50 12.4 Prediction
Recall this example from earlier content:
A random sample of 11 statistics students produced the following data, where
x is the third exam score out of 80, and y is the final exam score out of 200. Can you predict the final exam score of a random student if you know the third exam score?
| x (third exam score) | y (final exam score) |
|---|---|
| 65 | 175 |
| 67 | 133 |
| 71 | 185 |
| 71 | 163 |
| 66 | 126 |
| 75 | 198 |
| 67 | 153 |
| 70 | 163 |
| 71 | 159 |
| 69 | 151 |
| 69 | 159 |
Table showing the scores on the final exam based on scores from the third exam.
Scatter plot showing the scores on the final exam based on scores from the third exam.
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]\displaystyle\hat{{y}}=-{173.51}+{4.83}{({73})}={179.08}[/latex]
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 1
Use the data above for this example:
- What would you predict the final exam score to be for a student who scored a 66 on the third exam?
Show Answer
When third exam is 66, final exam score = -17351 + 4.83(66) = 145.27
- What would you predict the final exam score to be for a student who scored a 90 on the third exam?
Show Answer
The x values in the data are between 65 and 75.
90 is outside of the domain of the observed x values in the data (independent variable), so you cannot reliably predict the final exam score for this student. (Even though it is possible to enter 90 into the equation for x and calculate a corresponding y value, the y value that you get will not be reliable.)
To understand really how unreliable the prediction can be outside of the observed x values observed in the data, make the substitution x= 90 into the equation.
[latex]\displaystyle\hat{{y}}=-{173.51}+{4.83}{({90})}={261.19}[/latex]
The final-exam score is predicted to be 261.19. The largest the final-exam score can be is 200.
Note:The process of predicting inside of the observed x values observed in the data is called interpolation. |
Try It
Data are collected on the relationship between the number of hours per week practicing a musical instrument (x) and scores on a math test (y). 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?
Show Answer
Score = 72.5 + (2.8)(5) = 86.5
References
Data from the Centers for