The Linear Model
Univariate Statistics and Methodology using R
Where we Were
...
(Intercept) 321.24 91.05 3.53 0.00093 ***
BloodAlc 32.28 8.88 3.64 0.00067 ***
...Intercept and Slope
...
(Intercept) 321.24 91.05 3.53 0.00093 ***
BloodAlc 32.28 8.88 3.64 0.00067 ***
...What If We Know Nothing
there is no relationship between our independent and dependent variables
- equivalent to “not knowing” the independent variable
- if we’ve measured a bunch of RTs but nothing else, our best estimate of a new RT is the mean RT
y ~ 1 or RT ~ 1
The Null Model
- how much better is
RT ~ BloodAlcthanRT ~ 1?
Simplify the Data
- to make the next few graphs less busy
ourDat <- dat |>
slice_sample(n = 20)
Total Sum of Squares
\[ \textrm{total SS}=\sum{(y - \bar{y})^2} \]
sum of squared differences between observed \(y\) and mean \(\bar{y}\)
= total amount of variance in the data
= total unexplained variance from the null model
Residual Sum of Squares
\[\textrm{residual SS} = \sum{(y - \hat{y})^2}\]
sum of squared differences between observed \(y\) and predicted \(\hat{y}\)
= total unexplained variance from a given model in the data
Model Sum of Squares
\[ \textrm{model SS} = \sum{(\hat{y} - \bar{y})^2} \]
sum of squared differences between predicted \(\hat{y}\) and observed \(\bar{y}\)
total variance explained by a given model
Testing the Model: R2
\[ R^2 = \frac{\textrm{model SS}}{\textrm{total SS}} = \frac{\sum{(\hat{y}-\bar{y})^2}}{\sum{(y-\bar{y})^2}} \]
how much the model improves over the null
\(0 \le R^2 \le 1\)
we want \(R^2\) to be large
Testing the Model: F
\(F\) ratio depends on mean squares
\[(\textrm{MS}_x=\textrm{SS}_x/\textrm{df}_x)\]
\[F=\frac{\textrm{model MS}}{\textrm{residual MS}}=\frac{\sum{(\hat{y}-\bar{y})^2}/\textrm{df}_m}{\sum{(y-\hat{y})^2}/\textrm{df}_r}\]
how much the model ‘explains’
\(0<F\)
we want \(F\) to be large
Hold On
- we’ve made a lot of assumptions
The Most Important Rule
- look at your data, not just the statistics
Anscombe’s Quartet
Anscombe (1973)
Residuals
\[y_i=b_0+b_1\cdot{}x_i+\epsilon_i\]
\[\color{red}{\epsilon\sim{}N(0,\sigma)~\text{independently}}\]
- normally distributed (mean should be \(\simeq{}\) zero)
homogeneous (differences from \(\hat{y}\) shouldn’t be systematically smaller or larger for different \(x\))
independent (residuals shouldn’t influence other residuals)
In More Detail
linearity
plot(mod, which = 1)plots residuals \(\epsilon_i\) against fitted values \(\hat{y}_i\)
the ‘average residual’ is roughly zero across \(\hat{y}\), so relationship is likely to be linear
In More Detail
normality
hist(resid(mod))
In More Detail
normality
plot(density(resid(mod)))check that residuals \(\epsilon\) are approximately normally distributed
in fact there’s a better way of doing this…
In More Detail
normality
plot(mod, which = 2)Q-Q plot compares the residuals \(\epsilon\) against a known distribution (here, normal)
observations close to the straight line mean residuals are approximately normal
Q-Q Plots
y axis
for a normal distribution, what values should (say) 5%, or 10% of the observations lie below?
expressed in “standard deviations from the mean”
qnorm(c(0.05, 0.1))[1] -1.645 -1.282Q-Q Plots
x axis
from our residuals, what values are (say) 5%, or 10%, of observations found to be less than?
