Question: Using SPSS, cross-tabulate experiences of falls by treatment group and carry out Fisher's exact test.
We choose Analyze, Descriptive Statistics, Crosstabs. Put the variable 'Experience of falls' into Row(s): and 'Group' into Column(s):. Click Statistics, Chisquare, Continue. Click OK.
You should get the case processing summary and then the following:
Experience of falls * Group Crosstabulation | ||||
---|---|---|---|---|
Count | ||||
Group | Total | |||
Control | Exercise | |||
Experience of falls | No fall | 110 | 119 | 229 |
Fall | 9 | 2 | 11 | |
Total | 119 | 121 | 240 |
Chi-Square Tests | |||||
---|---|---|---|---|---|
Value | df | Asymp. Sig. (2-sided) | Exact Sig. (2-sided) | Exact Sig. (1-sided) | |
Pearson Chi-Square | 4.7926^{a} | 1 | .029 | ||
Continuity Correction^{b} | 3.536 ^{ } | 1 | .060 | ||
Likelihood Ratio | 5.155 ^{ } | 1 | .023 | ||
Fisher's Exact Test | .033 | .028 | |||
Linear-by-Linear Association | 4.772 ^{ } | 1 | .029 | ||
N of Valid Cases | 240 ^{ } | ||||
^{a} 0 cells (.0%) have expected count less than 5. The minimum expected count is 5.45. | |||||
^{b} Computed only for a 2x2 table |
For a two by two table, we automatically get Fisher's exact test and chi-squared with Yates' continuity correction.
We have Fisher's exact test: P = 0.03.
As usual, I have added a leading zero and for P values given only the first significant (i,e, non-zero) figure. When we do this we round as for decimal places. Here we round down from 0.033 to 0.03.
The chi-squared test is valid and gives us the same probablity, P = 0.03, after rounding up from 0.029.
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Last updated: 20 February, 2012.