Chi-squared test of independence

Definition taught in Chi-squared tests, part of Statistical inference

Chi-squared test of independence

Deaths during follow-up by anaemia status: counts (left) and the proportion who died in each group (right). From Descriptive statistics.
Type Definition
Taught in Chi-squared tests
Part Statistical inference
Properties 1
Worked examples 4

For an r\times c contingency table with fitted counts E_{ij}=R_iC_j/n, the statistic

T=\sum_{i=1}^r\sum_{j=1}^c\frac{(O_{ij}-E_{ij})^2}{E_{ij}}

is approximately \chi^2_{(r-1)(c-1)} when the row and column factors are independent. Large values are evidence against independence.

Properties

Pearson test with estimated parameters

Main article: Pearson test with estimated parameters

Let (X_1,\ldots,X_k) be multinomial with cell probabilities p_i(\theta), where \theta has r free coordinates. If \widehat\theta is the maximum likelihood estimator of \theta from the counts and \widehat p_i=p_i(\widehat\theta), then

\sum_{i=1}^k\frac{(X_i-n\widehat p_i)^2}{n\widehat p_i}

is approximately \chi^2_{k-1-r} for large n.

Examples

Belief in ESP

Worked example from Chi-squared tests

A survey asks 114 artists and 344 non-artists how strongly they believe in extra-sensory perception.

Strong Moderate None Total
Artists 67 41 6 114
Non-artists 129 183 32 344
Total 196 224 38 458

Question. Is belief in ESP independent of being an artist?

Solution. The fitted counts are R_iC_j/458. For artists with strong belief, 114\times196/458=48.79; the six fitted counts are

Strong Moderate None
Artists 48.79 55.76 9.46
Non-artists 147.21 168.24 28.54

The six contributions (O_{ij}-E_{ij})^2/E_{ij} are 6.80, 3.91, 1.27, 2.25, 1.29 and 0.42, so T=15.94 on (2-1)(3-1)=2 degrees of freedom. The p-value is 0.0003. Independence is firmly rejected. The largest contributions come from artists: more of them hold strong belief, and fewer moderate belief, than independence predicts.

Anaemia and death

Worked example from Descriptive statistics

Question. Did patients with anaemia die more often?

Solution.

table(heart$anaemia, heart$death, dnn = c("anaemia", "death"))
proportions(table(heart$anaemia, heart$death), margin = 1)
Survived Died Proportion died
No anaemia 120 50 0.294
Anaemia 83 46 0.357

Row proportions, margin = 1, compare the death rate within each anaemia group: 29.4% without anaemia and 35.7% with it. The difference is six percentage points. Whether that is more than chance would produce with groups this size is a chi-squared test of independence, and it is not: the p-value is 0.25. High blood pressure gives 29.4% against 37.1%, with p-value 0.17, while for diabetes, sex and smoking the rates in the two groups differ by less than two percentage points.

A two-by-two table is a two-proportion test

Question page: A two-by-two table is a two-proportion test

Of 70 viewers of one channel, 40 are left-wing; of 100 viewers of another, 50 are. Compute the Pearson statistic for independence of channel and allegiance, and check that it equals the square of the pooled two-proportion z statistic, 0.918.

The table and its fitted counts R_iC_j/170 are

Left Not left Total
Channel 1 40 (37.06) 30 (32.94) 70
Channel 2 50 (52.94) 50 (47.06) 100
Total 90 80 170

Every cell differs from its fitted count by 2.94 in absolute value, so

T=2.94^2\Bigl(\frac1{37.06}+\frac1{32.94}+\frac1{52.94}+\frac1{47.06}\Bigr)=0.843.

The pooled two-proportion statistic is z=0.918, and 0.918^2=0.843. The two statistics agree, and \chi^2_1 is the law of the square of a standard normal. The chi-squared p-value, 0.36, is the two-sided z test’s; the one-sided z test halves it.

Counting degrees of freedom

Question page: Counting degrees of freedom

Give the degrees of freedom of the Pearson statistic in each case. (a) A 4\times5 contingency table tested for independence. (b) Counts in 8 classes compared with a binomial distribution \operatorname{Bin}(7,p), with p estimated. (c) Counts in 10 classes compared with a normal distribution whose mean and variance are both estimated. (d) The same, with the mean and variance known.

  1. (4-1)(5-1)=12.

  2. 8-1-1=6: eight classes, one estimated parameter.

  3. 10-1-2=7.

  4. 10-1=9: nothing is estimated.

Used in

See also

Pearson statistic · Pearson chi-squared approximation · Pearson test with estimated parameters

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