# 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?

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.