Section 4.2: Least-Squares Regression

The equation T = 6x + 53 roughly approximates the tuition per credit at ECC since 2001.   In this case, x represents the number of years since 2001 and T represents the tuition amount for that year.

The graph below illustrates the relationship.

In this example, we can see that both the 6 and the 53 have very specific meanings:

The 6 is the increase per year. In other words, for every additional year, the tuition increases $6.

The 53 represents the initial tuition, or the tuition per credit hour in 2001.

In Example 1 from section 4.1, we talked about the relationship between student heart rates (in beats per minute) before and after a brisk walk.

Let's highlight a pair of points on that plot and use those two points to find an equation for a line that might fit the scatter diagram.

Using the points (52, 56) and (90, 116), we get a slope of

So an equation for the line would be:

y - y₁ = m(x - x₁)

y - 56 = 1.58(x - 52)

y - 56 = 1.58x - 82.16

y = 1.58x - 26.16

It's interesting to note the meanings behind the slope and y-intercept for this example. A slope of 1.58 means that for every additional beat per minute before the brisk walk, the heart rate after the walk was 1.58 faster.

The y-intercept, on the other hand, doesn't apply in this case. A y-intercept of -26.16 means that if you have 0 beats per minute before the walk, you'll have -26.16 beats per minute after the walk. ?!?!

This brings up a very important point - models have limitations. In this case, we say that the y-intercept is outside the scope of the model.

Let's again use the data from Example 1 from section 4.1. In Example 2 from earlier this section, we found the model:

= 1.58x - 30.16

Let's use this model to predict the "after" heart rate for a particular students, the one whose "before" heart rate was 86 beats per minute.

The predicted heart rate, using the model above, is:

= 1.58(86) - 26.16 = 109.72

Using that predicted heart rate, the residual is then:

residual = = 98 - 109.72 = -11.72

Here's that residual if we zoom in on that particular student:

Notice here that the residual is negative, since the predicted value was more than the actual observed "after" heart rate.

Let's again use the data from Example 1 in Section 4.1, but instead of just using two points to get a line, we'll use the method of least squares to find the Least-Squares Regression line.

Using computer software, we find the following values:

≈72.16667
s_x ≈10.21366
= 90.75
s_y ≈17.78922
r ≈ 0.48649

Note: We don't want to round these values here, since they'll be used in the calculation for the correlation coefficient - only round at the very last step.

Using the formulas for the LSR line, we have

= 0.8473x + 29.60

(A good general guideline is to use 4 digits for the slope and y-intercept, though there is no strict rule.)

One thought that may come to mind here is that this doesn't really seem to fit the data as well as the one we did by picking two points! Actually, it does do a much better job fitting ALL of the data as well as possible - the previous line we did ourselves did not address most of the points that were above the main cluster. In the next section, we'll talk more about how outliers like the (58, 128) point far above the rest can affect a model like this one.