Chi-Square Goodness of Fit

196. Chi-Square Goodness of Fit

Compares an observed categorical distribution to a theoretical categorical distribution.

Χ^{2} = \sum_{𝑖 = 1}^{𝑘} \frac{{(𝑜_{𝑖} - 𝑒_{𝑖})}^{2}}{𝑒_{𝑖}}

$𝑜_{𝑖}$ : observed frequency in category $𝑖$
$𝑒_{𝑖}$ : expected frequency in category $𝑖$

df = 𝑘 - 1

$𝑘$ : number of categories

Example

Your professor gave a hard test. He said 45% got A’s, 20% B’s, 15% C’s, 10% D’s and 10% F’s. You don’t believe him. You take a simple random sample of 60 students and find 20 got A’s, 15 B’s, 8 O’s, 12 D’s and 5 F’s. At $𝛼 = 0.05$ , test.

Category	A	B	C	D	F	Total
Expected % ( $𝑂_{𝑖}$ )	45%	20%	15%	10%	10%	100%
Observed # ( $𝐸_{𝑖}$ )	20	15	8	12	5	60
Expected #	$60 \cdot 0.45 = 27$	$60 \cdot 0.20 = 12$	$60 \cdot 0.15 = 9$	$60 \cdot 0.10 = 6$	$60 \cdot 0.10 = 6$	60
$\frac{{(𝑂 - 𝐸)}^{2}}{𝐸}$	$1.815$	$0.750$	$0.111$	$6.000$	$0.167$

Hypothesis

\begin{aligned} 𝐻_{0} & : 𝑝_{𝐴} = 0.45, 𝑝_{𝐵} = 0.20, 𝑝_{𝐶} = 0.15, 𝑝_{𝐷} = 0.10, 𝑝_{𝐹} = 0.10 \\ (the professor's claimed distribution is correct) \\ 𝐻_{𝑎} & : at least one 𝑝_{𝑖} differs from its claimed value \end{aligned}

Test Statistic & P-Value

\begin{aligned} 𝜒^{2} & = \sum_{𝑖 = 1}^{𝑛} \frac{{(𝑜_{𝑖} - 𝑒_{𝑖})}^{2}}{𝑒_{𝑖}} \\ = 𝑧_{𝐴}^{2} + 𝑧_{𝐵}^{2} + 𝑧_{𝐶}^{2} + 𝑧_{𝐷}^{2} + 𝑧_{𝐹}^{2} \\ = \frac{{(20 - 27)}^{2}}{27} + \frac{{(15 - 12)}^{2}}{12} + \frac{{(8 - 9)}^{2}}{9} + \frac{{(12 - 6)}^{2}}{6} + \frac{{(5 - 6)}^{2}}{6} \\ = 1.81 + 0.75 + 0.11 + 6.00 + 0.17 \\ = 8.84 \end{aligned}

\begin{aligned} 𝑝 -value & = 0.0652 \\ 𝛼 & = 0.05 \\ critical value & = 𝜒_{0.05, 4}^{2} = 9.49 \\ 𝜒^{2} & = 8.84 \end{aligned}

Decision

Since $𝜒^{2} = 8.84 < 9.49$ (equivalently $𝑝 = 0.0652 > 𝛼 = 0.05$ ), we fail to reject $𝐻_{0}$ .

Conclusion

At the $5 %$ significance level there is not enough evidence to conclude that the professor’s claimed grade distribution differs from the true one.

Example

Complaints by Quarter (testing for a uniform distribution)

	Q1	Q2	Q3	Q4	$\sum$
$𝑜_{𝑖}$	43	52	54	40	189

Under $𝐻_{0}$ all four quarters are equally likely, so $𝑒_{𝑖} = 𝑁 / 𝑘 = 189 / 4$ for every $𝑖$ .

	Q1	Q2	Q3	Q4	$\sum$
$𝑜_{𝑖}$	43	52	54	40	189
$𝑒_{𝑖}$	$\frac{189}{4}$	$\frac{189}{4}$	$\frac{189}{4}$	$\frac{189}{4}$	189

	Q1	Q2	Q3	Q4	$\sum$
$𝑜_{𝑖}$	43	52	54	40	189
$𝑒_{𝑖}$	$47.25$	$47.25$	$47.25$	$47.25$	189

	$\frac{{(43 - 47.25)}^{2}}{47.25}$	$\frac{{(52 - 47.25)}^{2}}{47.25}$	$\frac{{(54 - 47.25)}^{2}}{47.25}$	$\frac{{(40 - 47.25)}^{2}}{47.25}$	$\sum$
	$0.38$	$0.48$	$0.96$	$1.11$	$𝟐.𝟗𝟒$

Χ^{2} = 0.38 + 0.48 + 0.96 + 1.11 = 2.94

$df = 𝑘 - 1 = 3$ , and $Χ_{0.05, 3}^{2} = 7.815$ . Since $2.94 < 7.815$ we fail to reject $𝐻_{0}$ ( $𝑝 \approx 0.40$ ): the complaints are consistent with a uniform distribution across quarters.

chi2_got.py

from scipy import stats
f_obs = np.array([43, 52, 54, 40])
f_exp = np.array([47, 47, 47, 47])
stats.chisquare(f_obs=f_obs, f_exp=f_exp)