# Modelling and Regression on the Classpad

CAS calculators can perform regression for a number of different curves. We will use Linear, Quadratic and Cubic Regression.

.

#### Example 1

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

.

**Solution**

- 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.

.

- Looking at the graph we can see that the graph curves upwards so it is probably a quadratic or cubic.

.

- If the graph doesn’t draw automatically, you may need to click in the table area then go to the SetGraph menu, select Settings, then give it the following shown here:

.

For this demonstration, I will start by doing a Linear Regression (for comparison purposes).

**Linear Regression**

- Click in the table area then go to Calc menu and select Linear Reg.

.

- You should get a screen like the one shown here:

.

- XList and YList should match the lists where you entered your data (List1 and List2).

- If you set Copy Formula away from OFF you can store the resulting regression equation in Y1 (or Y2, etc)

- Select OK and you should get a screen like the one shown here:

.

- Using the values of a and b that are supplied, we get:

… … … … $y = 10.96x - 4.24$

This is the linear regression equation or the “line of best fit.”

I have circled the **r ^{2}** value on the Linear Regression screen.

- The
**r**value is a measure of how closely the line fits the data. The^{2}**r**values vary from 0 to 1 where:^{2}- 0 means the data bears no relationship to the line
- more than 0.9 means a strong correlation exists
- 1 means a perfect fit – every data point is on the line.

The r^{2} value of 0.92 shown here is pretty good.

But looking at the graph suggests a quadratic or cubic function might be better.

Repeating the process with Quadratic Regression and Cubic Regression gives the following results:

.

**Quadratic Regression:**

.

… … $y = 2.20x^2 - 0.05x + 3.10$

… … $r^2 = 0.999993$

.

**Cubic Regression:**

.

… … $y = 0.005x^3 +2.17x^2 - 0.01x + 3.08$

… … $r^2 = 0.999994$

We can see that both the Quadratic and Cubic Regression Lines are effectively a perfect fit.

.

Since the cubic regression has an **x ^{3}** term coefficient of less than 0.005,

we can use the Quadratic Regression for all practical purposes.

.

## 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:

**Solution**

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

**Linear Regression**

.

… … $y = -4.54x + 33.4$

… … $r^2 = 0.8035$

It is clear from the **r ^{2}** value that the correlation is good, but could be better.

**Hyperbola Regression**

To model a hyperbola, $X = \dfrac{1}{x}$, 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

.

The graph of X against y shows a very strong linear correlation

… … $y = 30.20X + 5.17$

… … $r^2 = 0.9959$

If we then substitute $X = \dfrac{1}{x}$ back in, this would give us the hyperbola:

… … $y = \dfrac{30.20}{x} + 5.17$

**Truncus Regression**

To model a truncus, $X = \dfrac{1}{x^2}$, 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

.

The graph of X against y shows a strong correlation, but not as good as for the Hyperbola

… … $y = 24.49X + 11.41$

… … $r^2 = 0.9477$

If we then substitute $X = \dfrac{1}{x^2}$ back in, this would give us the truncus:

… … $y = \dfrac{24.49}{x^2} + 11.41$

Comparison of the results shows that the hyperbola is the best fit (biggest r^{2} value).

Here is the graph of the truncus against the original data:

**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)

.