Last time, we discussed the use of categorical variables in multivariate regression. Often, these are encoded as indicator columns in the design matrix.

In [ ]:

```
url = 'http://stats191.stanford.edu/data/salary.table'
salary.table = read.table(url, header=T)
salary.table$E = factor(salary.table$E)
salary.table$M = factor(salary.table$M)
salary.lm = lm(S ~ X + E + M, salary.table)
head(model.matrix(salary.lm))
```

Often, especially in experimental settings, we record

*only*categorical variables.Such models are often referred to

*ANOVA (Analysis of Variance)*models.These are generalizations of our favorite example, the two sample $t$-test.

Suppose we want to understand the relationship between recovery time after surgery based on an patient's prior fitness.

We group patients into three fitness levels: below average, average, above average.

If you are in better shape before surgery, does it take less time to recover?

In [ ]:

```
url = 'http://stats191.stanford.edu/data/rehab.csv'
rehab.table = read.table(url, header=T, sep=',')
rehab.table$Fitness <- factor(rehab.table$Fitness)
head(rehab.table)
```

In [ ]:

```
attach(rehab.table)
boxplot(Time ~ Fitness, col=c('red','green','blue'))
```

First generalization of two sample $t$-test: more than two groups.

Observations are broken up into $r$ groups with $n_i, 1 \leq i \leq r$ observations per group.

Model: $$Y_{ij} = \mu + \alpha_i + \varepsilon_{ij}, \qquad \varepsilon_{ij} \sim N(0, \sigma^2).$$

Constraint: $\sum_{i=1}^r \alpha_i = 0$. This constraint is needed for “identifiability”. This is “equivalent” to only adding $r-1$ columns to the design matrix for this qualitative variable.

This is not the same

*parameterization*we get when only adding $r-1$ 0-1 columns, but it gives the same*model*.The estimates of $\alpha$ can be obtained from the estimates of $\beta$ using

`R`

's default parameters.For a more detailed exploration into

`R`

's creation of design matrices, try reading the following tutorial on design matrices.

Model is easy to fit: $$\widehat{Y}_{ij} = \frac{1}{n_i} \sum_{j=1}^{n_i} Y_{ij} = \overline{Y}_{i\cdot}$$ If observation is in $i$-th group: predicted mean is just the sample mean of observations in $i$-th group.

Simplest question: is there any group (main) effect? $$H_0:\alpha_1 = \dots = \alpha_r= 0?$$

Test is based on $F$-test with full model vs. reduced model. Reduced model just has an intercept.

Other questions: is the effect the same in groups 1 and 2? $$H_0:\alpha_1=\alpha_2?$$

In [ ]:

```
rehab.lm <- lm(Time ~ Fitness)
summary(rehab.lm)
```

In [ ]:

```
print(predict(rehab.lm, list(Fitness=factor(c(1,2,3)))))
c(mean(Time[Fitness == 1]), mean(Time[Fitness == 2]), mean(Time[Fitness == 3]))
```

Recall that the rows of the `Coefficients`

table above do not
correspond to the $\alpha$ parameter. For one thing, we would see
three $\alpha$'s and their sum would have to be equal to 0.

Also, the design matrix is the indicator coding we saw last time.

In [ ]:

```
head(model.matrix(rehab.lm))
```

There are ways to get

*different*design matrices by using the`contrasts`

argument. This is a bit above our pay grade at the moment.Upon inspection of the design matrix above, we see that the

`(Intercept)`

coefficient corresponds to the mean in`Fitness==1`

, while`Fitness==2`

coefficient corresponds to the difference between the groups`Fitness==2`

and`Fitness==1`

.

Much of the information in an ANOVA model is contained in the ANOVA table.

