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 | 7.376^{a} | 1 | .007 | ||
Continuity Correction^{b} | 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.