R help - May 2012 - ANOVA question

Hello all,

I'm very satisfied to say that my grip on both R and statistics is
showing the first hints of firmness, on a very greenhorn level.

I'm faced with a problem that I intend to analyze using ANOVA, and to
test my understanding of a primitive, one-way ANOVA I've written the
self-contained practice script below. It works as expected.

But here's my question: How can I not only get the values of the
coefficients for the different levels of the explanatory factor(s),
but also the corresponding standard errors and confidence levels?
Below I have started doing that "on foot" by looping over the levels
of my single factor, but I suppose this gets complicated and messy
with more complex models. Any ideas?

Thanks,
robert


set.seed(0)

N <- 100 # sample size

MEAN <- c(10, 20, 30, 40, 50)
VAR <- c(20,20,1, 20, 20)
LABELS <- c("A", "B", "C", "D",
"E")

# create a data frame with labels
df <- data.frame(Label=rep(LABELS, each=N))
df$Value <- NA
# fill in random data for each factor level
for (i in 1:length(MEAN)) {
    df$Value[(1+N*(i-1)):(N*i)] <- rnorm(N, MEAN[i], sqrt(VAR[i]))
}



par(mfrow=c(2,2))
plot(df)          # Box plot of the data
plot(df$Value)    # scatter plot

mod_aov <- aov(Value ~ Label, data=df)

print(summary(mod_aov))
print(mod_aov$coefficients)

rsd <- mod_aov$residuals

plot(rsd)

# find and print mean() and var() for each level
for (l in levels(df$Label)) {
    index <- df$Label == l

    # Method 1: directly from data
    smp <- df$Value[index]  # extract sample for this label
    ssq_smp <-  var(smp)*(length(smp)-1) # sum of squares is variance
                                         # times d.f.
    # Method 2: from ANOVA residuals
    rsd_grp <- rsd[index]                # extract residuals
    ssq_rsd <- sum(rsd_grp **2)          # compute sum of squares

    # print mean, variance, and difference between SSQs from the two
    # methods.
    write(sprintf("%s: mean=%5.1f var=%5.1f (%.2g)", l,
        mean(smp), var(smp),
        ssq_smp-ssq_rsd), "")
# ...and it works like expected! But is there a shortcut that would give me
# the same result in a one-liner?
}

Dear Robert,

It is easier to use lm instead of aov if you want coefficients for each group.
Note that you can use rnorm vectorised.

set.seed(0)
N <- 100 # sample size
MEAN <- c(10, 20, 30, 40, 50)
VAR <- c(20,20,1, 20, 20)
LABELS <- factor(c("A", "B", "C",
"D", "E"))

# create a data frame with labels
df <- data.frame(Label=rep(LABELS, each=N))
df$Value <- rnorm(nrow(df), mean = MEAN[df$Label], sd = sqrt(VAR[df$Label]))

mod_aov <- aov(Value ~ Label, data=df)
mod_lm <- lm(Value ~ Label, data = df)
all.equal(anova(mod_aov), anova(mod_lm))

summary(mod_aov)
summary(mod_lm)

summary(mod_lm)$coef
confint(mod_lm)

#without intercept
mod_lm0 <- lm(Value ~ 0 + Label, data = df)
summary(mod_lm0)$coef
confint(mod_lm0)

Best regards,

Thierry


ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54 43 61 85
Thierry.Onkelinx at inbo.be
www.inbo.be

To call in the statistician after the experiment is done may be no more than
asking him to perform a post-mortem examination: he may be able to say what the
experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not ensure
that a reasonable answer can be extracted from a given body of data.
~ John Tukey

-----Oorspronkelijk bericht-----
Van: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
Namens Robert Latest
Verzonden: vrijdag 11 mei 2012 9:37
Aan: r-help at r-project.org
Onderwerp: [R] ANOVA question

Hello all,

I'm very satisfied to say that my grip on both R and statistics is showing
the first hints of firmness, on a very greenhorn level.

I'm faced with a problem that I intend to analyze using ANOVA, and to test
my understanding of a primitive, one-way ANOVA I've written the
self-contained practice script below. It works as expected.

But here's my question: How can I not only get the values of the
coefficients for the different levels of the explanatory factor(s), but also the
corresponding standard errors and confidence levels?
Below I have started doing that "on foot" by looping over the levels
of my single factor, but I suppose this gets complicated and messy with more
complex models. Any ideas?

Thanks,
robert


set.seed(0)

N <- 100 # sample size

MEAN <- c(10, 20, 30, 40, 50)
VAR <- c(20,20,1, 20, 20)
LABELS <- c("A", "B", "C", "D",
"E")

# create a data frame with labels
df <- data.frame(Label=rep(LABELS, each=N)) df$Value <- NA # fill in
random data for each factor level for (i in 1:length(MEAN)) {
    df$Value[(1+N*(i-1)):(N*i)] <- rnorm(N, MEAN[i], sqrt(VAR[i])) }



par(mfrow=c(2,2))
plot(df)          # Box plot of the data
plot(df$Value)    # scatter plot

mod_aov <- aov(Value ~ Label, data=df)

print(summary(mod_aov))
print(mod_aov$coefficients)

rsd <- mod_aov$residuals

plot(rsd)

# find and print mean() and var() for each level for (l in levels(df$Label)) {
    index <- df$Label == l

    # Method 1: directly from data
    smp <- df$Value[index]  # extract sample for this label
    ssq_smp <-  var(smp)*(length(smp)-1) # sum of squares is variance
                                         # times d.f.
    # Method 2: from ANOVA residuals
    rsd_grp <- rsd[index]                # extract residuals
    ssq_rsd <- sum(rsd_grp **2)          # compute sum of squares

    # print mean, variance, and difference between SSQs from the two
    # methods.
    write(sprintf("%s: mean=%5.1f var=%5.1f (%.2g)", l,
        mean(smp), var(smp),
        ssq_smp-ssq_rsd), "")
# ...and it works like expected! But is there a shortcut that would give me #
the same result in a one-liner?
}

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