Source | SS | df | $\mathbb{E}(MS)$ |

Treatment | $SSTR=\sum_{i=1}^r n_i \left(\overline{Y}_{i\cdot} - \overline{Y}_{\cdot\cdot}\right)^2$ | r-1 | $\sigma^2 + \frac{\sum_{i=1}^r n_i \alpha_i^2}{r-1}$ |

Error | $SSE=\sum_{i=1}^r \sum_{j=1}^{n_i}(Y_{ij} - \overline{Y}_{i\cdot})^2$ | $\sum_{i=1}^r (n_i - 1)$ | $\sigma^2$ |

In [ ]:

```
anova(rehab.lm)
```

Note that $MSTR$ measures “variability” of the “cell” means. If there is a group effect we expect this to be large relative to $MSE$.

We see that under $H_0:\alpha_1=\dots=\alpha_r=0$, the expected value of $MSTR$ and $MSE$ is $\sigma^2$. This tells us how to test $H_0$ using ratio of mean squares, i.e. an $F$ test.

Rows in the ANOVA table are, in general, independent.

Therefore, under $H_0$ $$F = \frac{MSTR}{MSE} = \frac{\frac{SSTR}{df_{TR}}}{\frac{SSE}{df_{E}}} \sim F_{df_{TR}, df_E}$$ the degrees of freedom come from the $df$ column in previous table.

Reject $H_0$ at level $\alpha$ if $F > F_{1-\alpha, df_{TR}, df_{E}}.$

In [ ]:

```
F = 336.00 / 19.81
pval = 1 - pf(F, 2, 21)
print(data.frame(F,pval))
```

- Suppose we want to ``infer'' something about $$ \sum_{i=1}^r a_i \mu_i$$ where $\mu_i = \mu+\alpha_i$ is the mean in the $i$-th group. For example: $$ H_0:\mu_1-\mu_2=0 \qquad \text{(same as $H_0:\alpha_1-\alpha_2=0$)}?$$
For example:

Is there a difference between below average and average groups in terms of rehab time?

We need to know $$ \text{Var}\left(\sum_{i=1}^r a_i \overline{Y}_{i\cdot} \right) = \sigma^2 \sum_{i=1}^r \frac{a_i^2}{n_i}.$$

After this, the usual confidence intervals and $t$-tests apply.

In [ ]:

```
head(model.matrix(rehab.lm))
```

This means that the coefficient Fitness2 is the estimated difference between the two groups.

In [ ]:

```
detach(rehab.table)
```

Often, we will have more than one variable we are changing.

After kidney failure, we suppose that the time of stay in hospital depends on weight gain between treatments and duration of treatment.

We will model the `log`

number of days as a function of the other
two factors.

Variable | Description |

Days | Duration of hospital stay |

Weight | How much weight is gained? |

Duration | How long under treatment for kidney problems? (two levels) |

In [ ]:

```
url = 'http://statweb.stanford.edu/~jtaylo/stats191/data/kidney.table'
kidney.table = read.table(url, header=T)
kidney.table$D = factor(kidney.table$Duration)
kidney.table$W = factor(kidney.table$Weight)
kidney.table$logDays = log(kidney.table$Days + 1)
attach(kidney.table)
head(kidney.table)
```

Second generalization of $t$-test: more than one grouping variable.

Two-way ANOVA model:

- $r$ groups in first factor
- $m$ groups in second factor
- $n_{ij}$ in each combination of factor variables.

Model: $$Y_{ijk} = \mu + \alpha_i + \beta_j + (\alpha \beta)_{ij} + \varepsilon_{ijk} , \qquad \varepsilon_{ijk} \sim N(0, \sigma^2).$$

In kidney example, $r=3$ (weight gain), $m=2$ (duration of treatment), $n_{ij}=10$ for all $(i,j)$.

Two-way ANOVA: main questions of interest

Are there main effects for the grouping variables? $$H_0:\alpha_1 = \dots = \alpha_r = 0, \qquad H_0: \beta_1 = \dots = \beta_m = 0.$$

Are there interaction effects: $$H_0:(\alpha\beta)_{ij} = 0, 1 \leq i \leq r, 1 \leq j \leq m.$$

We've already seen these interactions in the IT salary example.

