The Chi-Square Test

Used when both outcome and exposure variables are binary (dichotomous) or even multichotomousAllows the researcher to calculate a p-value associated with relative risks (RR) and odds ratio (OR)Simplest for is a 2 x 2 contingency tableObtain a set of expected values and compare observed to expected

How does chi-square work?

Compares the observed 2 x 2 table to a theoretical table (“expected”) that would occur if the null hypothesis (H0) were true

Diseased Not diseased

Exposed O11 O12

Not exposed O21 O22

Diseased Not diseased

Exposed E11 E12

Not exposed E21 E22

?

i i

iiobs E

EO 22 )(

Observed Expected under H0

Calculating the “expected” table

Diseased Not diseasedExposed O11 O12

Not exposed O21 O22

O11 + O12 = R1

O21 + O22 = R2

O12 + O22 = C2O11 + O21 = C1 ΣOi = N

Start with observed table

Diseased Not diseasedExposed E11 = R1C1/N E12 = R1C2/N

Not exposed E21 = R2C1/N E22 = R2C2/N

Calculate the expected table

Chi-square test from breast cancer study

Set alpha value: let’s use 0.05State H0 and HA.H0: There is no association between race and breast cancer status.(The proportion of Whites is the same in cases and controls)

pcases = pcontrolswhere p = proportion white

Null and alternate hypotheses

Null hypothesis (H0):There is no association between race and breast cancer status.(The proportion of Whites is the same in cases and controls.)

pcases = pcontrols

Alternate hypothesis (HA):There is an association between race and breast cancer status.(The proportion of Whites is the different in cases and controls.)

pcases ≠ pcontrols

Calculating expected table

Cases ControlsWhite 677 905Black 272 619

Start with observed table

Calculate the expected table

1582

Cases ControlsWhite 1582*949/2473 = 607.1 1582*1524/2473 = 974.9Black 891*949/2473 = 341.9 891*1524/2473 = 549.1

891

1524949 2473

Calculate chi-square statistic

i i

iiobs E

EO 22 )(

Cases ControlsWhite 677 905Black 272 619

Cases ControlsWhite 607.1 974.9Black 341.9 549.1

χ2 = (677-607.1)2/607.1 + (905-974.9)2/974.9+ (272-341.9)2/341.9 + (619-549.1)2/549.1

χ2 = 36.2

( O11 – E11 )2 / E11

Chi-square distributions

Test statistic = 36.2Compare to a known χ2 distribution to get p-valueNeed to know degrees of freedom

= (rows – 1)*(columns – 1) In this case, d.f. = (2 – 1)(2 – 1) = 1*1 = 1

Chi-square value

Prob

abili

ty

2 4

Critical value for 1 d.f. = 3.84 (where alpha = 0.05) Rejection

region for H0

Our chi square statistic = 36.2

Chi-square table and interpretation

Examine the list of values and associated p-values from column with 1 degree of freedomFor an observed χ2 of 36.2, the associated p-value is < 0.001.

Since p < α, we reject H0 and conclude that there is an

association between race and outcome (case/control) status.

Chi-square caveat #1

What about tables that are larger than 2x2?

If there is a statistically significant association (i.e. p < α), we don’t know among which group(s) the association(s) occur(s).Need to conduct individual 2 x 2 χ2 analyses to determine if/where the association(s) exist.

(…which is < α)

What about very small sample sizes?Answer: Chi-square is not an appropriate test if 20% or more of the cells (i.e. 1 or more cells in a 2x2 table) have expected counts of less than 5.Solution: Use Fisher’s exact test

Based on permutations, not on a theorized distribution

Chi-square caveat #2