Tests of association in contingency tables

Review

Do asthma rates differ by sex?

From a subsample of NHANES data:

	asthma	no asthma
male	30	769
female	49	781

Test for a difference in prevalence: ${\begin{cases} H_{0} : p_{F} = p_{M} \\ H_{A} : p_{F} \neq p_{M} \end{cases}$

table(asthma$sex, asthma$asthma) |>
  prop.test(alternative = 'two.sided', 
            conf.level = 0.95)


    2-sample test for equality of proportions with continuity correction

data:  table(asthma$sex, asthma$asthma)
X-squared = 3.6217, df = 1, p-value = 0.05703
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.0434742223  0.0004958005
sample estimates:
    prop 1     prop 2 
0.03754693 0.05903614

The data provide evidence that asthma prevalence differs between men and women (Z = 2.108, p = 0.0436). With 95% confidence, the prevalence among women is estimated to be between 0.07 and 4.22 percentage points higher than that among men.

$χ^{2}$ tests of association

Association and independence

Consider the more general hypothesis that smoking and cancer are independent:

${\begin{cases} H_{0} : & smoking ⊥ cancer \\ H_{A} : & \neg (smoking ⊥ cancer) \end{cases}$

Cell-wise and marginal proportions:

	Smokers	NonSmokers	total
Cancer	0.4826	0.01744	0.5
Control	0.4186	0.0814	0.5
total	0.9012	0.09884	1

Under independence we’d expect:

$\underset{cell proportion}{\underset{⏟}{p_{i j}}} \approx \underset{marginal proportions}{\underset{⏟}{p_{i} \times p_{j}}}$

For example (shown in bold):

50% of subjects are controls
9.8% of subjects are nonsmokers
so roughly 50% of 9.8% (= 4.9%) would be healthy nonsmokers
but actually 8.1% are healthy nonsmokers – almost twice what we’d expect!

Basis for a test: expected counts

Expected proportions translate directly to expected counts: $p_{i j} = p_{i} \times p_{j} ⟺ n_{i j} = \frac{n_{i \cdot} \times n_{\cdot j}}{n}$

Actual counts:

	O1	O2	total
G1	$n_{11}$	$n_{12}$	$n_{1 \cdot}$
G2	$n_{21}$	$n_{22}$	$n_{2 \cdot}$
total	$n_{\cdot 1}$	$n_{\cdot 2}$	$n$

Expected counts under independence:

	O1	O2	total
G1	${\hat{n}}_{11} = \frac{n_{1 \cdot} \times n_{\cdot 1}}{n}$	${\hat{n}}_{12} = \frac{n_{1 \cdot} \times n_{\cdot 2}}{n}$	$n_{1 \cdot}$
G2	${\hat{n}}_{21} = \frac{n_{2 \cdot} \times n_{\cdot 1}}{n}$	${\hat{n}}_{22} = \frac{n_{2 \cdot} \times n_{\cdot 2}}{n}$	$n_{2 \cdot}$
total	$n_{\cdot 1}$	$n_{\cdot 2}$	$n$

Idea for a test: reject $H_{0}$ if actual and expected counts differ enough across the table.

This is more general than inference of proportions because it doesn’t depend on $p_{i}$ or $p_{j}$ being meaningful estimates of population proportions.

Computing expected counts

Actual counts:

	O1	O2	total
G1	$n_{11}$	$n_{12}$	$n_{1 \cdot}$
G2	$n_{21}$	$n_{22}$	$n_{2 \cdot}$
total	$n_{\cdot 1}$	$n_{\cdot 2}$	$n$

Expected counts under independence:

	O1	O2	total
G1	${\hat{n}}_{11} = \frac{n_{1 \cdot} \times n_{\cdot 1}}{n}$	${\hat{n}}_{12} = \frac{n_{1 \cdot} \times n_{\cdot 2}}{n}$	$n_{1 \cdot}$
G2	${\hat{n}}_{21} = \frac{n_{2 \cdot} \times n_{\cdot 1}}{n}$	${\hat{n}}_{22} = \frac{n_{2 \cdot} \times n_{\cdot 2}}{n}$	$n_{2 \cdot}$
total	$n_{\cdot 1}$	$n_{\cdot 2}$	$n$

For the case-control study:

Actual
	Smokers	NonSmokers	total
Cancer	83	3	86
Control	72	14	86
total	155	17	172

Expected
	Smokers	NonSmokers	total
Cancer	77.5	8.5	86
Control	77.5	8.5	86
total	155	17	172

