02Jmodellingcpdb

= Changing to Linear Regression on the Classpad =

If the calculator can’t automatically do the regression curve you want, you’ll have to change the x-values (in List1) using the appropriate function the same way we did in Modelling and then check using Linear Regression.

Example 2

Use the Classpad to find the best model (equation) for the following data:



Go to the Statistics Page. Enter the x values into List1 and the y values into List2

Tap the left icon to draw the graph

The shape of the graph appears to be either a hyperbola or a truncus. The Classpad doesn’t offer regression for those shapes so we need to do a little manual processing.

Again, for the purposes of this exercise only, I’ve included Linear Regression on the original data for comparison

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... ... ... y = –4.54x + 33.4 ... ... ... r 2 = 0.8035
 * Linear Regression**:

To model a **hyperbola**, X = 1/x, replace the x-values in list1 with ... ... ... 1, 1/2, 1/3, 1/4, 1/5, 1/6 and do linear regression on those values compared to the y-values in list2

... ... ... y = 30.20X + 5.17 ... ... ... r 2 = 0.9959
 * Linear Regression with X = 1/x**

To model a **truncus**, X = 1/x 2, replace the x-values in list1 with ... ... ... 1, 1/4, 1/9, 1/16, 1/25, 1/36 and do linear regression on those values compared to the y-values in list2

... ... ... y = 24.49X + 11.41 ... ... ... r 2 = 0.9477
 * Linear Regression with X = 1/x 2 **.

Comparison of the results shows that the hyperbola is the best fit (biggest r 2 value), so the answer to the question is:

math \\ . \qquad y=30.20X + 5.17 \qquad. \\ . \\ . \text{Replace the X with } \dfrac{1}{x} \qquad. \\ . \\ . \qquad y = \dfrac{30.20}{x} + 5.17 math

Note: We could have done Example 1 this way to force the quadratic regression to stick to the form y = ax 2 + b (instead of y = ax 2 + bx + c)