How to calculate an ANOVA table
Calculations by Hand
We look at the following example: Let us say we measure the height of some plants under the effect of 3 different fertilizers. Treatment X Y Z Overall mean STEP 0: The model: Yij = µ + Ai + i ij

Measures Mean 1 2 2 ... 5 6 5 ... 2 1 ... // . . .

ˆ Ai ... ... ...

(0.1) (0.2)

n i Ai = 0

Interpretation: An observation yij is given by: the average height of the plants (µ), plus the effect of the fertilizer (Ai ). and an ”error” term ( ij ), i.e. every seed is different and therefore any plant will be different. All these values (µ, Ai , ij ) are UNKNOWN! Our GOAL is to test if the hypothesis A1 = A2 = A3 = 0 is plausible1 . Remark 1 If we have a control group (for example treatment ”X” is ”without any fertilizer”, then we assume that the values of X are in some way the best approximation for µ, therefore we can choose A1 = 0 is spite of condition (0.2). STEP 1: complete the first table. For the treatment means it is enough to calculate the mean of the values M eanX M eanY M eanZ
1

= = =

1+2+2 = 1.667 3 5+6+5 = 5.333 3 1+2 = 1.5 2

We DO NOT find ”the correct value” for the Ai We WILL NOT find which factor (treatment) has an effect, we just look if in general treatments has effect on the results.

1

The (estimated) overall mean (ˆ, which is an estimation of the exact, unµ known overall mean µ) is calculated as follows2 : µ = ˆ 1+2+2+5+6+5+2+1 =3 8

ˆ The estimated effects Ai are the difference between the ”estimated treatment mean” and the ”estimated overall mean”, i.e. ˆ Ai = M eani − µ ˆ So ˆ A1 = 1.667 − 3 = −1.333 ˆ A2 = 5.333 − 3 = 2.333 ˆ A3 = 1.5 − 3 = −1.5

Then: Treatment X Y Z Overall mean STEP 2: The ANOVA table. Cause of the variation Treatment Residuals Total F Krit ... Measures Mean 1 2 2 1.667 5 6 5 5.333 2 1 1.5 // 3 ˆ Ai -1.333 2.333 -1.5

df ... ... ...

SS ... ... ...

MS ... ...

F ...

For the column df (degrees of freedom) just remember the rule ”minus one”: We have 3 different Treatments ⇒ dftreat = 3 − 1 = 2 We have 8 different measurements ⇒ dftot = 8 − 1 = 7 dftreat + dfres = dftot ⇒ dfres = 7 − 2 = 5 For the column SS (sum of squares) we can proceed as follows:
2

Remark that the overall mean does not necessary coincide with the mean of the yi. !

2

SStreat = ”sum of squares between treatment groups” ˆi A2 · #measures = = (−1.33)2 · 3 + (2.33)2 · 3 + (1.5)2 · 2 = 26.17 SSres = ”sum of squares within treatment groups” = i j

(yij − yi. )2 = i SSrowi

=

(1 − 1.667)2 + (2 − 1.667)2 + (2 − 1.667)2 + [0.667] + [0.5]

= 1.83 SStot = ”Total sum of squares” = i,j (yij − µ)2 ˆ

= (1 − 3)2 + (2 − 3)2 + . . . + (1 − 3)2 = 28 Remark 2 The total ”SS” is always equal to the sum of the other ”SS”! SStot = SStreat + SSres 28 = 26.17 + 1.83 For the column M S (mean square) just remember the rule M S = SS/df , then: M Streat = M Sres SStreat dftreat SSres = dfres = = 26.17 = 13.08 2 1.83 = 0.37 5

The F-value is just given by: F = M Streat 13.08 = = 35.68 M Sres 0.37

Interpretation: The F − value says us how far away we are from the hypothesis ”we can not distinguish between error and treatment”, i.e. ”Treatment is not relevant according to our data”! A big F −value implies that the effect of the treatment is relevant! Remark 3 A small F −value does NOT imply that the hypothesis Ai = 0 ∀i is true. (We just can not conclude that it is false!) 3

STEP 3: The decision: Similar as for a T-test we calculate the critical value for the level α = 5% with degrees of freedom 2 and 5 (just read off the values from the appropriate table)3 . krit α = 5% ⇒ F2,5 (5%) = 5.79 krit We have calculated F = 35.68 > F2,5 (5%). Consequently we REJECT THE HYPOTHESIS A1 = A2 = A3 = 0!!! Similarly we could obtain the same result by calculating the p − value

p = 0.11% ⇐ F2,5 (p) = 35.68 0.11% is less than 5%. Consequently we reject the hypothesis A1 = A2 = A3 = 0!!!

Calculations with R
STEP 0: Insert the data v