The chi-square ( $χ^{2}$ ) statistic

A measure of the amount by which actual counts differ from expected counts under independence is the chi (pronounced /ˈkaɪ ) square statistic:

$χ^{2} = \sum_{i j} \frac{{(n_{i j} - {\hat{n}}_{i j})}^{2}}{{\hat{n}}_{i j}} (\sum_{all cells} \frac{(observed - expected)^{2}}{expected})$

Cell-wise calculation:

	smokers	nonsmokers
cancer	$\frac{(83 - 77.5)^{2}}{77.5}$	$\frac{(3 - 8.5)^{2}}{8.5}$
control	$\frac{(72 - 77.5)^{2}}{77.5}$	$\frac{(14 - 8.5)^{2}}{8.5}$

Result:

	Smokers	NonSmokers
Cancer	0.3903	3.559
Control	0.3903	3.559

Chi-square statistic: $χ^{2} = 0.3903 + 0.3903 + 3.5588 + 3.5588 = 7.8983$

Sampling distribution for $χ^{2}$

Under $H_{0}$ , the $χ^{2}$ statistic has a sampling distribution that can be approximated by a $χ_{1}^{2}$ model.

subscript indicates degrees of freedom parameter

The model assumes no expected counts are too small.

rule of thumb: at least 5 ( ${\hat{n}}_{i j} \geq 5$ )
consequences: if ${\hat{n}}_{i j}$ are too small, the statistic is inflated relative to the model, leading to a higher type I error rate

Computing $p$ values

${\begin{cases} H_{0} : & smoking ⊥ cancer \\ H_{A} : & \neg (smoking ⊥ cancer) \end{cases}$

To determine the test outcome, find the $p$ -value:

$P (χ_{1}^{2} > χ_{obs}^{2}) = P (χ_{1}^{2} > 7.898) = 0.0049$

table(smoking$group, smoking$smoking) |> 
  chisq.test(correct = F)


    Pearson's Chi-squared test

data:  table(smoking$group, smoking$smoking)
X-squared = 7.8983, df = 1, p-value = 0.004948

The data provide evidence of an association between smoking and lung cancer ( $χ^{2}$ = 7.989 on 1 degree of freedom, p = 0.0049).

If smoking and cancer were independent, only 0.49% of random samples would produce a table that deviates from expected counts by more than what we observed.

Residuals in $χ^{2}$ tests

The residual for each cell is defined as a standardized difference between the observed and expected count:

$r_{i j} = \frac{n_{i j} - {\hat{n}}_{i j}}{\sqrt{{\hat{n}}_{i j}}}$

Examining residuals can indicate the source(s) of an inferred association.

$r_{i j} > 0$ : observation exceeds expectation
$r_{i j} < 0$ : observation is under expectation
large $| r_{i j} |$ explain the association

# store test result
rslt <- table(smoking$group, smoking$smoking) |>
  chisq.test(correct = F)

# examine residuals
rslt$residuals

	Smokers	NonSmokers
Cancer	0.6248	-1.886
Control	-0.6248	1.886

Look for the largest residuals to explain the result:

more nonsmokers among controls
fewer nonsmokers among cases

Continuity correction

The $χ^{2}$ test for independence is typically applied with Yates’ continuity correction.

This consists in using a modified version of the test statistic:

$χ_{Yates}^{2} = \sum_{i j} \frac{{(| n_{i j} - {\hat{n}}_{i j} | - 0.5)}^{2}}{{\hat{n}}_{i j}}$

every other detail of the test is the same
doesn’t change expected counts
residuals are still computed as $\frac{n_{i j} - {\hat{n}}_{i j}}{\sqrt{{\hat{n}}_{i j}}}$

Implementation:

# construct table and pass to chisq.test
table(smoking$group, smoking$smoking) |> 
  chisq.test(correct = T)


    Pearson's Chi-squared test with Yates' continuity correction

data:  table(smoking$group, smoking$smoking)
X-squared = 6.5275, df = 1, p-value = 0.01062

Note the larger $p$ -value – the test is a bit more conservative.

