Section 4.4: Summary of Bivariate Data

This section puts all the ideas we have seen in chapter 4 together in one example.

Table 5: Critical Values Table for Correlation

Solution:To begin, either type, or copy and paste the data into two columns of STATBELT Utility (Table 6 has four pairs of columns, so paste each pair under the one before). Make sure that you label the columns HS GPA and College GPA. You can save the data as well (Data → Import or export data → Export data as CSV) in case you wish to use it later.

Select Graphs → Scatterplot, and choose HS GPA for “Explanatory variable (x)” and College GPA for “Response variable (y).” Click “Open graph in a new window.” The scatterplot is shown in figure 4.37.

Figure 4.37: HS GPA and College GPA scatterplot.

It appears that there is a moderate positive correlation between high school grade point average and college grade point average.

Select Descriptive Statistics → Correlation. Check both variables and click “Show results in a new window.” The correlation is shown in figure 4.38.

Figure 4.38: Correlation between HS GPA and College GPA.

The correlation is $r\approx 0.7785$. There were 48 data pairs, so we look at the critical values chart to find the critical value, CV. For n = 48, the critical value is CV = 0.285

Since $\left|r\right|=\left|0.7785\right|=0.7785$ is larger than CV = 0.285, the correlation is significant. Therefore, using regression is appropriate. We can use a linear model to make predictions.

Select Descriptive Statistics → Regression. Then choose HS GPA for “Explanatory variable (x)” and College GPA for “Response variable (y),” and click “Show results in a new window.” The regression output is shown in figure 4.39.

Figure 4.39: Regression results for HS GPA and College GPA.

The least squares regression line is $\widehat{y}=0.866+0.6695x$ where $\widehat{y}$ represents the predicted college GPA for a student with high school GPA equal to x.

The y-intercept is 0.866, which is the predicted college GPA for a student who had a high school GPA equal to zero. That is outside the scope of the model since all the students had high school GPAs that were between 1.49 and 4.0.

The slope of the regression line is 0.6695, or about 0.67. If a high school GPA were to increase by 1 grade point, the corresponding college GPA would increase by 0.67 grade points, on average. In other words, the difference between a 2.5 and a 3.5 high school GPA is one grade point, which will correspond to a 0.67 point difference in college GPA.

To predict the college GPA for student with a 3.5 high school GPA, plug $x=3.5$ into the equation.

$$\begin{array}{cc}\widehat{y}& =0.866+0.6695x\\ \widehat{y}& =0.866+0.6695\left(3.5\right)\\ \widehat{y}& =0.866+2.34325\\ \widehat{y}& \approx 3.20925\end{array}$$

So, students who earn a 3.5 GPA in high school will get, on average, a 3.21 GPA in college.

No. The minimum x-value is 1.49, and the maximum is 4.0. We can use the model to make predictions for college grade point averages only for high school grade point averages between 1.49 and 4.0, or very close.

This question is asking us to find the residual for a data value with HS GPA = 1.9 and College GPA = 3.23. The observed value is 3.23. We must use the regression equation again to find the predicted college GPA for a HS GPA = 1.9. So, plug $x=1.9$ into the equation.

$$\begin{array}{cc}\widehat{y}& =0.866+0.6695x\\ \widehat{y}& =0.866+0.6695\left(1.9\right)\\ \widehat{y}& =0.866+1.27205\\ \widehat{y}& \approx 2.13805\end{array}$$

So, students who earn a 1.9 GPA in high school will get, on average, a 2.14 GPA in college. That means that Freddy did much better than predicted! How much better?

The residual is $\text{Residual}=\text{Observed College GPA}-\text{Predicted College GPA}=y-\widehat{y}$ .

So, $\text{Residual}=3.23-2.14=1.09$

Since the residual is positive, Freddy had a college GPA that was 1.09 grade points higher than the average college GPA for students with a 1.9 high school GPA. See figure 4.40.

Figure 4.40: Scatterplot, regression line, and residual for Freddy McDanger.

Again, we are asked to find a residual, this time for a data value with HS GPA = 3.72 and College GPA = 3.15. The observed value is 3.15. We must use the regression equation again to find the predicted college GPA for a HS GPA = 3.72. So, plug

$x=3.72$ into the equation.

$$\begin{array}{cc}\widehat{y}& =0.866+0.6695x\\ \widehat{y}& =0.866+0.6695\left(3.72\right)\\ \widehat{y}& =0.866+2.49054\\ \widehat{y}& \approx 3.35654\end{array}$$

So, students who earn a 3.72 GPA in high school will get, on average, a 3.36 GPA in college. So, Irving did a little worse than predicted, but not much worse.

The residual is $\text{Residual}=3.15-3.36=-0.21$

Since the residual is negative, Irving has a slightly lower than average college GPA for students who had a 3.72 high school GPA. But is sounds like he was making money and having fun, so that small drop in GPA might be okay with him! See figure 4.41.

Figure 4.41: Scatterplot, regression line, and residual for Irving VonChucklebutte.

From the regression output in figure 4.39, we can see that the coefficient of determination $R^2$ (given as r² in the output) is approximately $R^2\approx 0.606$. That means that high school GPA explains about $100\times 0.606=60.6\%$ of the variation in college grade point averages. That means that 39.4% of the variation in college grade point average is not explained by high school grades. The remaining variation occurs for other reasons, such as content of courses, textbook or materials cost, instructors, work schedules, living situations, potential distractions, etc.