Chi-Squared Goodness of Fit Tests

The chi-squared ($\chi^2$χ2) goodness-of-fit test answers a question that runs through all of applied statistics: does my data actually follow the model I assumed? You have a set of observed frequencies and a candidate distribution — uniform, binomial, Poisson, normal — that predicts expected frequencies. The test measures the total discrepancy with a single statistic and compares it against the $\chi^2$χ2 distribution. This lesson covers the full method with correct degrees-of-freedom bookkeeping, the all-important class-pooling rule, and three fully-worked tests with mark schemes.

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1. Where this sits in AQA 7367

This is Paper 3 Statistics option (7367/3S) content (per-paper weighting AO1 40% / AO2 25% / AO3 35%) — the first of two hypothesis-testing lessons (Lesson 10 covers contingency tables). Computing expected frequencies and the test statistic is AO1; the reasoning — choosing the degrees of freedom, deciding when to pool classes, and writing a conclusion in context — is heavily AO2/AO3. The prerequisites are the Poisson and binomial distributions (Lessons 2–3) for generating expected frequencies, and the general logic of hypothesis testing (null/alternative, significance level, critical value) from A-Level Mathematics.

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2. Core theory

The idea and the statistic

If the model is correct, observed frequencies $O_i$Oi​ should sit close to the expected frequencies $E_i$Ei​; large gaps are evidence against the model. The test statistic measures the total relative gap:

$X^2 = \sum_i \frac{(O_i - E_i)^2}{E_i}.$X2=∑i​Ei​(Oi​−Ei​)2​.

Squaring makes every term non-negative (so over- and under-estimates do not cancel); dividing by $E_i$Ei​ weights each gap relative to what was expected (a gap of 5 matters more where 10 was expected than where 100 was). Under $H_0$H0​, $X^2$X2 is approximately distributed as chi-squared with $\nu$ν degrees of freedom, written $\chi^2_\nu$χν2​. The test is one-tailed (upper): only a large $X^2$X2 discredits the model.

It is worth pausing on why the statistic takes this particular form. Each term $\tfrac{(O_i - E_i)^2}{E_i}$Ei​(Oi​−Ei​)2​ is, to a good approximation, a squared standardised residual: if the count in class $i$i behaves like a Poisson (or binomial) variable with mean $E_i$Ei​, its standard deviation is roughly $\sqrt{E_i}$Ei​​, so $\tfrac{O_i - E_i}{\sqrt{E_i}}$Ei​​Oi​−Ei​​ is a standardised quantity and its square is $\tfrac{(O_i - E_i)^2}{E_i}$Ei​(Oi​−Ei​)2​. Summing $k$k such squared standardised residuals gives, under $H_0$H0​, something close to a sum of squared standard normals — and a sum of squared independent standard normals is exactly a chi-squared variable. This is the intuition behind the reference distribution, and it also explains the $E_i \ge 5$Ei​≥5 rule: the normal approximation to each count is poor when the expected frequency is small, so the chi-squared approximation to the sum breaks down unless the expected frequencies are reasonably large. Understanding this makes the otherwise mysterious pooling rule feel inevitable rather than arbitrary.

Degrees of freedom

$\boxed{\,\nu = (\text{number of classes used}) - 1 - (\text{number of parameters estimated from the data})\,}$ν=(number of classes used)−1−(number of parameters estimated from the data)​

The "$-1$−1" is the constraint that $\sum E_i = \sum O_i = n$∑Ei​=∑Oi​=n (the totals are forced to match). Each parameter you estimate from the sample (rather than being told) costs a further degree of freedom.

Model being tested	Parameters estimated	$\nu$ν (with $k$k classes)
Uniform / given distribution	0	$k - 1$k−1
Poisson, $\lambda$λ given	0	$k - 1$k−1
Poisson, $\lambda$λ estimated by $\bar x$xˉ	1	$k - 2$k−2
Binomial, $p$p given	0	$k - 1$k−1
Binomial, $p$p estimated	1	$k - 2$k−2
Normal, $\mu$μ and $\sigma^2$σ2 estimated	2	$k - 3$k−3

⚠️ Count $k$k after any pooling (next), and remember an estimated parameter only counts if you estimated it from this data set.

The pooling rule

The chi-squared approximation is unreliable when an expected frequency is small. The standard A-Level rule is: every $E_i$Ei​ must be at least 5; if any $E_i < 5$Ei​<5, combine it with an adjacent class (and add the corresponding observed frequencies) until all pooled expected frequencies reach 5. Then recount $k$k for the degrees of freedom.

