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The question I have is how to test if two proportions from the same sample of patients are significantly different from each other, indicated by a p-value?

Example: 100 people, 1 has Disease A and 2 have Disease B. Is the occurrence of Disease A more frequent than Disease B, ie. occurrence of Disease A given by 1/100 compared to 2/100 for B?

Keyword is 'same' sample of patients. Any help would be much appreciated. :)

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    $\begingroup$ Can the same patient have both diseases? $\endgroup$ Commented Aug 16, 2012 at 11:24
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    $\begingroup$ Since these samples are the same standard inference assuming independent random samples form a given population is violated. What is the inference that you want to make? Are you wanting to test whether or not the data indicate that the incidence for B is greater than for A in the population that the 100 patients were sampled from? $\endgroup$ Commented Aug 16, 2012 at 11:32
  • $\begingroup$ @Michael Chernick: yes patient can have both diseases.I do want to test whether or not the data indicate that the incidence for B is greater than for A in the population that the 100 patients were sampled from $\endgroup$ Commented Aug 16, 2012 at 11:42

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What you're asking about is a test of dependent (or paired) proportions. Please see this article on McNemar's test or this calculator site (its language isn't the clearest, but it will help you calculate the result you're looking for).

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  • $\begingroup$ The problem with McNemar is, however, that it takes into account mutual exclusive cases only, i.e. those persons who has A only and B only. Significance of differences depends only on these figures and does not depends on total sample. For example, you have only total 10 cases "A only" plus "B only". Calculated p will depends on how A differ from B and does not depends that these 10 cases over 30 persons total (who have both A and B or nothing) or over 3000. The p will be the same. This is a problem, $\endgroup$
    – Niksr
    Commented Nov 23, 2015 at 20:38
  • $\begingroup$ So whats the solution. I still think that the normal test for proportion, the z test, is usable. $\endgroup$
    – user108806
    Commented Mar 16, 2016 at 15:35

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