Unit 14 : Chi squared, logistic regression and
      the normal approximation to the binomial

In this unit, we look at R programs to analyse dichotomous ("yes/no", "success/failure") data. We compare three approaches

  1. Chi-squared test of association in the 2x2 contingency table
  2. The t-test using the normal approximation to the binomial, and
  3. logistic regression
These three methods of testing the difference between two treatment groups are investigated in the video:



Review of and comments on the video:

In the video, we looked at 100 subjects per group (treatment A or treatment B), with a binary (dichotomous) outcome, success or failure.
Since the chi-squared test is not recommended for cell numbers less than 5, our comparisons started with 5 success compared with 6 successes and continued up to 95 successes.
Of course, we don't need to test 5 versus 5, 6 versus 6, etc. up to 95 versus 95, so those comparisons were eliminated.
So, how many comparisons are there? Using the notation (5,6) as shorthand for comparing results 5 to 6, then the set of all possible comparisons can be represented by:
(5,5) (5,6) (5,7)... (5,95)
(6,5) 56,6) (6,7)... (6,95)
...
(95,5) (95,6) (95,7).. (95,95)
Now, we only want the upper triangular part of this array since the diagonal compares equal outcomes and the lower triangular part are the same comparisons as the upper triangular part, but in the reverse order.
There are 91 columns in the array, which means the total number of entries in the array is 912. If we subtract the number on the diagonal (91), then the comparisons left are twice as many as I want to count (since there are equally many in the lower and upper triangular parts).
So, the number of comparisons is (912 - 91)/2 = 91x90/2 = 4095, the number of comparisons shown in the video.

Finally, here is the link to the programmes and word doc used in the video, as promised.

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