113 I1.05: Section 3 Part 2
Example 3: Using Models.xls to fit a linear model to a dataset
The table to the right gives data on the measured rate at which sediment built up in a factory holding tank during routine operation, after a cleaning process that is repeated a couple of times per year. The factory operators want to use this information to make a formula to predict sediment depth at any chosen time after a cleaning.
A preliminary graph of the data shows that the pattern of the points is reasonably close to a straight line. Therefore, the “Linear Model” worksheet in Models.xls is the appropriate one to use.
| Days since cleaning |
Depth (mm) |
| 10 | 29.9 |
| 20 | 48.0 |
| 30 | 60.5 |
| 40 | 88.6 |
| 50 | 102.9 |
| 60 | 114.1 |
| 70 | 141.1 |
| 80 | 149.5 |
Solution:
In an earlier topic, you used a spreadsheet to adjust the intercept and slope of a linear equation and saw the resulting changes in the position of the straight-line graph. We will now use that same technique to make a good linear model for this data with Models.xls.
- Insert a new worksheet into Models.xls, labeling its tab “Linear Sediment Model”. Then copy into the new worksheet the contents of the read-only worksheet labeled “Linear Model Template”.
- In this case we want to predict sediment depth for any given number of days since the last cleaning. This means that we want to use day as the input variable x and depth as the output variable y.
- Copy the data to the spreadsheet (columns A and B, rows 3 to 10 for the numbers), then label the top of the data columns with “Days” in A2 and “Depth” in B2.
- Select C3 (which contains a preset linear formula based on the values in G3 and G4) and spread the formula down to row 10, matching the data. At first, these model values will be zeros.
- Also select and spread D3 and E3 down to row 10. The values in columns D and E will not be very meaningful until you adjust the model to be a good fit.
- Make a scatter plot of the data and model columns together (that is, the rectangle A1:C10). At first, the model points will lie on a horizontal line along the x-axis.
- Adjust the parameters in G3 and G4 so that the model points are as close as you can get them to the data points.
For a linear model, here is a good parameter-adjustment strategy:
- Set the intercept to approximately where the data trend crosses the y axis (about 10 in this case, although you do not need to be exactly right since you will adjust the intercept again in [c] below),
- Adjust the slope to make the model line parallel to the data trend (in this case, 1 is too low a value for the slope, and 2 is too high; 1.8 seems about right).
- Now adjust the intercept to its best value, moving the model line without changing its slope until the model goes right through the data (in this case, a value of 11 for the intercept works well).
- Check to see if the model is good. In this case, the model points are close to the data points over the whole data range, showing that a linear model is the correct type to use for this data.
- Write the mathematical formula for the model you have found: y = 1.8 x + 11
The Linear Sediment Model worksheet should look about like this at the end of this solution process:
| A | B | C | D | E | F | G | ||
| 1 | x | y data | y model | Data-Model | Linear model: y = m * x + b | |||
| 2 | Days | Depth | Prediction | deviation | y = 1.8x+11 | |||
| 3 | 10 | 29.9 | 29 | 0.9 | 11 | b: Intercept | ||
| 4 | 20 | 48.0 | 47 | 1 | 1.8 | m: Slope | ||
| 5 | 30 | 60.5 | 65 | -4.5 | ||||
| 6 | 40 | 88.6 | 83 | 5.6 | ||||
| 7 | 50 | 102.9 | 101 | 1.9 | ||||
| 8 | 60 | 114.1 | 119 | -4.9 | ||||
| 9 | 70 | 141.1 | 137 | 4.1 | ||||
| 10 | 80 | 149.5 | 155 | -5.5 | ||||
| 11 | ||||||||
| 12 | ||||||||
| 13 | ||||||||