An

*additive model*says that the effects of the two factors occur additively -- such a model has no interactions.An

*interaction*is present whenever the additive model does not hold.

When these broken lines are not parallel, there is evidence of an interaction. The one thing missing from this plot are errorbars. The above broken lines are clearly not parallel but there is measurement error. If the error bars were large then we might consider there to be no interaction, otherwise we might.

In [ ]:

```
interaction.plot(W, D, logDays, type='b', col=c('red',
'blue'), lwd=2, pch=c(23,24))
```

Many constraints are needed, again for identifiability. Let’s not worry too much about the details

Constraints:

$\sum_{i=1}^r \alpha_i = 0$

$\sum_{j=1}^m \beta_j = 0$

$\sum_{j=1}^m (\alpha\beta)_{ij} = 0, 1 \leq i \leq r$

$\sum_{i=1}^r (\alpha\beta)_{ij} = 0, 1 \leq j \leq m.$

We should convince ourselves that we know have exactly $r*m$ free parameters.

Easy to fit when $n_{ij}=n$ (balanced) $$\widehat{Y}_{ijk}= \overline{Y}_{ij\cdot} = \frac{1}{n}\sum_{k=1}^{n} Y_{ijk}.$$

Inference for combinations $$\text{Var} \left(\sum_{i=1}^r \sum_{j=1}^m a_{ij} \overline{Y}_{ij\cdot}\right) = \frac{ \sigma^2}{n} \cdot \sum_{i=1}^r \sum_{j=1}^m{a_{ij}^2}.$$

Usual $t$-tests, confidence intervals.

In [ ]:

```
kidney.lm = lm(logDays ~ D*W, contrasts=list(D='contr.sum', W='contr.sum'))
summary(kidney.lm)
```

Suppose we are interested in comparing the mean in $(D=1,W=3)$ and $(D=2,W=2)$ groups. The difference is $$ E(\bar{Y}_{13\cdot}-\bar{Y}_{22\cdot}) $$

By independence, its variance is $$\text{Var}(\bar{Y}_{13\cdot}) + \text{Var}(\bar{Y}_{22\cdot}) = \frac{2 \sigma^2}{n}. $$

In [ ]:

```
estimates = predict(kidney.lm, list(D=factor(c(1,2)), W=factor(c(3,2))))
print(estimates)
sigma.hat = 0.7327 # from table above
n = 10 # ten observations per group
fit = estimates[1] - estimates[2]
upper = fit + qt(0.975, 54) * sqrt(2 * sigma.hat^2 / n)
lower = fit - qt(0.975 ,54) * sqrt(2 * sigma.hat^2 / n)
data.frame(fit,lower,upper)
```

In [ ]:

```
head(model.matrix(kidney.lm))
```

The most direct way to compute predicted values is using the `predict`

function

In [ ]:

```
predict(kidney.lm, list(D=factor(1),W=factor(1)), interval='confidence')
```

In the balanced case, everything can again be summarized from the ANOVA table

Source | SS | df | $\mathbb{E}(MS)$ |

A | $SSA=nm\sum_{i=1}^r \left(\overline{Y}_{i\cdot\cdot} - \overline{Y}_{\cdot\cdot\cdot}\right)^2$ | r-1 | $\sigma^2 + nm\frac{\sum_{i=1}^r \alpha_i^2}{r-1}$ |

B | $SSB=nr\sum_{j=1}^m \left(\overline{Y}_{\cdot j\cdot} - \overline{Y}_{\cdot\cdot\cdot}\right)^2$ | m-1 | $\sigma^2 + nr\frac{\sum_{j=1}^m \beta_j^2}{m-1}$ |

A:B | $SSAB = n\sum_{i=1}^r \sum_{j=1}^m \left(\overline{Y}_{ij\cdot} - \overline{Y}_{i\cdot\cdot} - \overline{Y}_{\cdot j\cdot} + \overline{Y}_{\cdot\cdot\cdot}\right)^2$ | (m-1)(r-1) | $\sigma^2 + n\frac{\sum_{i=1}^r\sum_{j=1}^m (\alpha\beta)_{ij}^2}{(r-1)(m-1)}$ |

Error | $SSE = \sum_{i=1}^r \sum_{j=1}^m \sum_{k=1}^{n}(Y_{ijk} - \overline{Y}_{ij\cdot})^2$ | (n-1)mr | $\sigma^2$ |

Rows of the ANOVA table can be used to test various of the hypotheses we started out with.