convert to “standard deviations from the mean”
quantile(
scale(resid(mod)),
c(0.05, 0.1)
) 5% 10%
-1.405 -1.280 
- Q-Q Plot shows these values plotted against each other
In More Detail
normality
plot(mod,which=2)Q-Q plot compares the residuals \(\epsilon\) against a known distribution (here, normal)
observations close to the straight line mean residuals are approximately normal
- numbered points refer to row numbers in the original data
In More Detail
homogeneity of variance
plot(mod, which = 3)graph shows \(\sqrt{|\textrm{standardized residuals|}}\)
the size of the residuals is approximately the same across values of \(\hat{y}\)
Influence
even substantial outliers may only have small effects on the model
here, only the intercept is affected
Influence
- observations with high leverage are inconsistent with other data, but may not be distorting the model
Influence
- we care about observations with high influence (outliers with high leverage)
Cook’s Distance
- a standardised measure of “how much the model differs without observation \(i\)”
![]()
Cook’s Distance
Cook’s Distance\[D_i=\frac{\sum_{j=1}^{n}{(\hat{y}_j-\hat{y}_{j(i)})^2}}{(p+1)\hat{\sigma}^2}\]
- \(\hat{y}_j\) is the \(j\)th fitted value
- \(\hat{y}_{j(i)}\) is the \(j\)th value from a fit which doesn’t include observation \(i\)
- \(p\) is the number of regression coefficients
- \(\hat{\sigma}^2\) is the estimated variance from the fit, i.e., mean squared error
Cook’s Distance
plot(bad, which = 4)observations labelled by row
various rules of thumb, but start looking when Cook’s Distance > 0.5
Cook’s Distance
plot(mod, which = 4)observations labelled by row
various rules of thumb, but start looking when Cook’s Distance > 0.5
So Now We Know…
- the relationship is approximately linear
- the residuals are approximately normal
- the variance is approximately homogeneous
- we believe the observations are independent
- there are no overly-influential observations
Call:
lm(formula = RT ~ BloodAlc, data = dat)
Residuals:
Min 1Q Median 3Q Max
-115.92 -40.42 1.05 42.93 126.64
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 321.24 91.05 3.53 0.00093 ***
BloodAlc 32.28 8.88 3.64 0.00067 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 55.8 on 48 degrees of freedom
Multiple R-squared: 0.216, Adjusted R-squared: 0.2
F-statistic: 13.2 on 1 and 48 DF, p-value: 0.000673- we are ready to report our observations
What We Know
Adding a predictor of blood alcohol improved the amount of variance explained over the null model (R2=0.22, F(1,48)=13.2, p=.0007). Reaction time slowed by 32.3 ms for every additional 0.01% blood alcohol by volume (t(48)=3.64, p=.0007).
Learning to Read
the Playmo School has been evaluating its reading programmes, using 50 students
ages of students
hours per week students spend reading of their own volition
whether they are taught using phonics or whole-word methods
outcome: “reading age”
Learning to Read
| age | hrs_wk | method | R_AGE |
|---|---|---|---|
| 10.0 | 5.1 | phonics | 14.1 |
| 8.0 | 4.4 | phonics | 11.8 |
| 9.3 | 5.8 | phonics | 13.8 |
| 8.7 | 5.4 | phonics | 13.3 |
| 10.2 | 4.6 | phonics | 14.3 |
| 10.5 | 4.6 | word | 9.6 |
| 9.4 | 4.6 | word | 8.0 |
| 8.6 | 3.6 | word | 6.6 |
| 8.1 | 4.4 | word | 7.5 |
| 6.1 | 5.1 | word | 5.5 |
Learning to Read
| age | hrs_wk | method | R_AGE |
|---|---|---|---|
| 10.0 | 5.1 | phonics | 14.1 |
| 8.0 | 4.4 | phonics | 11.8 |
| 9.3 | 5.8 | phonics | 13.8 |
| 8.7 | 5.4 | phonics | 13.3 |
| 10.2 | 4.6 | phonics | 14.3 |
| 10.5 | 4.6 | word | 9.6 |
| 9.4 | 4.6 | word | 8.0 |
| 8.6 | 3.6 | word | 6.6 |
| 8.1 | 4.4 | word | 7.5 |
| 6.1 | 5.1 | word | 5.5 |
Does Practice Affect Reading Age?
p <- reading |>
ggplot(aes(x = hrs_wk,
y = R_AGE)) +
geom_point(size = 3) +
ylab("reading age") +
xlab("hours reading/week")
p
hours per week is correlated with reading age: r=0.3812, p=0.0063
we can use a linear model to say something about the effect size
Does Practice Affect Reading Age?
p + geom_smooth(method = "lm")
- each extra hour spent reading a week adds 1.193 years to reading age
Assumptions Not Met!
it seems that the assumptions aren’t met for this model
(another demonstration on the right)
one reason for this can be because there’s still systematic structure in the residuals
i.e., more than one thing which can explain the variance
