SME
Grade 12 Math · Statistics
Analysis of variance (ANOVA) and the F distribution
Use this for “Complete the ANOVA table and find F and its P-value from the sums of squares.”
Chart
The F curve with df₁ = 2 on top and df₂ = 12 underneath.
The test statistic is F = 5, and the area past it is P = 0.0263.
The critical value for α = 0.05 is 3.885: F is in the rejection region, so P < α and we reject H₀.
This is an example. Type your own values to start a new problem.
MSB
=
SSB
df₁
MSW
=
SSW
df₂
F
=
MSB
MSW
SST
Total sum of squares
worked out (not typed)
P
p-value
worked out (not typed)
α
Significance level
example
Formulas
MSB
=
SSB
df₁
30
=
60
2
MSW
=
SSW
df₂
6
=
72
12
SST = SSB + SSW
132 = 60 + 72
F
=
MSB
MSW
5
=
30
6
P
=
Fcdf(
F
,
∞,
df₁,
df₂)
0.02634 = Fcdf(5, ∞, 2, 12)
Assumptions
•
H₀: every group has the same mean; Hₐ: at least one mean differs.
•
df₁ = k − 1 for k groups and df₂ = N − k for N values in all; each mean square is SS ÷ df.
•
The groups are independent random samples from normal populations with equal spreads.
Step-by-step
WE KNOW
SSB = 60, df₁ = 2, SSW = 72, df₂ = 12, α = 0.05
FIND
mean square between groups (MSB), mean square within groups (MSW), total sum of squares (SST), F statistic (F), p-value (P)
Step 1 · Find mean square between groups (MSB)
Use MSB = SSB ÷ df₁
A mean square is a sum of squares over its df.
MSB
=
60
2
MSB = 30
Step 2 · Find mean square within groups (MSW)
Use MSW = SSW ÷ df₂
The spread inside the groups, pooled: SSW over its df.
MSW
=
72
12
MSW = 6
Step 3 · Find total sum of squares (SST)
Use SST = SSB + SSW
The total spread of every value about the grand mean splits into between and within.
SST = 60 + 72
SST = 132
Step 4 · Find F statistic (F)
Use F = MSB ÷ MSW
How much the group means spread compared with the spread inside the groups.
F
=
30
6
F
=
5
Step 5 · Find p-value (P)
Use P = Fcdf(F, ∞, df₁, df₂)
The
area
under
the
F
curve
past
the
statistic:
a
large
F
,
means
far
apart
for
their
spread,
leaves
little.
P
=
Fcdf(5,
∞,
2,
12)
P
=
0.02634
→
0.02634
<
α
=
0.05:
reject
H₀:
convincing
evidence
that
at
least
one
group
mean
differs
Check
Put the numbers back in. Both sides should match.
30
=
60
2
✓
6
=
72
12
✓
132 = 60 + 72 ✓
5
=
30
6
✓
0.02634 = Fcdf(5, ∞, 2, 12) ✓
Topic check
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