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One-way ANOVA with Tukey's test

H₀: μ₁ = … = μ₄ · H₁: at least one mean differs

Your result lies beyond the edge of the chart; it is marked at the edge.

F = 278.1959 · df = (3, 11996) · p = 1.40E-174

p = 1.40E-174 ≤ α = 0.05: reject H₀. At least one group's mean differs (F = 278.1959, η² = 0.06505); Tukey's test finds 5 of 6 pairs that differ.

How to read the picture

The curve is the null world of F: how the ratio of between to within would vary if every group had the same mean.
The vertical line is your F; the shaded tail is the p-value.
The ANOVA table shows the two sums of squares that make F.
Tukey's intervals: a pair whose interval excludes 0 differs, with the familywise error held at α.

Your data

Groups: n, mean, standard deviation

One group per line (or separated by ;): its size, mean and standard deviation, separated by commas.

Significance level α

α = 0.1

α = 0.05

α = 0.01

α = 0.001

ANOVA asks one question about several groups at once: do their means differ by more than the spread inside the groups would explain? F is the variation between the group means over the variation within the groups. If ANOVA rejects, Tukey's test says which pairs differ, with intervals that hold for all the pairs together.

Not sure this is the right test? Ask the test chooser

The estimate and what the data show

The ANOVA table
Source	SS	df	MS	F	p
between	9467.7615	3	3155.9205	278.1959	1.40E-174
within	136085.4584	11996	11.3442

Tukey's test for every pair (Tukey–Kramer)
Pair	x̄ᵢ − x̄ⱼ	Interval	q	p	p ≤ α
1 − 2	-0.8892	[-1.096, -0.6825]	15.6279	< 1.00E-12	✓
1 − 3	-2.3478	[-2.57, -2.1256]	38.396	< 1.00E-12	✓
1 − 4	-0.1985	[-0.4276, 0.03056]	3.1489	0.1161	✗
2 − 3	-1.4586	[-1.6833, -1.2338]	23.5829	< 1.00E-12	✓
2 − 4	0.6907	[0.4592, 0.9222]	10.8396	< 1.00E-12	✓
3 − 4	2.1493	[1.9038, 2.3947]	31.8222	< 1.00E-12	✓

The worked steps

The grand mean
x̄ = Σ nᵢ x̄ᵢ / N = 14.8972, N = 12000
Between groups
SS_between = Σ nᵢ (x̄ᵢ − x̄)² = 9467.7615
Within groups
SS_within = Σ (nᵢ − 1) sᵢ² = 136085.4584
Mean squares
MS_between = 9467.7615 / 3 = 3155.9205 · MS_within = 136085.4584 / 11996 = 11.3442
The F statistic
F = MS_between / MS_within = 3155.9205 / 11.3442 = 278.1959
The p-value
p = P(F(3, 11996) ≥ 278.1959) = 1.40E-174
Critical value at α
α = 0.05: F* = 2.6056
Effect size (η²)
η² = SS_between / SS_total = 9467.7615 / 145553.2198 = 0.06505
Tukey's critical q
q* (4 groups, df = 11996) = 3.6337

The same in code

scipy, from the data

from scipy import stats

stats.f_oneway(g1, g2, g3, g4)
stats.tukey_hsd(g1, g2, g3, g4)

statsmodels, from the data (a value column and a group column)

from statsmodels.stats.multicomp import pairwise_tukeyhsd

pairwise_tukeyhsd(values, groups, alpha=0.05)

scipy, from the summaries

from scipy import stats

F = 3155.9205 / 11.3442
p = stats.f.sf(F, 3, 11996)
q_crit = stats.studentized_range.isf(0.05, 4, 11996)

Excel

Data Analysis ToolPak: Anova: Single Factor
=F.DIST.RT(278.1959, 3, 11996)

Excel has no Tukey test; scipy's tukey_hsd and statsmodels' pairwise_tukeyhsd both use the Tukey–Kramer form for unequal group sizes.

Formulas

The between sum of squares measures how far the group means sit from the grand mean; the within sum measures the spread inside the groups. Tukey's q compares one pair's gap with the studentized range for k groups.

Try this

Between against within

Open the three machines and make every standard deviation 4 instead of about 2.

Look for: The means have not moved, so the between sum of squares is the same; the within sum grows, F shrinks and the verdict can flip. Differences between groups are judged against the noise inside them.
Size matters

Open Helion's regions with 10 customers each, then Helion's full regions.

Look for: With 10 per region only the largest gap stands out; with thousands, even North and South differ. With thousands, read η² to see how much of the variation the regions explain.
Why not six t tests?

Open Helion's regions and count Tukey's pairs: six comparisons among four groups.

Look for: Six t tests at 5 % each would give a false alarm somewhere about 26 % of the time if nothing differed. Tukey's intervals are wider so that all six together keep the 5 %.