2.4 Fitting linear models to data  (Page 2/14)

[link] compares the two processes for the cricket-chirp data addressed in [link] . We can see that interpolation would occur if we used our model to predict temperature when the values for chirps are between 18.5 and 44. Extrapolation would occur if we used our model to predict temperature when the values for chirps are less than 18.5 or greater than 44.

There is a difference between making predictions inside the domain and range of values for which we have data and outside that domain and range. Predicting a value outside of the domain and range has its limitations. When our model no longer applies after a certain point, it is sometimes called model breakdown    . For example, predicting a cost function for a period of two years may involve examining the data where the input is the time in years and the output is the cost. But if we try to extrapolate a cost when $x=\mathrm{50,}$ that is in 50 years, the model would not apply because we could not account for factors fifty years in the future.

Understanding interpolation and extrapolation

Use the cricket data from [link] to answer the following questions:

1. Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable.
2. Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable.

1. The number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpolation. Using our model:
   $$\begin{array}{l}T(30)=30+1.2(30)\hfill \\ =66\text{degrees}\hfill \end{array}$$
   
   Based on the data we have, this value seems reasonable.
2. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation because 40 is outside the range of our data. Using our model:
   $$\begin{array}{l}40=30+1.2c\hfill \\ 10=1.2c\hfill \\ c\approx 8.33\hfill \end{array}$$

We can compare the regions of interpolation and extrapolation using [link] .

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Finding the line of best fit using a graphing utility

While eyeballing a line works reasonably well, there are statistical techniques for fitting a line to data that minimize the differences between the line and data values Technically, the method minimizes the sum of the squared differences in the vertical direction between the line and the data values. . One such technique is called least squares regression    and can be computed by many graphing calculators, spreadsheet software, statistical software, and many web-based calculators For example, http://www.shodor.org/unchem/math/lls/leastsq.html . Least squares regression is one means to determine the line that best fits the data, and here we will refer to this method as linear regression.