Question

A clinical trial is conducted comparing a new (investigational) analgesic drug to one already on the market (Tylenol) for arthritis pain. Participants were randomly assigned to only one of the two treatment groups and the outcome was self-reported pain relief within 30 minutes. The 2 x 2 crosstab analysis with SPSS produced the following results: Chi-Square Tests Value df Asymp. Sig. (2-sided) Exact Sig. (2-sided) Exact Sig. (1-sided) Pearson Chi-Square 11.161a 1 .001 Continuity Correctionb 10.212 1 .001 Likelihood Ratio 11.349 1 .001 Fisher's Exact Test .001 .001 Linear-by-Linear Association 11.115 1 .001 N of Valid Cases 240 a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 32.50. b. Computed only for a 2x2 table Note: Footnote “a” is very important because it tells us about a potential violation of a Pearson chi-square test assumption. 25. Is there a violation of the Pearson chi-square test assumption?

Answer 1

In order to run a chi square test, we need to check whether the data satisfies the following assumptions:

a. The two study variables are categorical in nature, Here, we find that our variables satisfy this assumption. Variable 1: Whether the drug administered is investigational. Variable 2. Whether the drug administered is Tylenol ......................(Both with categories Yes / No))

b. Each observation is independent and represents a single subject.

c. No more than 20% of the expected counts are less than 5.

Here, the footnote “a” tells us about a potential violation of the third assumption. Here, from the output, we find that the minimum expected count for our data is 32.50, which is greater than 5.We have zero cells (no cells) with expected count less than 5  Hence, we may safely conclude that there is no violation of the Pearson chi-square test assumption based on footnote a.