Question 3: Carry out a chi-squared test of the null hypothesis that survival is not related to treatment. Is the test valid?
We choose Analyze, Descriptive Statistics, Crosstabs. Put the variable 'Survival' into Row(s): and 'Treatment' into Column(s):. Click Statistics, Chisquare, Continue. Click OK.
You should get the case processing summary and then the following:
| Survival * Treatment Crosstabulation | ||||
|---|---|---|---|---|
| Count | ||||
| Treatment | Total | |||
| Streptomycin | Control | |||
| Survival | Survived | 51 | 38 | 389 |
| Died | 4 | 14 | 18 | |
| Total | 55 | 52 | 107 | |
| Chi-Square Tests | |||||
|---|---|---|---|---|---|
| Value | df | Asymp. Sig. (2-sided) | Exact Sig. (2-sided) | Exact Sig. (1-sided) | |
| Pearson Chi-Square | 7.376a | 1 | .007 | ||
| Continuity Correctionb | 6.039 | 1 | .014 | ||
| Likelihood Ratio | 7.705 | 1 | .006 | ||
| Fisher's Exact Test | .009 | .006 | |||
| Linear-by-Linear Association | 7.307 | 1 | .007 | ||
| N of Valid Cases | 107 | ||||
| a 0 cells (.0%) have expected count less than 5. The minimum expected count is 8.75. | |||||
| b Computed only for a 2x2 table | |||||
We automatically get Fisher's exact test and chi-squared with Yates' continuity correction. The linear-by-linear association test is meaningless with only two rows and columns.
Is the chi-squared test valid? Yes, because there are no cells with expected frequencies less than five. It does not matter that an observed frequency is less than five.
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Last updated: 20 February, 2012.