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Chi-square tests: goodness of fit and independence

H₀: the counts follow the expected mix · H₁: they do not

χ² = 1.3493 · df = 3 · p = 0.7175

p = 0.7175 > α = 0.05: H₀ is not rejected. The counts fit the expected mix within chance (χ² = 1.3493). Failing to reject is not proof that the mix is exactly as expected.

How to read the picture

The curve is the null world: how χ² would vary if the counts only wobbled around the expected ones.
The vertical line is your χ²: the sum of every cell's (O − E)²/E.
The shaded tail is the p-value; χ² tests look only at the upper tail, since any departure makes χ² larger.
The tables below show where the departure lives: the contributions, or the residuals (O − E)/√E.

Your data

Test

Goodness of fit: counts against an expected mix

Independence: a table of counts

Observed counts

One count per category, separated by commas.

Expected mix

Shares or counts in the same order; they are scaled to the observed total.

Parameters estimated from the data (ddof)

Significance level α

α = 0.1

α = 0.05

α = 0.01

α = 0.001

A chi-square test compares counts with the counts a hypothesis expects. For goodness of fit, the expected counts come from a planned mix; for independence, from the table's own margins. Each cell adds (O − E)²/E to χ², and the total lands on the chi-square curve. Type counts, never percentages.

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

The estimate and what the data show

Observed, expected and each category's contribution
Category	O	E	(O − E)² / E
1	742	736	0.04891
2	388	384	0.04167
3	271	288	1.0035
4	199	192	0.2552

The worked steps

The total count
n = 742 + 388 + 271 + 199 = 1600
Expected counts
E = n × share = 736, 384, 288, 192
Each category's contribution
(O − E)² / E = 0.04891 + 0.04167 + 1.0035 + 0.2552
The χ² statistic
χ² = 1.3493
Degrees of freedom
df = k − 1 − ddof = 4 − 1 − 0 = 3
The p-value
p = P(χ² ≥ 1.3493) = 0.7175
Critical value at α
α = 0.05: χ²* = 7.8147
The smallest expected count
min E = 192
Effect size (Cohen's w)
w = √(χ² / n) = 0.02904

The same in code

scipy

from scipy import stats

stats.chisquare(f_obs=[742, 388, 271, 199], f_exp=[736, 384, 288, 192], ddof=0)

Excel

=CHISQ.TEST(A2:A5, B2:B5)
=CHISQ.DIST.RT(1.3493, 3)

Excel's CHISQ.TEST works out df itself as k − 1, which is wrong when a parameter was estimated from the data (ddof > 0); CHISQ.DIST.RT with your own df is then the way.

Formulas

Each category adds its squared miss, scaled by what was expected; df counts the categories that are free to vary.

Try this

Counts, never percentages

Open the contact mix and double every observed count (1484, 776, 542, 398).

Look for: The shares stay the same, but χ² doubles and the p-value falls: χ² depends on how many were counted, which is why it must be given counts, not percentages.
Yates on and off

Open Helion H1 as a 2 × 2 table and switch Yates' correction off.

Look for: Without the correction, χ² equals z² from the two-proportion test (4.095² = 16.77), and the p-values agree. Yates shrinks each |O − E| by a half, which matters only for small counts.
Where the dependence lives

Open segment × payment method and read the residuals table.

Look for: The largest residual, Corporate × bank transfer, says corporate customers pay by transfer far more than independence predicts. χ² says there is dependence; the residuals say where.