Suggested answers to exercise: Outcome of the MRC streptomycin trial, 3

Question 3: Carry out a chi-squared test of the null hypothesis that survival is not related to treatment. Is the test valid?

Suggested answer

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
Treatment Total
Streptomycin Control
Survival Survived 51 38 389
Died 4 14 18
Total 55 52 107

Chi-Square Tests
Value   df   Asymp. Sig.
Exact Sig.
Exact Sig.
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.

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