Practise Karl Pearson’s correlation coefficient, coefficient of determination, regression equation, prediction and interpretation using an IOE-style workbook problem.
In this question, performance is predicted from experience. So experience is the independent variable \(X\), and performance is the dependent variable \(Y\).
First find \(r\) and the coefficient of determination. Then fit the regression line of performance on experience.
The regression coefficient tells how much the predicted performance changes for one additional year of experience.
\[ r=\frac{n\sum XY-\sum X\sum Y} {\sqrt{\left[n\sum X^2-(\sum X)^2\right]\left[n\sum Y^2-(\sum Y)^2\right]}} \]
\[ \text{Coefficient of Determination}=r^2\times100\% \]
\[ b=\frac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2} \]
\[ \hat{Y}=a+bX \]
The following data gives the experience of machine operators in years and their performance as given by the number of good parts turned out per 100 pieces.
| Experience \(X\) | 16 | 12 | 18 | 4 | 3 | 10 | 5 | 12 |
|---|---|---|---|---|---|---|---|---|
| Performance \(Y\) | 87 | 88 | 89 | 68 | 78 | 80 | 75 | 83 |
(a) Find Karl Pearson’s correlation coefficient and interpret it.
(b) Determine the coefficient of determination and interpret it.
(c) Fit the regression equation of performance rating on experience and estimate the probable performance of an operator who has 8 years of experience.
(d) What does the regression coefficient indicate?
First find the required sums, then calculate \(r\), \(r^2\times100\%\), \(b\), \(a\), and finally \(\hat{Y}\) for \(X=8\).
Let experience be \(X\) and performance be \(Y\).
| \(n\) | \(\sum X\) | \(\sum Y\) | \(\sum X^2\) | \(\sum Y^2\) | \(\sum XY\) |
|---|---|---|---|---|---|
| 8 | 80 | 648 | 1018 | 52856 | 6727 |
Karl Pearson’s correlation coefficient:
\[ r=\frac{n\sum XY-\sum X\sum Y} {\sqrt{\left[n\sum X^2-(\sum X)^2\right]\left[n\sum Y^2-(\sum Y)^2\right]}} \]
\[ r=\frac{8(6727)-80(648)} {\sqrt{\left[8(1018)-80^2\right]\left[8(52856)-648^2\right]}} \]
\[ r=\frac{1976}{\sqrt{1744\times2944}} \]
\[ r=0.872 \]
Interpretation of \(r\):
Since \(r=0.872\) is positive and close to \(1\), there is a strong positive linear correlation between experience and performance.
In this data, operators with more experience tend to have higher performance.
Coefficient of determination:
\[ \text{Coefficient of Determination}=r^2\times100\% \]
\[ =(0.872)^2\times100\% \]
\[ =76.05\% \]
About \(76.05\%\) of the variation in performance is explained by experience according to this fitted linear model.
Now find the regression equation of performance on experience.
\[ b=\frac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2} \]
\[ b=\frac{8(6727)-80(648)}{8(1018)-80^2} \]
\[ b=\frac{1976}{1744} \]
\[ b=1.133 \]
Also,
\[ \bar{X}=\frac{80}{8}=10,\qquad \bar{Y}=\frac{648}{8}=81 \]
Using \(a=\bar{Y}-b\bar{X}\),
\[ a=81-1.133(10) \]
\[ a=69.670 \]
The regression equation is:
\[ \hat{Y}=69.670+1.133X \]
Equivalently, after finding \(b\), we may write directly:
\[ Y-\bar{Y}=b(X-\bar{X}) \]
\[ Y-81=1.133(X-10) \]
Simplifying gives the same regression line:
\[ \hat{Y}=69.670+1.133X \]
For an operator with \(8\) years of experience, put \(X=8\):
\[ \hat{Y}=69.670+1.133(8) \]
\[ \hat{Y}=78.73 \]
The probable performance of an operator with \(8\) years of experience is about \(78.73\) good parts per 100 pieces.
Meaning of the regression coefficient:
The regression coefficient \(b=1.133\) indicates that for each additional 1 year of experience, the predicted performance increases by about \(1.133\) good parts per 100 pieces.