For instance, we see that under $H_0:(\alpha\beta)_{ij}=0, \forall i,j$ the expected value of $SSAB$ and $SSE$ is $\sigma^2$ – use these for an $F$-test testing for an interaction.

Under $H_0$ $$F = \frac{MSAB}{MSE} = \frac{\frac{SSAB}{(m-1)(r-1)}}{\frac{SSE}{(n-1)mr}} \sim F_{(m-1)(r-1), (n-1)mr}$$

In [ ]:

```
anova(kidney.lm)
```

We can also test for interactions using our usual approach

In [ ]:

```
anova(lm(logDays ~ D + W, kidney.table), kidney.lm)
```

`R`

formulae¶While we see that it is straightforward to form the
interactions test using our usual `anova`

function approach, we generally
*cannot* test for main effects by this approach.

In [ ]:

```
lm_no_main_Weight = lm(logDays ~ D + W:D)
anova(lm_no_main_Weight, kidney.lm)
anova(lm(logDays ~ D), lm(logDays ~ D + W))
```

In fact, these models are identical in terms of their *planes* or their
*fitted values*. What has happened is that `R`

has
formed a different design matrix using its rules for `formula`

objects.

In [ ]:

```
lm1 = lm(logDays ~ D + W:D)
lm2 = lm(logDays ~ D + W:D + W)
anova(lm1, lm2)
```

So far, we have used `anova`

to compare two models. In this section,
we produced tables for just 1 model. This also works for
*any* regression model, though we have to be a little careful
about interpretation.

Let's revisit the job aptitude test data from last section.

In [ ]:

```
url = 'http://stats191.stanford.edu/data/jobtest.table'
jobtest.table <- read.table(url, header=T)
jobtest.table$MINORITY <- factor(jobtest.table$MINORITY)
jobtest.lm = lm(JPERF ~ TEST + MINORITY + TEST:MINORITY, jobtest.table)
summary(jobtest.lm)
```

Now, let's look at the `anova`

output. We'll see the results don't match.

In [ ]:

```
anova(jobtest.lm)
```

The difference is how the `Sum Sq`

columns is created. In the `anova`

output, terms in the
response are added sequentially.

We can see this by comparing these two models directly. The `F`

statistic doesn't agree
because the `MSE`

above is computed in the *fullest* model, but the `Sum of Sq`

is correct.

In [ ]:

```
anova(lm(JPERF ~ TEST, jobtest.table),
lm(JPERF ~ TEST + MINORITY, jobtest.table))
```

Similarly, the first `Sum Sq`

in `anova`

can be found by:

In [ ]:

```
anova(lm(JPERF ~ 1, jobtest.table), lm(JPERF ~ TEST, jobtest.table))
```

There are ways to produce an *ANOVA* table whose $p$-values agree with
`summary`

. This is done by an ANOVA table that uses Type-III sum of squares.

In [ ]:

```
library(car)
Anova(jobtest.lm, type=3)
```

In [ ]:

```
summary(jobtest.lm)
```

In kidney & rehab examples, the categorical variables are well-defined categories: below average fitness, long duration, etc.

In some designs, the categorical variable is “subject”.

Simplest example: repeated measures, where more than one (identical) measurement is taken on the same individual.

In this case, the “group” effect $\alpha_i$ is best thought of as random because we only sample a subset of the entire population.

A “group” effect is random if we can think of the levels we observe in that group to be samples from a larger population.

Example: if collecting data from different medical centers, “center” might be thought of as random.

