13 Multiple Linear Regression
Let's say we want to make a prediction about the following four people who have these scores on the five OCEAN scales:
| Openness | Conscientousness | Extraversion | Agreeableness | Neuroticism | |
|---|---|---|---|---|---|
| Participant 1 | 9 | 8 | 6 | 6 | 5 |
| Participant 2 | 8 | 9 | 5 | 5 | 6 |
| Participant 3 | 5 | 6 | 8 | 8 | 9 |
| Participant 4 | 5 | 6 | 9 | 9 | 8 |
And let's say we have this output from the multiple linear regression model we created using all of the five OCEAN scales as predictors:
| Unstandardised Coefficient | t-value | p-value | |
|---|---|---|---|
| Intercept | 197.65 | 22.43 | .0001 |
| Openness | -5.28 | -7.68 | .0001 |
| Conscientiousness | -5.68 | -7.44 | .0001 |
| Extraversion | -0.4 | -0.65 | .522 |
| Agreeableness | 0.64 | 0.69 | .495 |
| Neuroticism | 1.89 | 2.56 | .014 |
| --- | --- | --- | --- |
| Multiple R2 | .7619 | Adjusted R2 | .7348 |
| F-statistic | 28.16 on 5 and 44 DF | p-value | .001 |
13.1 Making a Prediction
Now based on the information we have gained above we can actually use that to make predictions about a person we have not measured based on the formula:
\[\hat{Y} = b_{0} + b_{1}X_{1} + b_{2}X_{2} + b_{3}X_{3} + b_{4}X_{4} + b_{5}X_{5}\]
Well technically it is
\[\hat{Y} = b_{0} + b_{1}X_{1} + b_{2}X_{2} + b_{3}X_{3} + b_{4}X_{4} + b_{5}X_{5} + error\]
But we will disregard the error for now. Also you might have noticed the small hat about the \(Y\) making it \(\hat{Y}\) (pronounced Y-hat). This means we are making a prediction of Y (\(\hat{Y}\)) as opposed to an actually measured (i.e. observed) value (\(Y\)).
And to break that formula down a bit we can say:
- \(b_{0}\) is the intercept of the model. In this case \(b_{0}\) = 197.65
- \(b_{1}X_{1}\), for example, is read at the non-standardised coeffecient of the first predictor \(b_{1}\) multiplied by the measured value of the first predictor \(X_{1}\)
- \(b_{1}\), \(b_{2}\), \(b_{3}\), \(b_{4}\), \(b_{5}\) are the non-standardised coefficient values of the different predictors in the model. There is one non-standardised coefficient for each predictor. For example, let's say Openness is our first predictor and as such \(b_{1}\) = -5.28
- \(X_{1}\), \(X_{2}\), \(X_{3}\), \(X_{4}\), \(X_{5}\) are the measured values of the different predictors for a participant. For example, for Participant 1, again assuming Openness is our first predictor, then \(X_{1}\) = 9. If Conscientousness is our second predictor, Extraversion our third predictor, Agreeableness our fourth predictor, and Neuroticism our fifth predictor, then \(X_{2}\) = 8, \(X_{3}\) = 6, \(X_{4}\) = 6 and \(X_{5}\) = 5.
Then, using the information above, we know can start to fill in the information for Participant 1 as follows:
\[\hat{Y} = 197.65 + (-5.28 \times 9) + (-5.68 \times 8) + (-0.4 \times 6) + (0.64 \times 6) + (1.89 \times 5)\]
And if we start to work that through, dealing with the multiplications first, we see:
\[\hat{Y} = 197.65 + (-47.52) + (-45.44) + (-2.4) + (3.84) + (9.45)\]
Which then becomes:
\[\hat{Y} = 115.58\]
Giving a predicted value of \(\hat{Y}\) = 115.58, to two decimal places.
Likewise for Participant 2 we would have:
\[\hat{Y} = 197.65 + (-5.28 \times 8) + (-5.68 \times 9) + (-0.4 \times 5) + (0.64 \times 5) + (1.89 \times 6)\]
And if we start to work that through, dealing with the multiplications first, we see:
\[\hat{Y} = 197.65 + (-42.24) + (-51.12) + (-2) + (3.2) + (11.34)\]
Which then becomes:
\[\hat{Y} = 116.83\]
Giving a predicted value of \(\hat{Y}\) = 116.83, to two decimal places.
Likewise for Participant 3 we would have:
\[\hat{Y} = 197.65 + (-5.28 \times 5) + (-5.68 \times 6) + (-0.4 \times 8) + (0.64 \times 8) + (1.89 \times 9)\]
And if we start to work that through, dealing with the multiplications first, we see:
\[\hat{Y} = 197.65 + (-26.4) + (-34.08) + (-3.2) + (5.12) + (17.01)\]
Which then becomes:
\[\hat{Y} = 156.1\]
Giving a predicted value of \(\hat{Y}\) = 156.1, to two decimal places.
And finally for Participant 4 we would have:
\[\hat{Y} = 197.65 + (-5.28 \times 5) + (-5.68 \times 6) + (-0.4 \times 9) + (0.64 \times 9) + (1.89 \times 8)\]
And if we start to work that through, dealing with the multiplications first, we see:
\[\hat{Y} = 197.65 + (-26.4) + (-34.08) + (-3.6) + (5.76) + (15.12)\]
Which then becomes:
\[\hat{Y} = 154.45\]
Giving a predicted value of \(\hat{Y}\) = 154.45, to two decimal places.
And so in summary, based on our above model, and the above measured values, we would predict the following values:
- Participant 1, \(\hat{Y}\) = 115.58
- Participant 2, \(\hat{Y}\) = 116.83
- Participant 3, \(\hat{Y}\) = 156.1
- Participant 4, \(\hat{Y}\) = 154.45