### Cross tabulation

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

Phi = 0.408248

Pearson’s contingency = 0.377964

Cramér’s V = 0.408248

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