Menu location: Analysis_Crosstabs.
This a two or three way cross tabulation function. If you have two columns of numbers that correspond to different classifications of the same individuals then you can use this function to give a two way frequency table for the cross classification. This can be stratified by a third classification variable.
For two way crosstabs, StatsDirect offers a range of analyses appropriate to the dimensions of the contingency table. For more information see chi-square tests and exact tests.
For three way crosstabs, StatsDirect offers either odds ratio (for case-control studies) or relative risk (for cohort studies) meta-analyses for 2 by 2 by k tables, and generalised Cochran-Mantel-Haenszel tests for r by c by k tables.
Example
A database of test scores contains two fields of interest, sex (M=1, F=0) and grade of skin reaction to an antigen (none = 0, weak + = 1, strong + = 2). Here is a list of those fields for 10 patients:
Sex | Reaction |
0 | 0 |
1 | 1 |
1 | 2 |
0 | 2 |
1 | 2 |
0 | 1 |
0 | 0 |
0 | 1 |
1 | 2 |
1 | 0 |
In order to get a cross tabulation of these from StatsDirect you should enter these data in two workbook columns. Then choose crosstabs from the analysis menu.
For this example:
Reaction | ||||
0 | 1 | 2 | ||
Sex: | 0 | 2 | 2 | 1 |
1 | 1 | 1 | 3 |
We could then proceed to an r by c (2 by 3) contingency table analysis to look for association between sex and reaction to this antigen:
Contingency table analysis
Observed | 2 | 2 | 1 | 5 |
% of row | 40% | 40% | 20% | |
% of col | 66.67% | 66.67% | 25% | 50% |
Observed | 1 | 1 | 3 | 5 |
% of row | 20% | 20% | 60% | |
% of col | 33.33% | 33.33% | 75% | 50% |
Total | 3 | 3 | 4 | 10 |
% of n | 30% | 30% | 40% |
TOTAL number of cells = 6
WARNING: 6 out of 6 cells have EXPECTATION < 5
NOMINAL INDEPENDENCE
Chi-square = 1.666667, DF = 2, P = 0.4346
G-square = 1.726092, DF = 2, P = 0.4219
Fisher-Freeman-Halton exact P = 0.5714
ANOVA
Chi-square for equality of mean column scores = 1.5
DF = 2, P = 0.4724
LINEAR TREND
Sample correlation (r) = 0.361158
Chi-square for linear trend (M²) = 1.173913
DF = 1, P = 0.2786
NOMINAL ASSOCIATION
Phi = 0.408248
Pearson's contingency = 0.377964
Cramér's V = 0.408248
ORDINAL
Goodman-Kruskal gamma = 0.555556
Approximate test of gamma = 0: SE = 0.384107, P = 0.1481, 95% CI = -0.197281 to 1.308392
Approximate test of independence: SE = 0.437445, P = 0.2041, 95% CI = -0.301821 to 1.412932
Kendall tau-b = 0.348155
Approximate test of tau-b = 0: SE = 0.275596, P = 0.2065, 95% CI = -0.192002 to 0.888313
Approximate test of independence: SE = 0.274138, P = 0.2041, 95% CI = -0.189145 to 0.885455
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