Testing a difference in hits (ones versus zeros) between two independent groups $X$ and $Y$ should be possible with a t-test, according to the following considerations:

  • $x_i\in\{0,1\}$ is the measurement for the $i$-th item im group $X$, and $y_i\in\{0,1\}$ the same for group $Y$
  • the proportion in each group is the mean value of the measurements, i.e. $\mu_X=\sum_i x_i/n$ and $\mu_Y=\sum_i y_i/n$
  • the difference of mean values $\mu_X$ and $\mu_Y$ in two groups can be tested with a t-test

In this particular case, a proportion test (function prop.test in R) is an alternative test option. Interestingly, the results are quite different:

> x <- c(rep(1, 10), rep(0, 90))
> y <- c(rep(1, 20), rep(0, 80))
> t.test(x,y,paired=FALSE)
t = -1.99, df = 183.61, p-value = 0.04808
> prop.test(c(10,20), c(100,100))
X-squared = 3.1765, df = 1, p-value = 0.07471

Note the higher p-value of the prop.test. Does this mean that a t-test has a higher power, i.e., can distinguish between $H_0$ and its alternative already for smaller $n$? Is there a reason why a t-test should not be used in this case?

Addition (Edit: resolved in a comment under the answer by Thomas Lumley below): The result of the t-test is even more surprising in the light of the observation that even the asymptotic ("Wald") 95% confidence intervals of both measurements overlap (0.1587989 > 0.1216014):

> library(binom)
> binom.confint(10, 100, method="asymptotic")
      method  x   n mean      lower     upper
1 asymptotic 10 100  0.1 0.04120108 0.1587989
> binom.confint(20, 100, method="asymptotic")
      method  x   n mean     lower     upper
1 asymptotic 20 100  0.2 0.1216014 0.2783986

As confidence intervals based on the t-distribution should be even wider than those based on the normal distribution (i.e. $z_{1-\alpha/2}$), I do not understand why the t-test reports a significant difference at the 5% level.

  • 4
    $\begingroup$ Most of the difference is the continuity correction which was applied in the chi-squared but not in the t-test. The remainder of the difference is the same as the difference between z-test and t-test. If you use correct=FALSE in the chi-squared you will see the p-value is fairly close to that of the t-test, or if you compute the continuity-corrected t-test and compare with the above chi-squared, you'll see the two fairly close again. $\endgroup$
    – Glen_b
    Jul 23, 2021 at 13:44
  • $\begingroup$ @glen-b Ah yes, this makes the difference. And the continuity correction makes the test overly conservative. I have found simulations by Agostino et al. (Am. Stat. 42, pp. 199-201, 1988) that show that the t test or the uncorrected chi square test (correct=FALSE) has an actual $\alpha$ probability much closer to the nominal level than the corrected chi square test. The t test is thus perfectly ok in this case, and my question is answered. $\endgroup$
    – cdalitz
    Jul 25, 2021 at 9:19

3 Answers 3


You're correct that the tests should be more similar. They are tests of means, and for a light-tailed distribution, so you should expect them to agree. What's more, the estimated variance $\hat p(1-\hat p)/n$ for a binomial distribution is extremely close to $s^2/n$

> var(x)/100
[1] 0.0009090909
> .1*(.9)/100
[1] 9e-04
> .2*(.8)/100
[1] 0.0016
> var(y)/100
[1] 0.001616162

What you're seeing is the continuity correction. If you try it without, the $p$-values are almost identical

> t.test(x,y)

    Welch Two Sample t-test

data:  x and y
t = -1.99, df = 183.61, p-value = 0.04808
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.1991454034 -0.0008545966
sample estimates:
mean of x mean of y 
      0.1       0.2 

> prop.test(c(10,20),c(100, 100),correct=FALSE)

    2-sample test for equality of proportions without continuity correction

data:  c(10, 20) out of c(100, 100)
X-squared = 3.9216, df = 1, p-value = 0.04767
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.197998199 -0.002001801
sample estimates:
prop 1 prop 2 
   0.1    0.2 

The continuity correction for the chi-squared test is a bit controversial. It does dramatically reduce the number of situations where the test is anti-conservative, but at the price of making the test noticeably conservative. Not using the 'correction' gives p-values that are closer to a uniform distribution under the null hypothesis. And, as you see here, not using the correction gives you something closer to the t-test.

  • $\begingroup$ Thanks, this explains the difference! What I still do not understand is why the t-test reports a significant ($\alpha=5\%$)difference although the classic 95% confidence intervals $\pm z_{1-\alpha/2}\sqrt(\hat{p}(1-\hat{p})/n}$ (and, of course, also with $t$ instead of $z$) overlap. $\endgroup$
    – cdalitz
    Jul 24, 2021 at 10:59
  • 2
    $\begingroup$ Non-overlap of 95% confidence intervals is a much stronger criterion than a 95% confidence interval for the difference overlapping zero. With equal variance and sample size, a 95% ci for difference overlaps zero if the difference is smaller than $1.96\sqrt{2}\sigma/\sqrt{n}|$ and two intervals overlap if the difference is smaller than $2\times 1.96\sigma/\sqrt{n}$. Each interval overlapping the other point estimate is much closer to a 5% threshold. $\endgroup$ Jul 25, 2021 at 2:01
  • $\begingroup$ Ah yes, thanks.!This follows from $Var(X-Y)=Var(X)+Var(Y)$ for independent $X$ and $Y$. The continuity correction seems to be indeed controversial in this case: Agostino et al. ("The Appropriateness of Some Common Procedures for Testing the Equality of Two Independent Binomial Populations", Am. Stat. 1988) investigated its application in this particular use case of comparing proportions and came to the conclusion that the uncorrected chi squared test or the t test should be used. $\endgroup$
    – cdalitz
    Jul 25, 2021 at 9:05
  1. A difference between p-values of 0.048 and 0.074 is not large. This can easily happen between tests that don't do exactly the same but a similar thing.

