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From Design and Analysis of Experiments, by Douglas C Montgomery,

An example for ANOVA is given using the following data set,

enter image description here

With a conclusion being

enter image description here

But I am having a lot of trouble trying to do this myself.

I understand the results using manual calculations, but I want to do it in R.

What I did in R was as follows

data<-c(160,160,160,160,160,180,180,180,180,180,200,200,200,200,200,220,220,220,220,220) Etch160<-c(575,542,530,539,570) Etch180<-c(565,593,590,579,610) Etch200<-c(600,651,610,637,629) Etch220<-c(725,700,715,685,710) response<-c(Etch160,Etch180,Etch200,Etch220) dataframe<-data.frame(response,data) t<-aov(data~response,data=dataframe) summary(t)

For which I get a completely different answer and form.

I am wondering firstly, why? Secondly, how can I get a table like the book did using R? Is there some function I am missing? Where is my mistake?

Running my code gives enter image description here

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1 Answer 1

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The correct code you need to replicate the anova table you provided is as follows:

data<-c(160,160,160,160,160,180,180,180,180,180,200,200,200,200,200,220,220,220,220,220)

Etch160<-c(575,542,530,539,570)
Etch180<-c(565,593,590,579,610)
Etch200<-c(600,651,610,637,629)
Etch220<-c(725,700,715,685,710)

response<-c(Etch160,Etch180,Etch200,Etch220)

dataframe<-data.frame(response,data)

dataframe$data <- factor(dataframe$data)

t <- aov(response ~ data,data=dataframe)

summary(t)

Your first mistake was in mis-specifying the formula used inside the aov() function. This formula is of the form outcome ~ factor. In your case, the factor is RF Power - this factor has 4 levels, hence the 4 - 1 = 3 degrees of freedom listed next to it in the anova table. The 4 levels are 160, 180, 200 and 220.

Your second mistake was in not converting the RF Power variable (which you called "data") to a factor using the factor() function.

Addendum:

To check the normality assumption for the one-way ANOVA analysis you performed, simply use the R code below:

t <- aov(response ~ data,data=dataframe)

r <- residuals(t) 

par(mfrow=c(1,3))
hist(r, main="Histogram of Residuals", xlab="Residual") 
plot(density(r), main="Density Plot of Residuals")
qqnorm(r, main="Normal Quantile-Quantile Plot \nof Residuals")
qqline(r)

This code will produce the histogram of the residuals, the density plot of the residuals and the normal quantile-quantile plot of the residuals. You can examine these plots to see if the residuals look like they come from a normal distribution.

Rather than using aov() to conduct the one-way ANOVA, you could use the lm() command as follows:

m <- lm(response ~ data,data=dataframe)

anova(m) 

Then you could check the normality assumption for your linear model like so:

r <- residuals(m) 

par(mfrow=c(1,3))
hist(r, main="Histogram of Residuals", xlab="Residual") 
plot(density(r), main="Density Plot of Residuals")
qqnorm(r, main="Normal Quantile-Quantile Plot \nof Residuals")
qqline(r)

So both aov() and lm() use the same syntax in the case of one-way ANOVA: outcome ~ factor.

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    $\begingroup$ Thank you, that makes sense. By the way, I was wondering how to check normality assumptions in cases such as above. For example, what model do we fit using lm. Do we need to choose one to be the response, and the rest predictors? $\endgroup$
    – Quality
    Feb 6, 2019 at 6:12
  • $\begingroup$ Please see the Addendum to my original answer. $\endgroup$ Feb 6, 2019 at 15:19
  • 1
    $\begingroup$ Nice, helpful answer. As readers might be likely to encounter situations that aren't so nicely balanced as in this example, I'm including a link to issues that arise when you use anova() or aov() versus Anova() from the car package in such situations. People are often surprised to see that the apparent results can then differ depending on the order of entry of predictors into the model. $\endgroup$
    – EdM
    Feb 6, 2019 at 15:47

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