Post-hoc test for chi squared goodness-of-fit test: understanding Bonferroni

I have seen this, but it doesn't help me here.

Here is my setup:

I have four categories, and want to simultaneously test whether p_1 = 0.4, p_2 = 0.35, p_3 = 0.15, p_4 = 0.10. I am using a chi squared goodness-of-fit test here.

I find that I should reject the null hypothesis, meaning that at least one of these population proportions is indeed statistically significantly different from its hypothesized value.

Then, post-hoc, I want to find which one(s) is/are significantly different. To do this, I've used a confidence interval on each population proportion, but found that all of the confidence intervals contain the hypothesized value - so that means I shouldn't have been able to reject the original null hypothesis!

The formula I used for my confidence intervals was:

$$\hat{p}_i \pm z_{0.00625}\sqrt{\frac{\hat{p}_i(1-\hat{p}_i)}{n}}$$

where $$\hat{p}_i =$$ the sample proportion for category $$i$$, $$n =$$ the sample size, and $$z_{0.00625}$$ is the Bonferroni-adjusted value from the normal distribution. From a table I found, this is 2.50.

I think I must've done something wrong here, but I can't figure out what. I've looked at this source and I've also read this SE question, but I can't seem to make the connection to my case.

• What's the significance level $\alpha$ that you want the post-hoc tests to have? Oct 21, 2022 at 20:42

The idea of Bonferroni is to make sure the combined type I error probability, i.e., the probability of rejecting at least one $$H_0$$ if all tested $$H_0$$ are true, is lower than the desired significance level $$\alpha$$, say 0.05. This is done using a worst case computation, meaning that even if all tests are dependent in the worst possible way, error probability $$\le\alpha$$ is still respected. This is only possible if the probability to reject for each individual test is very low - in fact you can see this because for $$\alpha=0.05$$, the Bonferroni p-value threshold to reject (or, equivalently, one minus the adjusted confidence level) for $$k$$ tests is $$\alpha/k=0.05/4=0.0125$$. It is much harder to achieve $$p<0.0125$$ than $$p<0.05$$, and this translates into the Bonferroni-corrected test to have low power, i.e., a relatively low probability to reject even if the $$H_0$$ is actually false. In other words, it is quite conservative.

I assume your original $$\chi^2$$-test was carried out a level 0.05, just one test, no adjustment, so it is easier for data to reject the $$H_0$$, and it may well happen that it was easy enough to come out significant even if none of the four tests for single proportion, tested at effective level of 0.0125, were significant as the latter are conservative.

I guess that your original $$\chi^2$$ p-value may have been smaller than 0.05 but larger than 0.0125, which would illustrate the problem well. However this does not necessarily have to be so; there is a further possible explanation.

The $$\chi^2$$-test statistic sums up deviations from what is expected under $$H_0$$ over all four proportions. It being significant means that the combined variation is too big compared to what is expected under $$H_0$$. This can happen because one of the four individual deviations is far too big, in which case that individual proportion may well also come out significant when running the post-hoc tests (apart from conservativity of Bonferroni, see above). It may however also be that all four deviations are slightly too big, but there is no single one that stands out. It may then be that nothing significant can be seen when looking at the individual proportions in isolation; they may all deliver p-values that are maybe all small but bigger than 0.05 even if the overall $$\chi^2$$-test has $$p<0.05$$ from putting all four deviations together (which uses more information to find any problem with the $$H_0$$ than any single post hoc test; the same argument of course can apply to the Bonferroni threshold 0.0125).

Note by the way that, as the proportions have to sum up to 1, it is not possible that your null hypothesis is truly violated for just one proportion. If one proportion is too large, there must be at least one other that is too small. It is however possible that data will single out just one proportion as significantly different from $$H_0$$ in the post hoc test, which may happen if one proportion is strongly wrong in one direction, and all the others are "less wrong" (say all three by 1/3 of the difference) in the other direction. There may be enough data to find the problem with one proportion in such a situation, but not enough to find the problem with the others. (Keep in mind that not rejecting never means that the $$H_0$$ is true.)

• Since the probabilities (in a multinomial distribution on $k$ categories) are constrained to sum up to 1, why adjust the confidence level by $k$? It "feels intuitively" that it might be enough to adjust by $k-1$? Oct 21, 2022 at 21:56
• @dipetkov Fair point. What counts is the number of tests, and one of them is probably redundant. Though I don't think it'll be trivial to show that if you run these $k$ tests and adjust by $k-1$, the overall level is still fine. It does not hold that if $k-1$ tests are not significant, neither can be the $k$th one. So just making sure that the probability that any of $k-1$ of these tests is significant is low enough under $H_0$ isn't enough to guarantee this also for all $k$ tests. Oct 21, 2022 at 22:42
• By the way I'm not claiming that using Bonferroni in this way is the best thing to do here; I'm just explaining what the implications are of doing it. Oct 21, 2022 at 22:47
• Statistical significance of the joint test does not imply statistical significance of one of the four individual tests for the proportions (see, e.g., this question).
• Your Bonferroni correction seems fine and it ensures (in your case asymptotically) strong control of the FWER. But it doesn't ensure a low type II error rate. One more powerful alternative would be the Holm method.
• The denominator of the radicand should be $$n_i$$ in case $$n_i \neq n$$
• $$z_{0.00625} \approx -2.5$$.
• There are confidence intervals for binomial proportions, such as the Blyth-Still or the Agresti–Coull confidence interval, that are preferable to a Wald confidence interval.