Example: if surveying students on different campuses, “campus” may be a random effect.

How much sodium is there in North American beer? How much does this vary by brand?

Observations: for 6 brands of beer, we recorded the sodium content of 8 12 ounce bottles.

Questions of interest: what is the “grand mean” sodium content? How much variability is there from brand to brand?

“Individuals” in this case are brands, repeated measures are the 8 bottles.

In [ ]:

```
url = 'http://stats191.stanford.edu/data/sodium.table'
sodium.table = read.table(url, header=T)
sodium.table$brand = factor(sodium.table$brand)
sodium.lm = lm(sodium ~ brand, sodium.table)
anova(sodium.lm)
```

Assuming that cell-sizes are the same, i.e. equal observations for each “subject” (brand of beer).

Observations $$Y_{ij} \sim \mu+ \alpha_i + \varepsilon_{ij}, 1 \leq i \leq r, 1 \leq j \leq n$$

$\varepsilon_{ij} \sim N(0, \sigma^2_{\epsilon}), 1 \leq i \leq r, 1 \leq j \leq n$

$\alpha_i \sim N(0, \sigma^2_{\alpha}), 1 \leq i \leq r.$

Parameters:

$\mu$ is the population mean;

$\sigma^2_{\epsilon}$ is the measurement variance (i.e. how variable are the readings from the machine that reads the sodium content?);

$\sigma^2_{\alpha}$ is the population variance (i.e. how variable is the sodium content of beer across brands).

In random effects model, the observations are no longer independent (even if $\varepsilon$'s are independent $$ {\rm Cov}(Y_{ij}, Y_{i'j'}) = \left(\sigma^2_{\alpha} + \sigma^2_{\epsilon} \delta_{j,j'} \right) \delta_{i,i'}.$$

In more complicated models, this makes ``maximum likelihood estimation'' more complicated: least squares is no longer the best solution.

**It's no longer just a plane!**

This model has a very simple model for the

*mean*, it just has a slightly more complex model for the*variance*.Shortly we'll see other more complex models of the variance:

- Weighted Least Squares
- Correlated Errors

The *MLE (Maximum Likelihood Estimator)* is found by minimizing
$$
\begin{aligned}
-2 \log \ell (\mu, \sigma^2_{\epsilon}, \sigma^2_{\alpha}|Y) &= \sum_{i=1}^r \biggl[ (Y_i - \mu)^T (\sigma^2_{\epsilon} I_{n_i \times n_i} + \sigma^2_{\alpha} 11^T)^{-1} (Y_i - \mu) \\
& \qquad + \log \left( \det(\sigma^2_{\epsilon} I_{n_i \times n_i} + \sigma^2_{\alpha} 11^T) \right) \biggr].
\end{aligned}
$$

THe function $\ell(\mu, \sigma^2_{\epsilon}, \sigma^2_{\alpha})$ is called the *likelihood function*.

Only one parameter in the mean function $\mu.$

When cell sizes are the same (balanced), $$ \widehat{\mu} = \overline{Y}_{\cdot \cdot} = \frac{1}{nr} \sum_{i,j} Y_{ij}.$$ Unbalanced models: use numerical optimizer.

This also changes estimates of $\sigma^2_{\epsilon}$ -- see ANOVA table. We might guess that $df=nr-1$ and $$ \widehat{\sigma}^2 = \frac{1}{nr-1} \sum_{i,j} (Y_{ij} - \overline{Y}_{\cdot\cdot})^2.$$

**This is not correct.**

Again, the information needed can be summarized in an ANOVA table.

Source | SS | df | $\mathbb{E}(MS)$ |

Treatment | $SSTR=\sum_{i=1}^r n_i \left(\overline{Y}_{i\cdot} - \overline{Y}_{\cdot\cdot}\right)^2$ | r-1 | $\sigma^2_{\epsilon} + n \sigma^2_{\alpha}$ |

Error | $SSE=\sum_{i=1}^r \sum_{j=1}^{n_i}(Y_{ij} - \overline{Y}_{i\cdot})^2$ | $\sum_{i=1}^r (n_i - 1)$ | $\sigma^2_{\epsilon}$ |

ANOVA table is still useful to setup tests: the same $F$ statistics for fixed or random will work here.

