Many times, we want to summarize data by constructing a function which "fits" it. The original data might not represent a mathematical function, but we might find a function which models the data.
In this topic, we consider linear and exponential functions as models.
Example 6.1.a:
Consider the table of actual votes, in thousands (rounded), cast for U.S. President
Year |
1968 | 1972 | 1976 |
Total Votes (in thousands) |
74,000 | 78,000 | 82,000 |
Show that the relationship between variables is linear.
Solution:
With Year as the explanatory variable, and Total Votes
as the response variable, we check that the slope (rate of change) between successive
points is constant.
(78,000 - 74,000) / (1972 - 1968) = 4,000 / 4 = 1,000 votes per year
(82,000 - 78,000) / (1976 - 1972) = 4,000 / 4 = 1,000 votes per year
In some cases, values which grow (or decay) do not do so linearly, but exponentially.
Example 6.3:
End of Month |
Amount of Debt |
| 0 | $500.00 |
| 1 | $550.00 |
| 2 | $605.00 |
| 3 | $665.00 |
| 4 | $732.05 |
With Month as the explanatory variable, and Debt as the response, it is easy to see that the relationship between the variables is not linear. The rate of change between successive points is not constant, but there is a relationship between the data.
Let's look at the ratios of the current month's debt to the previous month's
debt, for successive points.
550 / 500 = 1.1
605 / 550 = 1.1
665 / 605 = 1.1
732.50 / 665 = 1.1
Since the ratios are constant, we can represent the amount of debt as a function:
Debt = 500 * (1.1month)
This is an exponential function since the explanatory variable is in the exponent of a constant.
If we write our constant as 1 + 0.10, we can see that the debt is growing at 10% per month.
If the constant in the exponential is less than 1, it would show exponential
decrease (or decay). For instance, a function representing
a 10% decrease would have a constant of 1 - 0.10 = 0.90, giving a function such
as:
Debt = 500 * (0.90month)