The procedure

1. State $H_0$H0​ (data follow the distribution) and $H_1$H1​ (they do not).
2. Compute expected frequencies $E_i = n\,P(\text{class }i)$Ei​=nP(class i) from the model.
3. Pool any classes with $E_i < 5$Ei​<5; recount $k$k.
4. Compute $X^2 = \sum \dfrac{(O_i - E_i)^2}{E_i}$X2=∑Ei​(Oi​−Ei​)2​.
5. Find $\nu = k - 1 - (\text{params estimated})$ν=k−1−(params estimated).
6. Read the critical value $\chi^2_\nu$χν2​ at the stated significance level (upper tail).
7. If $X^2 >$X2> critical value, reject $H_0$H0​; otherwise do not reject. Conclude in context.

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3. Worked examples with M1/A1 mark scheme

Example 1 — testing a uniform distribution (a fair die)

A die is rolled 120 times:

Face	1	2	3	4	5	6
Observed $O$O	15	23	18	25	17	22

$H_0$H0​: the die is fair (each face has probability $\tfrac16$61​). $H_1$H1​: the die is not fair. Expected $E = \tfrac{120}{6} = 20$E=6120​=20 per face (B1; all $E \ge 5$E≥5, no pooling).

Face	$O$O	$E$E	$(O-E)^2/E$(O−E)2/E
1	15	20	$25/20 = 1.25$25/20=1.25
2	23	20	$9/20 = 0.45$9/20=0.45
3	18	20	$4/20 = 0.20$4/20=0.20
4	25	20	$25/20 = 1.25$25/20=1.25
5	17	20	$9/20 = 0.45$9/20=0.45
6	22	20	$4/20 = 0.20$4/20=0.20

$X^2 = 1.25 + 0.45 + 0.20 + 1.25 + 0.45 + 0.20 = 3.80. \quad (\textbf{M1}\ \sum (O-E)^2/E;\ \textbf{A1}\ 3.80)$X2=1.25+0.45+0.20+1.25+0.45+0.20=3.80.(M1 ∑(O−E)2/E; A1 3.80)

Degrees of freedom $\nu = 6 - 1 = 5$ν=6−1=5 (B1). At the 5% level the critical value is $\chi^2_5 = 11.07$χ52​=11.07. Since $3.80 < 11.07$3.80<11.07 we do not reject $H_0$H0​ (M1 compare; A1 conclusion): there is no significant evidence at the 5% level that the die is unfair.

Example 2 — testing a Poisson distribution (parameter estimated)

Accidents per day at a junction over 100 days:

Accidents	0	1	2	3	4	5+
Days $O$O	30	36	20	8	4	2

$H_0$H0​: the data follow a Poisson distribution; $H_1$H1​: they do not.

Estimate $\lambda$λ by the sample mean:

$\bar x = \frac{0(30)+1(36)+2(20)+3(8)+4(4)+5(2)}{100} = \frac{0+36+40+24+16+10}{100} = \frac{126}{100} = 1.26. \quad (\textbf{B1})$xˉ=1000(30)+1(36)+2(20)+3(8)+4(4)+5(2)​=1000+36+40+24+16+10​=100126​=1.26.(B1)

Expected frequencies from $\text{Po}(1.26)$Po(1.26) (using $P(r) = \tfrac{\lambda}{r}P(r-1)$P(r)=rλ​P(r−1)):

$r$r	$P(X=r)$P(X=r)	$E = 100P$E=100P
0	$e^{-1.26} = 0.2837$e−1.26=0.2837	28.37
1	$1.26\times 0.2837 = 0.3575$1.26×0.2837=0.3575	35.75
2	$\tfrac{1.26}{2}\times 0.3575 = 0.2252$21.26​×0.3575=0.2252	22.52
3	$\tfrac{1.26}{3}\times 0.2252 = 0.0946$31.26​×0.2252=0.0946	9.46
$\ge 4$≥4	$1 - 0.9610 = 0.0390$1−0.9610=0.0390	3.90

The class $\ge 4$≥4 has $E = 3.90 < 5$E=3.90<5, so pool it with class 3: pooled $E = 9.46 + 3.90 = 13.36$E=9.46+3.90=13.36, pooled $O = 8 + (4+2) = 14$O=8+(4+2)=14 (M1 pooling shown). This leaves $k = 4$k=4 classes: $\{0, 1, 2, \ge 3\}${0,1,2,≥3}.

Class	$O$O	$E$E	$(O-E)^2/E$(O−E)2/E
0	30	28.37	$2.6569/28.37 = 0.0936$2.6569/28.37=0.0936
1	36	35.75	$0.0625/35.75 = 0.0017$0.0625/35.75=0.0017
2	20	22.52	$6.3504/22.52 = 0.2820$6.3504/22.52=0.2820
$\ge 3$≥3	14	13.36	$0.4096/13.36 = 0.0307$0.4096/13.36=0.0307

$X^2 = 0.0936 + 0.0017 + 0.2820 + 0.0307 = 0.408. \quad (\textbf{M1 A1})$X2=0.0936+0.0017+0.2820+0.0307=0.408.(M1 A1)

Degrees of freedom: $k = 4$k=4 classes, 1 parameter estimated ($\lambda$λ), so $\nu = 4 - 1 - 1 = 2$ν=4−1−1=2 (B1). At 5%, $\chi^2_2 = 5.991$χ22​=5.991. Since $0.408 < 5.991$0.408<5.991, do not reject $H_0$H0​ (A1): the data are consistent with a Poisson model.