Test for random effect: $H_0:\sigma^2_{\alpha}=0$ based on $$ F = \frac{MSTR}{MSE} \sim F_{r-1, (n-1)r} \qquad \text{under $H_0$}.$$

Why $r-1$ degrees of freedom?

Imagine we could record an infinite number of observations for each individu\ al, so that $\overline{Y}_{i\cdot} \rightarrow \mu + \alpha_i$.

To learn anything about $\mu_{\cdot}$ we still only have $r$ observations $(\mu_1, \dots, \mu_r)$.

Sampling more within an individual cannot narrow the CI for $\mu$.

Easy to check that $$ \begin{aligned} E(\overline{Y}_{\cdot \cdot}) &= \mu \\ \text{Var}(\overline{Y}_{\cdot \cdot}) &= \frac{\sigma^2_{\epsilon} + n\sigma^2_{\alpha}}{rn}. \end{aligned} $$

To come up with a $t$ statistic that we can use for test, CIs, we need to find an estimate of $\text{Var}(\overline{Y}_{\cdot \cdot})$.

- ANOVA table says $E(MSTR) = n\sigma^2_{\alpha}+\sigma^2_{\epsilon}$ which suggests $$ \frac{\overline{Y}_{\cdot \cdot} - \mu_{\cdot}}{\sqrt{\frac{MSTR}{rn}}} \sim t_{r-1}.$$

We have seen estimates of $\mu$ and $\sigma^2_{\epsilon}$. Only one parameter remains.

Based on the ANOVA table, we see that $$ \sigma^2_{\alpha} = \frac{1}{n}(\mathbb{E}(MSTR) - \mathbb{E}(MSE)). $$

This suggests the estimate $$ \hat{\sigma^2}_{\alpha} = \frac{1}{n} (MSTR-MSE). $$

However, this estimate can be negative!

Many such computational difficulties arise in random (and mixed) effects models.

The one-way random effects ANOVA is a special case of a so-called

*mixed effects*model: $$ \begin{aligned} Y_{n \times 1} &= X_{n \times p}\beta_{p \times 1} + Z_{n \times q}\gamma_{q \times 1} \\ \gamma &\sim N(0, \Sigma). \end{aligned} $$Various models also consider restrictions on $\Sigma$ (e.g. diagonal, unrestricted, block diagonal, etc.)

- Our multiple linear regression model is a (very simple) mixed-effects model with $q=n$, $$ \begin{aligned} Z &= I_{n \times n} \\ \Sigma &= \sigma^2 I_{n \times n}. \end{aligned} $$

`lme`

¶In [ ]:

```
library(nlme)
sodium.lme = lme(fixed=sodium~1,random=~1|brand, data=sodium.table)
summary(sodium.lme)
```

For reasons I'm not sure of, the degrees of freedom don't agree with our
ANOVA, though we do find the correct `SE`

for our estimate of $\mu$:

In [ ]:

```
MSTR = anova(sodium.lm)$Mean[1]
sqrt(MSTR/48)
```

The intervals formed by `lme`

use the 42 degrees of freedom, but
are otherwise the same:

In [ ]:

```
intervals(sodium.lme)
```

In [ ]:

```
center = mean(sodium.table$sodium)
lwr = center - sqrt(MSTR / 48) * qt(0.975,42)
upr = center + sqrt(MSTR / 48) * qt(0.975,42)
data.frame(lwr, center, upr)
```

Using our degrees of freedom as 7 yields slightly wider intervals

In [ ]:

```
center = mean(sodium.table$sodium)
lwr = center - sqrt(MSTR / 48) * qt(0.975,7)
upr = center + sqrt(MSTR / 48) * qt(0.975,7)
data.frame(lwr, center, upr)
```