Linear Regression
Calculaiton
| Week | Old process | Old | Calculations | Week | New process | New | Calculation |
| 1 | Slope | ERROR:#DIV/0! | 1 | Slope | ERROR:#DIV/0! | ||
| 2 | Intercept | ERROR:#DIV/0! | 2 | Intercept | ERROR:#DIV/0! | ||
| 3 | correlation coefficient (r) | ERROR:#DIV/0! | 3 | correlation coefficient (r) | ERROR:#DIV/0! | ||
| 4 | coefficient of determination (r^2) | ERROR:#DIV/0! | 4 | coefficient of determination (r^2) | ERROR:#DIV/0! | ||
| 5 | Average | ERROR:#DIV/0! | 5 | Average | ERROR:#DIV/0! | ||
| 6 | Lineat Regression | Y=b+mx | 6 | Linear Regression | Y=b+mx | ||
| 7 | 7 | ||||||
| 8 | 8 | ||||||
| 9 | 9 | ||||||
| 10 | 10 | ||||||
| 11 | 11 | ||||||
| 12 | 12 | ||||||
| EXAMPLE ON HOW TO INTERPRET RESULTS | |||||||
| Interpretations | These are for the new process | ||||||
| For every increase of one unit there is a decrease of 1.12 in predicted y | |||||||
| When there is an absence of x values the predicted y value is 56.0 | |||||||
| There is a negative and weak association between the variables | |||||||
| 11.2% of the variability present in predicted y can be explained by variability present in | |||||||
| the model. Another interpretation ends with th phrase: variability present in x. | |||||||
| In this example the equation for the LSRL is : y = -1.12x + 56.0 | |||||||
Old Process
Old process
New Process
New process