Exam Tip: Pool first, then count classes for $\nu$ν. Here pooling reduced six raw classes to four, and estimating $\lambda$λ removed one more degree of freedom — both steps are common mark-losers if skipped.

Example 3 — a two-class test with Yates' correction

A coin is tossed 100 times, giving 60 heads and 40 tails. Test at 5% whether the coin is fair.

$H_0$H0​: $p(\text{head}) = 0.5$p(head)=0.5; $H_1$H1​: $p \ne 0.5$p=0.5. Expected: 50 heads, 50 tails. Here $k = 2$k=2 and $\nu = 2 - 1 = 1$ν=2−1=1, so apply Yates' continuity correction:

$X^2 = \sum \frac{(|O - E| - 0.5)^2}{E} = \frac{(10 - 0.5)^2}{50} + \frac{(10 - 0.5)^2}{50} = \frac{90.25}{50} + \frac{90.25}{50} = 1.805 + 1.805 = 3.61. \quad (\textbf{M1 A1})$X2=∑E(∣O−E∣−0.5)2​=50(10−0.5)2​+50(10−0.5)2​=5090.25​+5090.25​=1.805+1.805=3.61.(M1 A1)

At 5%, $\chi^2_1 = 3.841$χ12​=3.841. Since $3.61 < 3.841$3.61<3.841, do not reject $H_0$H0​ (A1): no significant evidence the coin is biased. (Without Yates' correction $X^2 = 4.00 > 3.841$X2=4.00>3.841 would reject — showing why the correction matters when $\nu = 1$ν=1.)

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4. Specimen-style exam question

(Specimen-style — not from any real paper.)

A biologist counts the number of seeds germinating in each of 80 trays of 4 seeds, and proposes a $B(4, 0.5)$B(4,0.5) model.

Germinating	0	1	2	3	4
Trays $O$O	6	18	28	20	8

Test at the 5% level whether $B(4, 0.5)$B(4,0.5) fits.

Solution. $H_0$H0​: the data follow $B(4,0.5)$B(4,0.5); $H_1$H1​: they do not. The $B(4,0.5)$B(4,0.5) probabilities are $\tfrac{1}{16}, \tfrac{4}{16}, \tfrac{6}{16}, \tfrac{4}{16}, \tfrac{1}{16}$161​,164​,166​,164​,161​, so with $n = 80$n=80:

$r$r	$P$P	$E = 80P$E=80P	$O$O	$(O-E)^2/E$(O−E)2/E
0	1/16	5	6	$1/5 = 0.20$1/5=0.20
1	4/16	20	18	$4/20 = 0.20$4/20=0.20
2	6/16	30	28	$4/30 = 0.1333$4/30=0.1333
3	4/16	20	20	$0/20 = 0$0/20=0
4	1/16	5	8	$9/5 = 1.80$9/5=1.80

All $E \ge 5$E≥5, so no pooling. $X^2 = 0.20 + 0.20 + 0.1333 + 0 + 1.80 = 2.333$X2=0.20+0.20+0.1333+0+1.80=2.333. Here $p$p was given (not estimated), so $\nu = 5 - 1 = 4$ν=5−1=4; at 5%, $\chi^2_4 = 9.488$χ42​=9.488. Since $2.333 < 9.488$2.333<9.488, do not reject $H_0$H0​: $B(4, 0.5)$B(4,0.5) is an adequate model at the 5% level.

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5. Synoptic links

• Lessons 2–3 (Poisson/binomial): these supply the expected frequencies; the recurrence $P(r) = \tfrac{\lambda}{r}P(r-1)$P(r)=rλ​P(r−1) makes the Poisson table quick.
• Lesson 10 (contingency tables): the same statistic $\sum (O-E)^2/E$∑(O−E)2/E with a different degrees-of-freedom rule $(r-1)(c-1)$(r−1)(c−1) and a different $E_{ij} = R_iC_j/N$Eij​=Ri​Cj​/N.
• A-Level Maths hypothesis testing: identical logic of $H_0/H_1$H0​/H1​, significance level and critical value — but a goodness-of-fit test is always upper-tailed on $X^2$X2, because only a large discrepancy between observed and expected counts is evidence against the model.
• Estimation: using $\bar x$xˉ for $\lambda$λ, or the sample proportion for $p$p, connects to point estimation and is exactly what costs a degree of freedom; the more parameters you fit to the data, the fewer independent comparisons remain.

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6. Mark-scheme literacy