Example 4: Find a good linear model for this data, redefining the input parameter as needed, and use the model to predict sales in 2010.
Solution:
- Redefine the input parameter to “Years since 1990”, since that will cause the beginning of the data to have an input parameter of zero. To do this, copy the data from the table to a scratch-pad worksheet, make a column containing the Year data with 1990 subtracted, and copy the modified table to a copy of the linear model template.
- Since the first row now has a zero input variable, use the Sales figure for that row, 453, as the initial setting for the model intercept parameter in cell G3.
- Adjust the slope parameter in cell G4 until the model points are parallel to the trend of the data. A slope value of 26 is about right, but other nearby values would also be okay if the line looks correct.
- Readjust the intercept value to ensure that the model goes through the middle of the data. Increasing it to 460 improves the fit (but again, a nearby value is okay if the graph shows a good fit).
- The implied linear model is y = 26 x + 460.
- Thus the prediction of the model for sales in 2010 (20 years after 1990) is 980.
|
Year |
House
Sales |
1990 |
426 |
1991 |
517 |
1992 |
500 |
1993 |
558 |
1994 |
611 |
1995 |
558 |
1996 |
601 |
1997 |
596 |
1998 |
683 |
1999 |
693 |
2000 |
708 |
2001 |
761 |
2002 |
771 |
2003 |
831 |
2004 |
897 |
2005 |
822 |
2006 |
889 |
|