Lecture 9 The logic of an ANOVA

• Analysis of variance
• explanatory variables are called factors which have levels
  • sex could be a factor and has two levels (male and female)
• always more than two levels (otherwise you would use a t-test)
• Why not just use multiple t-tests
  • multiple testing
  • ANOVA does other cool things which we get into at the end of the lecture

• We say P < 0.05 is significant, but what does that mean
• If there is less than a 1 in 20 probability of getting this result or an even more extreme one by random chance, we accept it as true
• Now imagine we were comparing A to B to C, thats three t-tests (AB, BC, AC), now 0.05 is 3 in 20
• A:B:C:D 6 in 20
• A:B:C:D:E 10 in 20, 0.05 now is 50:50
• R.A. Fisher save us from this nightmare

• ANOVA is used to compare means by comparing variance (how?)

oneway <- read.csv("~/Dropbox/Teaching/old_teaching/zipped/oneway.csv")
names(oneway)

[1] "ozone"  "garden"

Calculate the residual of each point from the mean of y

• This overall variation is the total sum of squares SSY \[ SSY = \Sigma(y-\bar{y})^2 \]

Lets look at the data separated into its levels (garden)

The means are different but are they significantly different?

Calculate the residuals from the individual means

• This variation from the treatment means is the error sum of squares SSE \[ SSE = \Sigma_{j=1}^k\Sigma(y-\bar{y_j})^2 \]
• SSY = SSE + SSA
• SSA is the variation due to treatment

• imagine the means are not different
  • then the residuals would be the same as the previous graph (because the horizontal lines would not have moved) (SSE = SSY)
• imagine now that the means are different (the amount of ozone in the two gardens is different)
  • We would predict that the residuals should be smaller when computed from the individual means (SSE) compared to the residuals computed from the overall mean (SSY)
• We are back to signal versus noise (SSE vs SSA)
• How do we do that in a test?