8. The following is a table of homework averages and final course grades for selected students from a Finite Math class.

 HW CG 96 94 100 98 78 83 60 65 92 85

Find the line of best least squares fit for predicting the final average in the course as a function of the homework grade. What kind of a final average would this model predict for a person who got a 90 for their homework average. Compute Pearson's   r.   What does Pearson's   r   tell us about these distributions?

First, we need to find the means of the homework grades and the course averages. Let us call the homework grades   x   and the course averages   y.

 x y 96 94 100 98 78 83 60 65 92 85 Total 426 425 mean 85.2 85

Now we can compute the deviations of the   x's   and   y's

One way to check to make sure we got the means right is if the deviatins add up to 0.

Next we need columns for the squares of the deviations of the x's, squares of the deviations of the y's, and the product of these deviations

We now have all the information we need to be able to complete the problem.

The equation of the regression line is of the form

y = mx + b

where

and

So the regression equation is approximately

y = .755988 x + 20.58982

To predict the final average for a person who got a 90 average on their homework, substitute a 90 in for   x   in the regression equation

y = .755988(90) + 20.58982

= 88.6

If we plot the data and add the graph of the regression line, it looks like

The course average is pretty close to the homework average. The constant term of 20.5898 tells us that a person who did no homework could expect to get about 20 or 21 as a final course average. That would not be enough to pass.

We also have enough information to compute Pearson's r.

we can see from the graph that the points lie pretty close to the regression line. That would indicate that there is a high degree of correlation. That is reflected by the fact that Pearson's r is close to 1.

You will probably find it worth while to learn how to get these numbers on your calculator. Almost all graphing calculators and many scientific calculators have the capability to compute m, b, and r if you enter the data into the statistics package of the calculator. If your calculator will not give you Pearson's r, it can be computed by using the formula

where   sx   is the standard deviation of the x's and   sy is the standard deviation of the y's. It won't matter whether you use the population standard deviation or the sample standard deviation, so long as you are consistent. The differences in the standard deviations will cancel out when you take the ratio. If we clear denominators in this equation, we see that

rsy = msx

which says that r is the part of the variation in the y's, as measured by the standard deviation, which is explained by the variation in the x's. In this particular class, homework was 15% of the final grade, so you would expect that Pearson's r would be at least 15%. However, homework has more of an influence on the grade than the 15% which it contributes directly. Evidently scores on tests benefit from students doing the homework.