Learning Outcomes
- Make a prediction for a given value of the independent value using a regression equation
Recall the third exam/final exam example (example 2).
We examined the scatterplot and showed that the correlation coefficient is significant. We found the equation of the best-fit line for the final exam grade as a function of the grade on the third exam. We can now use the least-squares regression line for prediction.
Suppose you want to estimate, or predict, the mean final exam score of statistics students who received 73 on the third exam. The exam scores (x-values) range from 65 to 75. Since 73 is between the x-values 65 and 75, substitute x = 73 into the equation. Then:
[latex]\hat{y}[/latex] = -173.51 + 4.83(73) = 179.08
We predict that statistics students who earn a grade of 73 on the third exam will earn a grade of 179.08 on the final exam, on average.
Example
Use the third exam/final exam example (example 2).
- What would you predict the final exam score to be for a student who scored a 66 on the third exam?
- What would you predict the final exam score to be for a student who scored a 90 on the third exam?
Note
The process of predicting inside of the observed x values observed in the data is called interpolation. The process of predicting outside of the observed x values observed in the data is called extrapolation.
try it
Data are collected on the relationship between the number of hours per week practicing a musical instrument and scores on a math test. The line of best fit is as follows:
[latex]\displaystyle\hat{{y}}={72.5}+{2.8}{x}[/latex]
What would you predict the score on a math test would be for a student who practices a musical instrument for five hours a week?
