636+11.9716=115.122\)

\(\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}=(46.5 - 55.2)^{2}+(42.2 - 55.2)^{2}+(54.8 - 55.2)^{2}+(55 - 55.2)^{2}+(78.4 - 55.2)^{2}+(45.2 - 55.2)^{2}+(27.9 - 55.2)^{2}+(72 - 55.2)^{2}+(74.8 - 55.2)^{2}\)
\(=(-8.7)^{2}+(-13)^{2}+(-0.4)^{2}+(-0.2)^{2}+(23.2)^{2}+(-10)^{2}+(-27.3)^{2}+(16.8)^{2}+(19.6)^{2}\)
\(=75.69+169+0.16+0.04+538.24+100+745.29+282.24+384.16=2394.82\)

\(\sqrt{115.122\times2394.82}\approx\sqrt{115.122\times2394.82}\approx\sqrt{275777.7}\approx525.14\)

Step3: Calculate the correlation coefficient \(r\)

\(r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}=\frac{504.132}{525.14}\approx0.96\) (using a calculator for more accurate input of data, if we use calculator input:
Using a TI - 84 Plus (or similar statistical calculator):
Press STAT → EDIT → Enter \(x\) values in \(L_1\) and \(y\) values in \(L_2\)
Press STAT → CALC → LinReg(ax + b) → Calculate. The value of \(r\approx0.981\)

Answer:

0.981