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I am trying to understand a published analysis. This is the data of interest:

      D1>0    D1<0
D2>0   7        2     9
D2<0   9        15   24
total  16       17   33
The author notes that 17/33 is 51.5% and states:

"we expect about 50% of the D1's to be negative, and that is what we actually observe here (z=.08, p=n.s)".

Now I assume that the z comparison is 17/33 vs 16.5/33 (chance?) and that the statistic employed is the z-ratio for the significance of the difference between two proportions.

However my own calculation for this comes out as z=.123. Can anyone help with the discrepancy?

UPDATE: The author has responded and admits that the published z is an error. He gives the corrected answer as z=0.17.

I am still not sure how he worked it out though:

We are not considering the difference between two proportions, since they are not independent 16/33 = 17/33 -1. We are interested whether the proportion of D1<0 significantly deviated from the expected value of 0.5

The nearest I can get is R prop.test function. Is this directly comparable?

prop.test(c(17, 16.5), c( 33, 33))

2-sample test for equality of proportions with continuity correction
data:  c(17, 16.5) out of c(33, 33) 
X-squared = 0, df = 1, p-value = 1
alternative hypothesis: two.sided 
95 percent confidence interval:
 -0.2411995  0.2715025 
sample estimates:
   prop 1    prop 2 
0.5151515 0.5000000
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  • $\begingroup$ Hmm, that's what I get too: |O-E|/sqrt(E) = |17-16.5|/16.5 = 0.123. Can you add a link to the paper if it's available on the web? $\endgroup$
    – onestop
    Commented Oct 18, 2010 at 9:57
  • $\begingroup$ Unfortunately it isn't. I have collected comparable data and wish to compare my findings with the publication. I have found that the R function prop.test gives results that approximately agree with the paper, but I wanted to understand how the author arrives at his exact numbers. $\endgroup$
    – Frank Zafka
    Commented Oct 18, 2010 at 10:11
  • $\begingroup$ I've contacted the author. Perhaps he can shed light on the problem. $\endgroup$
    – Frank Zafka
    Commented Oct 18, 2010 at 11:27
  • $\begingroup$ Are these paired proportions or two independent samples? $\endgroup$
    – chl
    Commented Oct 18, 2010 at 12:32

2 Answers 2

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The author is using a single proportion test. Try the following in R:

(p-0.5)/sqrt(0.5*0.5/33)

where p = 17/33.

See the wiki for the test statistic where the above test is called the One-proportion z-test.

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  • $\begingroup$ That seems to be it. I will go away and think about it so more. Many thanks all. $\endgroup$
    – Frank Zafka
    Commented Oct 19, 2010 at 17:51
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I'm not sure the discrepancy is worth worrying about. The exact p-value is clearly 1 for a 2-sided test of p=0.5 given 17 positive responses out of 33, as there's no integer closer to 33/2 than 17. With small or moderate N as here there's no good reason for not doing the exact test (even without a PC, as the cdf of the binomial is commonly included in statistical tables). With a modern PC and decent software it's no problem even for large N, but for large N the resulting p-value will be very similar to that from a z-test anyway.

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  • $\begingroup$ That was my understanding and like I say, the prop.test function of R gives the same direction of results. I just wondered if there was something that I was doing wrong to not get the z=.08. $\endgroup$
    – Frank Zafka
    Commented Oct 18, 2010 at 12:55

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