SME
Grade 12 Math · Problem type
Three groups from their means and SDs
Use this for “Three groups of 5 have means 10, 14 and 12 and SDs 2, 3 and 2. Do the means differ?”
Chart
The F curve with df₁ = 2 on top and df₂ = 12 underneath.
The test statistic is F = 3.529, and the area past it is P = 0.0623.
The critical value for α = 0.05 is 3.885: F is not in the rejection region, so P ≥ α and we fail to reject H₀.
This is an example. Type your own values to start a new problem.
x̄
=
+
+
3
MSW
=
s₁
2
+
s₂
2
+
s₃
2
3
F
=
MSB
MSW
n
Size of each group
example
df₂
Degrees of freedom within groups
worked out (not typed)
P
p-value
worked out (not typed)
α
Significance level
example
Formulas
x̄
=
x̄₁
+
x̄₂
+
x̄₃
3
12
=
10 + 14 + 12
3
df₂
=
3
×
n
−
3
12 = 3 × 5 − 3
MSB
=
n
×
(
(
x̄₁
−
x̄
)
2
+
(
x̄₂
−
x̄
)
2
+
(
x̄₃
−
x̄
)
2
)
÷
2
20
=
5
×
(
(10 − 12)
2
+
(14 − 12)
2
+
(12 − 12)
2
)
÷
2
MSW
=
s₁
2
+
s₂
2
+
s₃
2
3
5.6667
=
2
2
+
3
2
+
2
2
3
F
=
MSB
MSW
3.5294
=
20
5.6667
P
=
Fcdf(
F
,
∞,
2,
df₂)
0.0623 = Fcdf(3.5294, ∞, 2, 12)
Assumptions
•
Three groups of the same size n: MSB = n × Σ(x̄ᵢ − x̄)² ÷ 2, where x̄ is the grand mean.
•
MSW = (s₁² + s₂² + s₃²) ÷ 3, the average of the three variances.
•
df₁ = 3 − 1 = 2 and df₂ = 3n − 3; the groups are independent, normal and equally spread.
Step-by-step
WE KNOW
n = 5, x̄₁ = 10, x̄₂ = 14, x̄₃ = 12, s₁ = 2, s₂ = 3, s₃ = 2, α = 0.05
FIND
grand mean (x̄), degrees of freedom within groups (df₂), mean square between groups (MSB), mean square within groups (MSW), F statistic (F), p-value (P)
Step 1 · Find grand mean (x̄)
Use x̄ = (x̄₁ + x̄₂ + x̄₃) ÷ 3
The groups are the same size, so the grand mean is the mean of the three means.
x̄
=
10 + 14 + 12
3
x̄
=
36
3
x̄
=
12
Step 2 · Find degrees of freedom within groups (df₂)
Use df₂ = 3 × n − 3
N
−
k:
3
n
values
in
all,
less
one
for
each
group’s
mean.
df₂ = 3 × 5 − 3
df₂ = 15 − 3
df₂ = 12
Step 3 · Find mean square between groups (MSB)
Use MSB = n × ((x̄₁ − x̄)² + (x̄₂ − x̄)² + (x̄₃ − x̄)²) ÷ 2
Each
group
mean
stands
for
its
n
values,
so
its
squared
distance
from
the
grand
mean
counts
n
times;
divide
by
df₁
=
2.
MSB
=
5
×
(
(10 − 12)
2
+
(14 − 12)
2
+
(12 − 12)
2
)
÷
2
MSB
=
5
×
(
(−2)
2
+
2
2
+
0
2
)
÷
2
MSB
=
5
×
4 + 4
2
MSB
=
5
×
8
2
MSB
=
40
2
MSB = 20
Step 4 · Find mean square within groups (MSW)
Use MSW = (s₁² + s₂² + s₃²) ÷ 3
Equal groups: the pooled variance is the average of the three variances.
MSW
=
2
2
+
3
2
+
2
2
3
MSW
=
4 + 9 + 4
3
MSW
=
17
3
MSW = 5.6667
Step 5 · Find F statistic (F)
Use F = MSB ÷ MSW
How much the group means spread compared with the spread inside the groups.
F
=
20
5.6667
F
=
3.5294
Step 6 · Find p-value (P)
Use P = Fcdf(F, ∞, 2, df₂)
The
area
under
the
F
curve
past
the
statistic:
a
large
F
,
means
far
apart
for
their
spread,
leaves
little.
P
=
Fcdf(3.5294,
∞,
2,
12)
P
=
0.0623
→
0.0623
≥
α
=
0.05:
fail
to
reject
H₀:
not
convincing
evidence
that
at
least
one
group
mean
differs
Check
Put the numbers back in. Both sides should match.
12
=
10 + 14 + 12
3
✓
12 = 3 × 5 − 3 ✓
20
=
5
×
(
(10 − 12)
2
+
(14 − 12)
2
+
(12 − 12)
2
)
÷
2
✓
5.6667
≈
2
2
+
3
2
+
2
2
3
✓
3.5294
≈
20
5.6667
✓
0.0623 = Fcdf(3.5294, ∞, 2, 12) ✓
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Analysis of variance (ANOVA) and the F distribution
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Comparing two variances
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