One-Way ANOVA F-Test

Simplest ANOVA


It's an $F$-Test

It's a parametric test of Variance:

  • it's parametric because it's based on Normality hypothesis


Comparing Means

Comparing means of several groups

  • We can compare means of two groups using Two-Sample $t$-test
  • But sometimes we want to compare means across many groups

First idea: do Pairwise comparison

  • but we may find the difference just by chance, even when there's no difference - because there'll be too many comparisons
  • need to do some correction


Test Of Independence

One-Way ANOVA can be used to analyze the relationships between two variables

  • numerical and categorical
  • we group the numerical one by associated categories
  • and find the means of each category
  • then we test if the means are the same
  • if yes - these two variables are independent


Post-ANOVA Comparison

suppose we reject $H_0$


The One-Way ANOVA Test

ANOVA: compare many means in a single hypothesis

  • are the means of several groups equal?
  • so it generalizes the $t$-test to more than two groups
  • used in Bivariate Analysis to test if a numerical and categorical variables are independent


ANOVA:

  • assessing the variability of the group mean relative to variability among individual observations
  • the questions answered: "is the variability on the sample means so large that it seems unlikely to be entirely due to chance?"


General Idea:

  • simultaneously consider many groups
  • evaluate if the sample means differ more than we'd expect from natural variation


Assumptions

  • observations are independent
  • data within each group is distributed normally
  • variability across the groups is approximately equal


ANOVA Test

  • $H_0: \mu_1 = \mu_2 = ... = \mu_k$
  • $H_A:$ at least one $\mu_i$ is different from the rest


Estimating variability:

  • this variability is called MSG: mean square between groups
  • $\text{df}_G = k - 1$ for $k$ groups


  • let $\bar{x}$ be the sample mean across all groups
  • $\text{MSG} = \cfrac{1}{\text{df}_G} \cdot \text{SSG} = \cfrac{1}{k - 1} \sum_{i=1}^k n_i (\bar{x}_i - \bar{x})^2$
  • $\text{SSG}$ - sum of squares between groups
  • $n_i$ - sample size of group $i$

this gives us a base point, and we compare it with $\text{MSE}$:


$\text{MSE}$: mean square error (pooled variance estimate)

  • $\text{df}_E = n - k$
  • $\text{SST} = \sum_{i=1}^n (x_i - \bar{x})^2$
  • sum of squares total
  • this is calculated over all observations in the data set
  • $\text{SSE} = \text{SST} - \text{SSG} = (n_1 - 1) \cdot s^2_1 + (n_2 - 1) \cdot s^2_2 + ... + (n_k - 1) \cdot s^2_k$
  • $s^2_i$ - sample variance of residuals in the group

sum of squared error standardized form of $\text{SSE}$: $\text{MSE} = 1 / \text{df}_E \text{SSE}$


if $H_0$ is true then differences are due to chance and MSG and MSE should be approximately equal

Then we can calculate the test statistics $F = \cfrac{\text{MSG}}{\text{MSE}}$


  • $\text{MSG}$ = between the group variability
  • $\text{MSE}$ = withing the group variability

$F$ is a $F$ statistics that follows $F$-distribution it has 2 associated parameters: $\text{df}_1$ and $\text{df}_2 $ for ANOVA it's $\text{df}_G$ and $\text{df}_E$

the larger $\text{MSG}$ relative to $\text{MSE}$, the larger $F$ is, and the stronger evidence against $H_0$


We use upper tail to compute the $p$-value


Test (From Data Mining (UFRT))

Given

  • $X$ with $k$ possible values $x_1, ..., x_k$ - categorical variable (e.g. Job)
  • and $Y$ - continuous variable (e.g. Age)

Let

  • $\mu = \text{mean}(Y)$: Mean value of $Y$
  • $\mu_k$: mean value of $Y$ for tuples such that $X = x_k$
  • $N_k$ : number of records such that that $X = x_k$
  • $N = \sum_k N_k$ : total number of records

Define:

  • Interclass Variance: $\text{Inter} = \cfrac{1}{K-1} \cdot \sum_k N_k \cdot (\mu_k - \mu)^2$
    • total variance
  • Intraclass Variance: $\text{Intra} = \cfrac{1}{N-K} \cdot \sum_k \sum_{j : X = x_k} ( y_j - \mu_k )^2$
    • variance inside each group

Test

  • to evaluate the correlation between $X$ and $Y$ calculate $F = \cfrac{\text{Inter}}{\text{Intra}}$
  • the null hypothesis $H_0$: all means $\mu_k$ are equal (i.e. assume independence),
  • under $H_0$ $F$-ratio follows $F_{K-1,N-K}$: $F$-distribution with $K-1,N-K$ degrees of freedom
  • if independent, all the means should be the same for all classes and $F$ should be 0


Examples

Example 1: Baseball

Batting Performance

we have 4 categories of baseball players:

  • outfielders: $\text{OF}$
  • infielders: $\text{IF}$
  • jilter: $\text{DH}$
  • catcher: $\text{C}$

Is there any difference in performance? (using on-base percentage OBP to measure it)


Test:

  • $H_0: \mu_\text{OF} = \mu_\text{IF} = \mu_\text{DH} = \mu_\text{C}$
  • $H_A$: at least one is different

we approximate each $\mu$ by $\bar{x}$


OF IF DH C
Summary statistics (source: table 5.27, OpenIntro)
Sample size ($n_i$) 120 154 14 39
Sample mean ($\bar{x}_i$) 0.334 0.332 0.348 0.323
Sample SD ($s_i$) 0.029 0.037 0.036 0.045


05a241ce52204838a53ad13554c3372d.png (source: fig 5.28, OpenIntro)


We see that DH and C look really different. Why don't we just check if $\mu_\text{DH} = \mu_\text{C}$?

