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Applied Biostatistics
Chi-squared tests
Martin Bland
Professor of Health Statistics
University of York
http://www-users.york.ac.uk/~mb55/
Analyses for qualitative data
Also called nominal, categorical.
Only two categories: dichotomous, attribute, quantal,
binary.
Methods:
Chi-squared test for association
Fisher’s exact test
Chi-squared test for linear association
Risk ratio, relative risk
Odds ratio
Contingency tables
Cross tabulation of two categorical variables:
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 50 849 899 Council tenant 29 229 258 Private tenant 11 164 175 Lives with parents 6 66 72 Other 3 36 39
Total 99 1344 1443
This kind of cross-tabulation of frequencies is also called a
contingency table or cross classification.
Want to test the null hypothesis that there is no relationship
or association between the two variables.
Contingency tables
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 50 849 899 Council tenant 29 229 258 Private tenant 11 164 175 Lives with parents 6 66 72 Other 3 36 39
Total 99 1344 1443
Want to test the null hypothesis that there is no relationship
or association between the two variables.
If the sample is large, we can do this by a chi-squared test.
If the sample is small, we must use Fisher’s exact test.
The chi-squared test for association
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 50 849 899 Council tenant 29 229 258 Private tenant 11 164 175 Lives with parents 6 66 72 Other 3 36 39
Total 99 1344 1443
Null hypothesis: no association between the two variables.
Alternative hypothesis: an association of some type.
The chi-squared test for association
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 899 Council tenant 258 Private tenant 175 Lives with parents 72 Other 39
Total 99 1344 1443
Proportion who are premature = 99/
Out of 899 owner occupiers, expect 899 × 99/1443 = 61.
to be premature deliveries if the null hypothesis were true.
The chi-squared test for association
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 61.7 837.3 899 Council tenant 258 Private tenant 175 Lives with parents 72 Other 39
Total 99 1344 1443
Proportion who are term = 1344/
Out of 899 owner occupiers, expect 899 × 1344/1443 = 837.
to be term deliveries if the null hypothesis were true.
Note that 61.7 + 837.3 = 899.
The chi-squared test for association
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 61.7 837.3 899 Council tenant 258 Private tenant 175 Lives with parents 72 Other 39
Total 99 1344 1443
Out of 258 council tenants, expect 258 × 99/1443 = 17.
to be premature deliveries if the null hypothesis were true.
Out of 258 council tenants, expect 258 × 1344/1443 = 240.
to be term deliveries if the null hypothesis were true.
The chi-squared test for association
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 61.7 837.3 899 Council tenant 17.7 240.3 258 Private tenant 175 Lives with parents 72 Other 39
Total 99 1344 1443
Out of 258 council tenants, expect 258 × 99/1443 = 17.
to be premature deliveries if the null hypothesis were true.
Out of 258 council tenants, expect 258 × 1344/1443 = 240.
to be term deliveries if the null hypothesis were true.
The chi-squared test for association
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 61.7 837.3 899 Council tenant 17.7 240.3 258 Private tenant 175 Lives with parents 72 Other 39
Total 99 1344 1443
Out of 258 council tenants, expect 258 × 99/1443 = 17.
to be premature deliveries if the null hypothesis were true.
Out of 258 council tenants, expect 258 × 1344/1443 = 240.
to be term deliveries if the null hypothesis were true.
Note that 17.7 + 240.3 = 258.
The chi-squared test for association
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 61.7 837.3 899 Council tenant 17.7 240.3 258 Private tenant 12.0 163.0 175 Lives with parents 72 Other 39
Total 99 1344 1443
The chi-squared test for association
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 61.7 837.3 899 Council tenant 17.7 240.3 258 Private tenant 12.0 163.0 175 Lives with parents 4.9 67.1 72 Other 39
Total 99 1344 1443
The chi-squared test for association
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 50 61.7 849 837.3 899 Council tenant 29 17.7 229 240.3 258 Private tenant 11 12.0 164 163.0 175 Lives with parents 6 4.9 66 67.1 72 Other 3 2.7 36 36.3 39
Total 99 1344 1443
Compare the observed and expected frequencies.
Add (observed – expected)^2 /expected for all cells.
If null hypothesis true and samples are large enough, this is an
observation from a Chi-squared distribution, often written ^2.
The Chi-squared distribution
Family of distributions, one parameter, called the degrees
of freedom.
5%
Probability density
Chi-squared with 4 degrees of freedom 0 5 10 15 20 25 30
0
.
.
Percentage points of the Chi-squared Distribution
Degrees Probability that the tabulated value of is exceeded freedom 10% 0.10 5% 0.05 1% 0.01 0.1% 0. 1 2.71 3.84 6.63 10. 2 4.61 5.99 9.21 13. 3 6.25 7.81 11.34 16. 4 7.78 9.49 13.28 18. 5 9.24 11.07 15.09 20. 6 10.64 12.59 16.81 22. 7 12.02 14.07 18.48 24. 8 13.36 15.51 20.09 26. 9 14.68 16.92 21.67 27. 10 15.99 18.31 23.21 29.
..... .....
The chi-squared test for association
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 50 849 899 Council tenant 29 229 258 Private tenant 11 164 175 Lives with parents 6 66 72 Other 3 36 39
Total 99 1344 1443
For a contingency table, the degrees of freedom are given by:
(number of rows – 1) × (number of columns – 1).
The chi-squared test for association
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 50 849 899 Council tenant 29 229 258 Private tenant 11 164 175 Lives with parents 6 66 72 Other 3 36 39
Total 99 1344 1443
For a contingency table, the degrees of freedom are given by:
(number of rows – 1) × (number of columns – 1).