$χ^{2}$ tests for $I \times J$ tables

FAMuSS data:

	CC	CT	TT	total
African Am	16	6	5	27
Asian	21	18	16	55
Caucasian	125	216	126	467
Hispanic	4	10	9	23
Other	7	11	5	23
total	173	261	161	595

Expected counts:

	CC	CT	TT	total
African Am	7.85	11.84	7.31	27
Asian	15.99	24.13	14.88	55
Caucasian	135.8	204.8	126.4	467
Hispanic	6.69	10.09	6.22	23
Other	6.69	10.09	6.22	23
total	173	261	161	595

expected counts and chi-square statistic are calculated exactly the same way
degrees of freedom are now $(I - 1) \times (J - 1)$
appropriate provided all ${\hat{n}}_{i j} > 1$ and most (~80%) ${\hat{n}}_{i j} \geq 5$

Extending to $I \times J$ tables

In detail:

	CC	CT	TT
African Am	$\frac{(16 - 7.85)^{2}}{7.85}$	$\frac{(6 - 11.84)^{2}}{11.84}$	$\frac{(5 - 7.306)^{2}}{7.306}$
Asian	$\frac{(21 - 15.99)^{2}}{15.99}$	$\frac{(18 - 24.13)^{2}}{24.13}$	$\frac{(16 - 14.88)^{2}}{14.88}$
Caucasian	$\frac{(125 - 135.8)^{2}}{135.8}$	$\frac{(216 - 204.9)^{2}}{204.9}$	$\frac{(126 - 126.4)^{2}}{126.4}$
Hispanic	$\frac{(4 - 6.687)^{2}}{6.687}$	$\frac{(10 - 10.09)^{2}}{10.09}$	$\frac{(9 - 6.224)^{2}}{6.224}$
Other	$\frac{(7 - 6.687)^{2}}{6.687}$	$\frac{(11 - 10.09)^{2}}{10.09}$	$\frac{(5 - 6.224)^{2}}{6.224}$

Then:

${\begin{cases} χ^{2} = \sum all cells above = 19.4 \\ P (χ_{8}^{2} > 19.4) = 0.01286 \end{cases} ⟹ reject hypothesis of no association$

Inference for $I \times J$ tables in R

The implementation is the same as for a $2 \times 2$ table:

# construct table and pass to chisq.test
table(famuss$race, famuss$actn3.r577x) |>
  chisq.test()


    Pearson's Chi-squared test

data:  table(famuss$race, famuss$actn3.r577x)
X-squared = 19.4, df = 8, p-value = 0.01286

The data provide evidence of an association between race and genotype ( $χ^{2}$ = 19.4 on 8 degrees of freedom, p = 0.01286).

Which genotype/race combinations are contributing most to this inferred association?

	CC	CT	TT
African Am	2.909	-1.698	-0.8531
Asian	1.252	-1.247	0.2897
Caucasian	-0.9254	0.7789	-0.03244
Hispanic	-1.039	-0.02804	1.113
Other	0.1209	0.2868	-0.4905

Malaria vaccine trial data

In a randomized trial for a malaria vaccine, 20 individuals were randomly allocated to receive a dose of the vaccine or a placebo.

	no infection	infection
placebo	0	6
vaccine	9	5

Recall the assumption for the $χ^{2}$ test that ${\hat{n}}_{i j} \geq 5$ in a 2x2 table. Here that’s not true.

# notice the warning message
table(malaria) |> chisq.test()

Warning in chisq.test(table(malaria)): Chi-squared approximation may be
incorrect

Not just a data artefact…

group sizes are 6 placebo and 14 vaccine
expected counts will always be small

Expected counts
	no infection	infection
placebo	2.7	3.3
vaccine	6.3	7.7

So what alternative do we have to test for association?

Fisher’s exact test

Fact 1: if you fix the margins, one table entry determines the rest.

Try it for yourself!

	no infection	infection	total
placebo			6
vaccine		5	14
total	9	11

Fact 2: under $H_{0}$ any entry is a completely random allocation from the marginal totals.

$P r (X = 5) = \frac{(\binom{11}{5}) (\binom{9}{14 - 5})}{(\binom{20}{14})}$

The probability above is simply: $\frac{ways to 5 infections to vaccine group}{ways to allocate subjects to vaccine group}$

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Tests of association in contingency tables

Tests of association in contingency tables
Today’s agenda
Review
Do asthma rates differ by sex?
Smoking and lung cancer
$χ^{2}$ tests of association
Association and independence
Basis for a test: expected counts
Computing expected counts
The chi-square ( $χ^{2}$ ) statistic
Sampling distribution for $χ^{2}$
Computing $p$ values
Residuals in $χ^{2}$ tests
Continuity correction
$χ^{2}$ tests for $I \times J$ tables
Extending to $I \times J$ tables
Inference for $I \times J$ tables in R
Residual analysis
Exact methods
Malaria vaccine trial data
Fisher’s exact test
Fisher’s exact test
Fisher’s exact test
Fisher’s exact test
Fisher’s exact test
Fisher’s exact test