  2. The theory of the t-test is for normally distributed data, which your data obviously are not. You're right that the t-test can be justified as an approximation, but there's no reason to use an approximation if a more precise test (namely the proportion test) is available. For sure there is no reason to expect the t-test to have a better power, or only in case that it is anticonservative, which is not a good thing (being an approximation, one would probably need to simulate what its finite sample characteristics are in this situation).

  3. Edited after looking up the reference Agostino et al. ("The Appropriateness of Some Common Procedures for Testing the Equality of Two Independent Binomial Populations", Am. Stat. 1988) given by cdalitz. This reference states that prop.test with continuity correction is too conservative whereas the t-test as well as the prop.test without continuity correction are normally closer to the nominal level, if occasionally anticonservative (which in may view not necessarily justifies an overall recommendation). This was also mentioned in the answer by Thomas Lumley.

    If we're ignoring the continuity correction for a moment, there are two differences between the t-test and prop.test (which is not fully documented but I think it does the z-test based on normal approximation).

    (a) prop.test uses the knowledge that the variance of the Binomial is $np(1-p)$ rather than using a sample variance based on normality. In my view what prop.test does here should clearly do better, as it is based on information about the specific setup used here.

    (b) prop.test uses a normal approximation whereas the t-test uses a t-approximation. Now both of these, applied to the Binomial situation, are asymptotic in nature (the t-distribution is only precise if the underlying data are normal which they aren't here), and actually they are asymptotically equivalent. Although the normal approximation looks more intuitive based on the Central Limit Theorem, this doesn't imply by any means than the normal works better than the t in the finite sample situation (and the t is as well justified by the CLT, if only indirectly). The t-distribution is motivated by the normal assumption, but in fact it may also be the case that the asymptotic normal distribution of prop.test underestimates the finite sample variability because it ignores the variability in the variance estimation, and the t-distribution, despite here not precisely justified, may do a better job at that.

    So I now believe that potentially (as could be confirmed by simulations, maybe somebody has done that?) the best thing to do could be using the test statistic of prop.test, i.e., the "correct" variance estimation, but replacing the asymptotic normal distribution by a t-distribution, which in some sense may put together the advantages of them both.

  • 1
    $\begingroup$ Fair enough, but the continuity issue is the same with the t-test, except there it's not "corrected". If somebody would use the t-test in a submission and I were reviewer, I wouldn't accept it unless the author would run the simulations themselves to convince the reader that the t-test is better. Until that happens I believe strongly that it isn't. $\endgroup$ Jul 23, 2021 at 23:50
  • 2
    $\begingroup$ –1 "A difference between p-values of 0.048 and 0.074 is not large." This statement depends on the context, and is not always true. $\endgroup$
    – Alexis
    Jul 24, 2021 at 4:29
  • 4
    $\begingroup$ +1 @Alexis A difference between p=0.048 and p=0.074 is trivial if you are using p-values as indices of evidence against the null hypothesis within the statistical model chosen. The difference may be interesting if you are using a cutoff of p=0.05 to distinguish between a 'significant' result and a 'not significant' result, but you probably shouldn't be doing that. See here: for a start stats.stackexchange.com/questions/16218/… $\endgroup$ Jul 24, 2021 at 7:06
  • 3
    $\begingroup$ @lewian "If somebody would use the t-test in a submission and I were reviewer, I wouldn't accept it" Although a reviewer has the power to do so, I would not reject an arcticle because the authors use a different test method than I prefer. After searching more about this problem, I found a paper by Agostino et al. ("The Appropriateness of Some Common Procedures for Testing the Equality of Two Independent Binomial Populations", Am. Stat. 1988), who made this investigations and eventually recommended the t-Test or the uncorrected chi squared test (prop.test(...,correct=FALSE). $\endgroup$
    – cdalitz
    Jul 25, 2021 at 9:00
  • 2
    $\begingroup$ "I don't know how to do paragraphs within an item of a numbered list" -- find a post with a numbered list that does it and look at the markdown for that post. There's an example here: stats.stackexchange.com/questions/485348/… $\endgroup$
    – Glen_b
    Jul 25, 2021 at 16:05

The t-test can be quite robust to deviations from the normality assumption, particularly when sample sizes are large, so I understand why one might want to use a t-test for this task.

However, you know the parametric family; since the outcome is either $0$ or $1$, the distribution is completely characterized by the relative proportion, thus Bernoulli. Consequently, you can rely on a parametric test designed for a Bernoulli variable, which the t-test is not.

Methods that are robust to deviations from parametric assumptions are wonderful, since we typically do not know the type of population distribution. (If we did, why did we not determine the population parameters when we had the chance!?) However, the case of a binary variable is unique in how it is completely defined by the relative proportion and must be Bernoulli (or easy to represent as Bernoulli, such as calling “heads” and “tails” of a coin $0$ and $1$, respectively).

  • 1
    $\begingroup$ I think for investigating the power one would need to sample from the alternative, but I advise against it because I'm pretty sure that the only way a better power can arise is bias/being anticonservative. (Edited after the earlier comment was deleted.) $\endgroup$ Jul 23, 2021 at 12:49

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