We have (5 – 1) × (2 – 1) = 4 degrees of freedom.
The chi-squared test for association
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 50 849 899 Council tenant 29 229 258 Private tenant 11 164 175 Lives with parents 6 66 72 Other 3 36 39
Total 99 1344 1443
For a contingency table, the degrees of freedom are given by:
(number of rows – 1) × (number of columns – 1).
We have (5 – 1) × (2 – 1) = 4 degrees of freedom.
^2 = 10.5, 4 d.f.
The chi-squared test for association
The chi-squared statistic is not an index of the strength of the
association.
If we double the frequencies, this will double chi-squared, but
the strength of the association is unchanged.
The chi-squared test for association
The test statistic follows the Chi-squared Distribution provided
the expected values are large enough.
This is a large sample test.
The smaller the expected values become, the more dubious
will be the test.
The conventional criterion for the test to be valid is this: the
chi-squared test is valid if at least 80% of the expected
frequencies exceed 5 and all the expected frequencies
exceed 1.
Also known as the Pearson chi-squared test.
Fisher’s exact test
Also called the Fisher-Irwin exact test.
Works for any sample size.
Used to be used only for small samples in 2 by 2 tables,
because of computing problems.
Calculate the probability of every possible table with the given
row and column totals.
Sum the probabilities for all the tables as or less probable
than the observed.
Fisher’s exact test
Time of delivery by housing tenure Housing tenure Premature Term Total
Owner-occupier 50 849 899 Council tenant 29 229 258 Private tenant 11 164 175 Lives with parents 6 66 72 Other 3 36 39
Total 99 1344 1443
^2 = 10.5, 4 d.f. Using a computer, P = 0.033.
Fishers’ exact test: P = 0.034.
Fisher’s exact test
Renal failure and mortality in peritonitis patients (Fennell, unpublished)
Status after | Renal failure | 3 months | Yes No | Total -------------+----------------+---------- Dead | 6 31 | 37 Alive | 2 62 | 64 -------------+----------------+---------- Total | 8 93 | 101
Fisher’s exact test
Renal failure and mortality in peritonitis patients (Fennell, unpublished)
Status after | Renal failure | 3 months | Yes No | Total -------------+----------------+---------- Dead | 6 Ex=2.9 31 | 37 Alive | 2 62 | 64 -------------+----------------+---------- Total | 8 93 | 101
The chi-squared test for linear association
Assessment of radiological appearance at six months as compared with appearance on admission (MRC 1948)
Radiological assessment Streptomycin Control
Considerable improvement 28 4 Moderate or slight improvement 10 13 No material change 2 3 Moderate or slight deterioration 5 12 Considerable deterioration 6 6 Deaths 4 14
Total 55 52
Chi-squared = 26.97, 5 d.f., P = 0.0001.
Does not take the ordering of the categories into account.
The chi-squared test for linear association
Assessment of radiological appearance at six months as compared with appearance on admission (MRC 1948)
Radiological assessment Streptomycin Control
Considerable improvement 28 4 Moderate or slight improvement 10 13 No material change 2 3 Moderate or slight deterioration 5 12 Considerable deterioration 6 6 Deaths 4 14
Total 55 52
Chi-squared = 26.97, 5 d.f., P = 0.0001.
Does not take the ordering of the categories into account.
Several tests do, including the Armitage chi-squared test for
trend, the Mantel-Haenszel linear-by-linear association,
Kendall’s tau b.
The chi-squared test for linear association
Assessment of radiological appearance at six months as compared with appearance on admission (MRC 1948)
Radiological assessment Streptomycin Control
Considerable improvement 28 4 Moderate or slight improvement 10 13 No material change 2 3 Moderate or slight deterioration 5 12 Considerable deterioration 6 6 Deaths 4 14
Total 55 52
Chi-squared = 26.97, 5 d.f., P = 0.0001.
Does not take the ordering of the categories into account.
SPSS does the Mantel-Haenszel linear-by-linear association
chi-squared test, whether you want it or not.
The chi-squared test for linear association
Assessment of radiological appearance at six months as compared with appearance on admission (MRC 1948)
Radiological assessment Streptomycin Control
Considerable improvement 28 4 Moderate or slight improvement 10 13 No material change 2 3 Moderate or slight deterioration 5 12 Considerable deterioration 6 6 Deaths 4 14
Total 55 52
Chi-squared = 26.97, 5 d.f., P = 0.0001.
Does not take the ordering of the categories into account.
Trend: chi-squared = 17.93, 1 d.f., P < 0.0001.
Mantel-Haenszel linear-by-linear: chi-squared = 17.76, 1 d.f.,
P < 0.0001.
The chi-squared test for linear association
Mantel-Haenszel linear-by-linear: chi-squared = 17.76, 1 d.f.,
P < 0.0001.
Assigns numerical values to categories. This must be
meaningful.
E.g.: Considerable improvement =1,
Moderate or slight improvement =2,
No material change = 3,
Moderate or slight deterioration = 4,
Considerable deterioration =5,
Death =6,
and
Streptomycin = 1,
Control =2.
The chi-squared test for linear association
Mantel-Haenszel linear-by-linear: chi-squared = 17.76, 1 d.f.,
P < 0.0001.
Assigns numerical values to categories. This must be
meaningful.
We then say, given these numerical scales, is there a
relationship of the form
improvement = constant + another constant × treatment
Applied Biostatistics
Chi-squared tests
Martin Bland
Professor of Health Statistics
University of York
http://www-users.york.ac.uk/~mb55/