  • the primary issue: we're inspecting the data before doing the check
  • this is called Data Snooping (or Data Fishing)
  • naturally we'd pick up the groups with largest differences and run the formal test
  • but it would lead to Type I Errors
  • it's also called Prosecutor's Fallacy


Calculate:

  • $\text{MSG} = 0.00252$
  • $\text{MSE} = 0.00121$
  • $k = 4$ groups, $\text{df_G} = 4 - 1 = 3$
  • $n = n_1 + n_2 + n_3 + n_4 = 327$, $\text{df}_E = n - k = 323$
  • $F = \cfrac{MSG}{MSE} = \cfrac{0.00252}{0.00121} \approx 1.994$


P-value

  • $p$-value is 0.115 > 0.05
  • 08a9f7d43c8c4d94a17e1512b49294f4.png
  • so we fail to reject $H_0$ at the significance level $\alpha=0.05$


Example 2: Statistics Class

We have high demand for a course, so run it several times in one semester

  • e.g. it's run 3 times: scores of each run are sets $A, B, C$
  • are these significant differences?

Test:

  • $H_0: \mu_A = \mu_B = \mu_C$
  • $H_A:$ scores may vary on average

b24e9fd275934951b45bc6114662b2db.png


Group $n$ Min. Mean Max. Std
$A$ 58 44 75.10 100 13.86867
$B$ 55 38 71.96 100 13.77056
$C$ 51 45 78.94 100 13.12008


We run ANOVA tests and obtain the following summary table:

Df Sum Sq Mean Sq $F$ value $Pr(>F)$
lecture 2 1290.11 645.06 3.48 0.0330
Residuals 161 29810.13 185.16


The $p$-value is greater than $\alpha=0.05$, so we reject $H_0$


library(openintro)
data(classData)

by(classData$m1, classData$lecture, summary)
by(classData$m1, classData$lecture, sd)
by(classData$m1, classData$lecture, length)


boxplot(classData$m1 ~ classData$lecture, col='skyblue', axes=F)
axis(side=2)
axis(side=1, at=1:3, labels=c('A', 'B', 'C'))


oneway.test(classData$m1 ~ classData$lecture, var.equal=T)
# or
aov1 = aov(classData$m1 ~ classData$lecture)
summary(aov1)


Post-ANOVA processing: use $t$-test to pairwise compare $A,B,C$

  • With Bonferroni Correction, $\alpha^* = \alpha / 3 = 0.05 / 3 = 0.017$
  • $A$ vs $B$: $p$-value is 0.228, don't reject
  • $A$ vs $C$: $p$-value is 0.148, don't reject
  • $B$ vs $C$: $p$-value is 0.01, reject


In R:

pairwise.t.test(classData$m1, classData$lecture,
                alternative='two.sided', p.adjust.method='bonferroni')

Or

TukeyHSD(aov1)


Example 3: Donuts

Example from [1]

  • study of donuts: the relationship between the amount of absorbed fat vs the type of fat
  • is there any relationship?

The data:

Fat1 Fat2 Fat3 Fat4
164 178 175 155
172 191 193 166
168 197 178 149
177 182 171 164
156 185 163 170
195 177 176 168

f46d8ce0d0e14e4795e96c2ca69760a1.png


We run ANOVA analysis and get the following:

  • $F = 5.4063$,
  • num $\text{df} = 3$, denom $\text{df} = 20$,
  • $p\text{-value} = 0.006876$


file = 'http://courses.statistics.com/software/data/donuts.txt'
donuts = read.table(file, header=T)
donuts = stack(donuts)
donuts

boxplot(donuts$values ~ donuts$ind)
oneway.test(donuts$values ~ donuts$ind, var.equal=TRUE)
# p\text{-value} is small, we reject the hypothesis of equal absorption. 


Same, done in steps:

groups = 4
# total variance
df.g = groups - 1
tot.mean = mean(donuts$values)

group.mean = tapply(donuts$values, donuts$ind, mean)
n = tapply(donuts$values, donuts$ind, length)

inter = sum(n * (group.mean - tot.mean) ^ 2) / df.g

# variance inside each group
df.e = length(donuts$values) - groups
intra.1 = tapply(donuts$values, donuts$ind, FUN=function(data) {
  m = mean(data)
  sum( (data - m)^ 2)
})
intra = sum(intra.1) / df.e

F.stat = inter/intra
F.stat

p = 1 - pf(F.stat, df1=df.g, df2=df.e)
p

27fdcec86c774679a8146a1ba4147b41.png

x = seq(0, 6, 0.05)
y = df(x, df1=df.g, df2=df.e)
plot(x, y, type='l')

abline(v=F.stat)


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