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I would like to know if it is correct to do the following statistical comparison: In a sample of $440$ patients, I have $241$ females ($54.8\%$) and $199$ males ($45.2\%$). Are these proportions statistically different?

I used $z$ test for independent samples. So: $z= 2.764$ (with Yates correction) and $p<0.01$.

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    $\begingroup$ I'm not sure I can see how 55% female can be independent of 45% male. You might want to use a binomial test to see if the true proportion is 50-50. $\endgroup$ Nov 8, 2014 at 23:06
  • $\begingroup$ Thank you for the information. And now another problema. When using the binomial test for these data. A one tail test has p<0,05 and a two tails test has p>0,05 (not significant). Which one must I use if I wanto to know if both proportions (54.8% and 45.2%) are different. My original interest was to know if there was statistical difference in sex proportions, but it doesn't matter which one. So I think I should use a two tails test, ¿Wright? $\endgroup$
    – sbonoli
    Nov 9, 2014 at 0:35
  • $\begingroup$ But if I think in terms of trying to determine if one of the proportions (whichever) is different from 0.5 I should use the one tail test. Really I'm confused. $\endgroup$
    – sbonoli
    Nov 9, 2014 at 0:41
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    $\begingroup$ You would run only one test. It would be two tailed. $\endgroup$ Nov 9, 2014 at 1:40

1 Answer 1

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(I'll turn my comments into an official answer.)

Since there are only females and males in reality, once the proportion female (male) has been determined the other is determined also. That is, these 'two' proportions are not independent of each other. You wonder if they are equal, which means you wonder if the proportion female is 50%. You can test this with a binomial test. Since you don't have a theory suggesting one percentage is greater apriori, you would use a two tailed test.

Using your data, here is a demonstration using R:

##### test if proportion female = proportion male
binom.test(x=c(241, 199), p=0.5, alternative="two.sided")
# 
#  Exact binomial test
# 
# data:  c(241, 199)
# number of successes = 241, number of trials = 440,
# p-value = 0.05051
# alternative hypothesis: true probability of success is not equal to 0.5
# 95 percent confidence interval:
#  0.4998970 0.5949117
# sample estimates:
# probability of success 
#              0.5477273 

##### test if sample is consistent with known population
binom.test(x=c(241, 199), p=0.513, alternative="greater")
# 
#  Exact binomial test
# 
# data:  c(241, 199)
# number of successes = 241, number of trials = 440,
# p-value = 0.07922
# alternative hypothesis: true probability of success is greater than 0.513
# 95 percent confidence interval:
#  0.5074098 1.0000000
# sample estimates:
# probability of success 
#              0.5477273
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  • $\begingroup$ Thank's for help. I've also seen that we can use binomial test if we know the proportion of the population from which we take the sample. According to the Venezuelan Census (2011) the proportion of females for the parrish to which belongs the town from which I took the sample is 51,3%. So I used the binomial test but it is said I've to use a one tail test, although I'm still interested in determining if the simple proportion is only different from the population proportion. I don't understand. Obviously the test performed in this way answers another question different from the original one. $\endgroup$
    – sbonoli
    Nov 10, 2014 at 0:03
  • $\begingroup$ But besides this disgression, I'm still interested in determining if male and female proportions in sample are different. So I used a test to determine if a sample proportion is different from a given one. Using z = (obs. prop. - given prop)/sample n. For the data I use: obs. prop. = 0,548; n = 440 and the given proportion I used is the male proportion (0,452). Then z = 2,002 and p=0,045 for a two tails test. Remember I'm interested in difference only, not if one is greater tan other. Is this approach right? $\endgroup$
    – sbonoli
    Nov 10, 2014 at 0:04
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    $\begingroup$ @sbonoli, the only way the 2 proportions can be different from each other is if they are not 50-50. To test that hypothesis, you use a (2-tailed) binomial test against p=0.5. It is what I show above. You can also test if your sample is consistent with the population (ie test if % female is 51.3%). In that case, you'd use p=0.513 in the function call above. (I can demonstrate it, if you need; it will be slightly less significant.) $\endgroup$ Nov 10, 2014 at 0:12
  • $\begingroup$ Not necessary, I've programmed the binomial test, and have found the answer. If I'm not wrong, testing 0,548 against 0,513 is not statistically significant (p = 0,079; 1 tail as suggested by SPSS. Or the double if we usea two-tailed test, as you stated above). $\endgroup$
    – sbonoli
    Nov 10, 2014 at 0:56

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