Take the 2-minute tour ×
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

I'm looking at clinical trial data where there are various numbers of patients on different arms (A, B, C, D). All arms receive a drug that is known to cause a toxicity, but some arms receive an additional drug which might contribute to the toxicity. I have been using 3 methods to identify the number of patients with the toxicity, and I was curious if the ratios of patients on the different arms differed between the methods and whether the ratios might have been higher on some arms (suggesting the additional toxicity of the added drug) than what is seen with the entire population.

For example, number of patients identified with the toxicity with the different methods:

             A     B    C    D   Total
method 1:    8    18   26   16      68
method 2:   45   111   53  113     322  
method 3:   68   158   69  153     448  
Overall:   695  1331  441  925    3392

To compare, I calculated the percentage of patients on each arm for each of the methods (Method 1 Arm A: 8/68) and compared it to the overall population (695/3392). However, after doing this, I was told I need to normalize by the patients on each arm. I am not sure what is meant by this. I can divide the numbers of patients identified for each arm by the total number of patients on the arm, but then how does this help me do the comparison I am interested in? I don't have a stats background, so I am not sure how to tackle this question in general.

share|improve this question
Welcome to the site, @Gally. I took the liberty of formatting your numbers so that they would display in a way that was easier for people to read. Please make sure that it is accurate. Note, eg, that in the original numbers 1A is 8, but in text in next paragraph it's 6. –  gung Oct 22 '12 at 16:42
Thank you, I fixed the discrepancy –  Gally Oct 22 '12 at 18:02
add comment

1 Answer

If I understand your problem correctly, then you can use the 2 by K Chi-Square test (without any standardization).


share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.