SME
Grade 12 Math · Problem type
Comparing two variances
Use this for “Two samples of 16 have SDs 6 and 4. Are the population variances different?”
Chart
The F curve with df₁ = 15 on top and df₂ = 15 underneath.
The test statistic is F = 2.25; the smaller tail past it is 0.0637, doubled for both tails: P = 0.1274.
The critical values for α = 0.05 are 0.3494 and 2.862: 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.
F
=
s₁
2
s₂
2
n₁
Size of the larger-SD sample
example
n₂
Size of the smaller-SD sample
example
df₁
Degrees of freedom on top
worked out (not typed)
df₂
Degrees of freedom underneath
worked out (not typed)
Hₐ
Alternative hypothesis
example
P
p-value
worked out (not typed)
α
Significance level
example
Formulas
df₁
=
n₁
−
1
15 = 16 − 1
df₂
=
n₂
−
1
15 = 16 − 1
F
=
s₁
2
s₂
2
2.25
=
6
2
4
2
P
=
the
F
tail
past
F
on
Hₐ’s
side,
with
df₁
and
df₂
degrees
of
freedom
0.1274
=
the
F
tail
past
2.25
on
σ₁
2
≠
σ₂
2
’s
side,
with
15
and
15
degrees
of
freedom
Assumptions
•
H₀: σ₁² = σ₂²; Hₐ is σ₁² ≠ σ₂², or σ₁² > σ₂² for a larger first variance. Put the larger SD first so F = s₁² ÷ s₂² ≥ 1.
•
F has df₁ = n₁ − 1 on top and df₂ = n₂ − 1 underneath; for ≠ the p-value doubles the tail past F, for > it is that tail.
•
Both populations must be normal: this test is very sensitive to skew.
Step-by-step
WE KNOW
s₁ = 6, s₂ = 4, n₁ = 16, n₂ = 16, Hₐ: σ₁² ≠ σ₂², α = 0.05
FIND
degrees of freedom on top (df₁), degrees of freedom underneath (df₂), F statistic (F), p-value (P)
Step 1 · Find degrees of freedom on top (df₁)
Use df₁ = n₁ − 1
The
first
sample’s
variance
has
n₁
−
1
degrees
of
freedom.
df₁ = 16 − 1
df₁ = 15
Step 2 · Find degrees of freedom underneath (df₂)
Use df₂ = n₂ − 1
The
second
sample’s
variance
has
n₂
−
1
degrees
of
freedom.
df₂ = 16 − 1
df₂ = 15
Step 3 · Find F statistic (F)
Use F = s₁² ÷ s₂²
The ratio of the variances: near 1 when the spreads match.
F
=
6
2
4
2
F
=
36
16
F
=
2.25
Step 4 · Find p-value (P)
Use P = the F tail past F on Hₐ’s side, with df₁ and df₂ degrees of freedom
Hₐ
≠
doubles
the
smaller
tail
of
the
F
curve
at
F
;
Hₐ
>
takes
the
right
tail
past
F
alone.
P
=
2
×
Fcdf(
F
,
∞,
df₁,
df₂)
P
=
2
×
Fcdf(2.25,
∞,
15,
15)
P
=
0.1274
→
Two
tails;
0.1274
≥
α
=
0.05:
fail
to
reject
H₀:
not
convincing
evidence
that
the
population
variances
differ
Check
Put the numbers back in. Both sides should match.
15 = 16 − 1 ✓
15 = 16 − 1 ✓
2.25
=
6
2
4
2
✓
0.1274 = 2 × Fcdf(2.25, ∞, 15, 15) ✓
Topic check
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