05-regression-c.Rmd

# 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:

```{r, echo = FALSE}

int <- 197.65
ope <- -5.28
con <- -5.68
ext <- -0.4
agr <- 0.64
neu <- 1.89

p1o <- 9
p1c <- 8 
p1e <- 6
p1a <- 6
p1n <- 5

p2o <- 8
p2c <- 9 
p2e <- 5
p2a <- 5
p2n <- 6

p3o <- 5
p3c <- 6 
p3e <- 8
p3a <- 8
p3n <- 9

p4o <- 5
p4c <- 6 
p4e <- 9
p4a <- 9
p4n <- 8

hy1 <- int + (ope * p1o) + (con * p1c) + (ext * p1e) + (agr * p1a) + (neu * p1n)
hy2 <- int + (ope * p2o) + (con * p2c) + (ext * p2e) + (agr * p2a) + (neu * p2n)
hy3 <- round(int + (ope * p3o) + (con * p3c) + (ext * p3e) + (agr * p3a) + (neu * p3n),2)
hy4 <- int + (ope * p4o) + (con * p4c) + (ext * p4e) + (agr * p4a) + (neu * p4n)

```


|    |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 R^2^| .7619| Adjusted R^2^|.7348|
|F-statistic|28.16 on 5 and 44 DF| p-value| .001|

## 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 $b_{0}$ = `r int`
* $b_{1}X_{1}$, 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 $b_{1}$ = `r ope`
* $X_{1}$, $X_{2}$, $X_{3}$, $X_{4}$, $X_{5}$ are the measured values of the different predictors for a participant. For example, for Participant 1, again assuming Openness is our first predictor, then $X_{1}$ = `r p1o`. If Conscientousness is our second predictor, Extraversion our third predictor, Agreeableness our fourth predictor, and Neuroticism our fifth predictor, then $X_{2}$ = `r p1c`, $X_{3}$ = `r p1e`, $X_{4}$ = `r p1a` and $X_{5}$ = `r p1n`.

Then, using the information above, we know can start to fill in the information for **Participant 1** as follows:

$$\hat{Y} = `r int` + (`r ope` \times `r p1o`) + (`r con` \times `r p1c`) + (`r ext` \times `r p1e`) + (`r agr` \times `r p1a`) + (`r neu` \times `r p1n`)$$

And if we start to work that through, dealing with the multiplications first, we see:

$$\hat{Y} = `r int` + (`r ope * p1o`) + (`r con * p1c`) + (`r ext * p1e`) + (`r agr * p1a`) + (`r neu * p1n`)$$

Which then becomes:

$$\hat{Y} = `r int + (ope * p1o) + (con * p1c) + (ext * p1e) + (agr * p1a) + (neu * p1n)`$$

Giving a predicted value of $\hat{Y}$ = `r hy1`, to two decimal places. 

Likewise for **Participant 2** we would have:

$$\hat{Y} = `r int` + (`r ope` \times `r p2o`) + (`r con` \times `r p2c`) + (`r ext` \times `r p2e`) + (`r agr` \times `r p2a`) + (`r neu` \times `r p2n`)$$

And if we start to work that through, dealing with the multiplications first, we see:

$$\hat{Y} = `r int` + (`r ope * p2o`) + (`r con * p2c`) + (`r ext * p2e`) + (`r agr * p2a`) + (`r neu * p2n`)$$

Which then becomes:

$$\hat{Y} = `r int + (ope * p2o) + (con * p2c) + (ext * p2e) + (agr * p2a) + (neu * p2n)`$$

Giving a predicted value of $\hat{Y}$ = `r hy2`, to two decimal places. 

Likewise for **Participant 3** we would have:

$$\hat{Y} = `r int` + (`r ope` \times `r p3o`) + (`r con` \times `r p3c`) + (`r ext` \times `r p3e`) + (`r agr` \times `r p3a`) + (`r neu` \times `r p3n`)$$

And if we start to work that through, dealing with the multiplications first, we see:

$$\hat{Y} = `r int` + (`r ope * p3o`) + (`r con * p3c`) + (`r ext * p3e`) + (`r agr * p3a`) + (`r neu * p3n`)$$

Which then becomes:

$$\hat{Y} = `r int + (ope * p3o) + (con * p3c) + (ext * p3e) + (agr * p3a) + (neu * p3n)`$$

Giving a predicted value of $\hat{Y}$ = `r hy3`, to two decimal places. 

And finally for **Participant 4** we would have:

$$\hat{Y} = `r int` + (`r ope` \times `r p4o`) + (`r con` \times `r p4c`) + (`r ext` \times `r p4e`) + (`r agr` \times `r p4a`) + (`r neu` \times `r p4n`)$$

And if we start to work that through, dealing with the multiplications first, we see:

$$\hat{Y} = `r int` + (`r ope * p4o`) + (`r con * p4c`) + (`r ext * p4e`) + (`r agr * p4a`) + (`r neu * p4n`)$$

Which then becomes:

$$\hat{Y} = `r int + (ope * p4o) + (con * p4c) + (ext * p4e) + (agr * p4a) + (neu * p4n)`$$

Giving a predicted value of $\hat{Y}$ = `r hy4`, 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, $\hat{Y}$ = `r hy1`  
* Participant 2, $\hat{Y}$ = `r hy2`
* Participant 3, $\hat{Y}$ = `r hy3`
* Participant 4, $\hat{Y}$ = `r hy4`