2
$\begingroup$

Medical scientists love p-values and in a publication, I found the following patient statistics table. I stumbled over the Number of patients line.

enter image description here

They divided their subject group into two subgroups TAV and BAV and I asked how they calculated the p-value, because you only have one characteristic BAV/TAV. If you look for instance at the Male row, you see that you can use a 2x2 contingency table consisting of 2 characteristics BAV/TAV and Male/Female and make e.g. a Fisher test.

What was done in the Number of patients row was to fake a 2x2 contingency table repeating and reversing the numbers:

$$\begin{array}{c|c} 347&32\\\hline 32&347 \end{array} $$

From the viewpoint that they want to express how similar the number of patients are in the BAV and TAV group, this kind of makes sense because it gives a p-value of 1 (two tailed Fisher) when BAV and TAV have the same number of patients. However, I have never seen this and I cannot find any reference to such an approach.

Question: Can someone tell me whether it is correct what they did?

$\endgroup$
3
  • $\begingroup$ Do you mean 2x2 table??? $\endgroup$
    – SmallChess
    Apr 15, 2016 at 11:32
  • $\begingroup$ In the title of the question, I really meant 1x2 because basically we have only 1x2 data. But I see your point that it might be confusing. Maybe we find a different title for the question that describes the problem better. Any suggestions? $\endgroup$
    – halirutan
    Apr 15, 2016 at 11:33
  • $\begingroup$ No need. If you mean 1x2, let it be. $\endgroup$
    – SmallChess
    Apr 15, 2016 at 11:34

2 Answers 2

3
$\begingroup$

The only "right" way I can think of to analyze that type of patient row is to test whether the proportions are equal. This can be done with a one way chi-square test, which is, indeed, a 1x2 table.

I've never seen the sort of procedure you mention - did they say in the article that that is what they did, or are you guessing/figuring out that that is what they did?

The next question is whether the two methods give the same results. The table you show doesn't give much, so let's test:

type <- c(TAV = 347, BAV = 32)
chisq.test(type)

and

x <- matrix(c(347,32,32,347),ncol = 2)
chisq.test(x)

give identical p values. The first is what I suggested, the second is what you said.

$\endgroup$
4
  • 1
    $\begingroup$ Thanks for pointing this out. I really know what they did because I asked. +1 $\endgroup$
    – halirutan
    Apr 15, 2016 at 12:09
  • 1
    $\begingroup$ Note, these are giving different p-values, it's just hard to tell since the p-value is so small. If you change 347 to 50, you'll see different p-values. I think this should be a one proportion z-test, please see my response $\endgroup$ Apr 21, 2016 at 7:02
  • $\begingroup$ I tried your solution on my data, however I got an error "all entries of 'x' must be nonnegative and finite". I guess it is because I have missing values/data. Would you be so kind and complete your answer for cases with missing values/data? $\endgroup$
    – pdeli
    Sep 15, 2023 at 15:38
  • 1
    $\begingroup$ @pdeli If you have missing data, that is a different question. Feel free to ask your own question. $\endgroup$
    – Peter Flom
    Sep 15, 2023 at 18:24
1
$\begingroup$

The p-value calculations are a little confusing. The p-values are trying to test whether the percentages are similar for the two subgroups. For the first row "Number of patients" the % in TAV has to be 1-% in BAV. Therefore, one should test whether p=0.5 for one proportion:

prop.test(221, 244, p=0.5, alternative="two.sided")

For the "Male Gender" row, the % of males in TAV does not have to be 1-% of males in BAV, so we have a different test. You can actually fill out the table:

       Male   Female  Total
TAV    221     126     347
BAV    23      9       32
Total  244     135

This can be tested using Fisher's exact test (although Barnard's exact test is superior) giving the p-value reported.

> data<-matrix(c(221,126,23,9),ncol=2,byrow=TRUE)
> fisher.test(data,alternative="two.sided")

    Fisher's Exact Test for Count Data

data:  data
p-value = 0.4418
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 0.2710563 1.5995457
sample estimates:
odds ratio 
  0.686984 
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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