Can ANOVA be significant when none of the pairwise t-tests is?

Question

Is it possible for one-way (with $N>2$ groups, or "levels") ANOVA to report a significant difference when none of the $N(N-1)/2$ pairwise t-tests does?

In this answer @whuber wrote:

It is well known that a global ANOVA F test can detect a difference of means even in cases where no individual [unadjusted pairwise] t-test of any of the pairs of means will yield a significant result.

so apparently it is possible, but I do not understand how. When does it happen and what the intuition behind such a case would be? Maybe somebody can provide a simple toy example of such a situation?

Some further remarks:

1. The opposite is clearly possible: overall ANOVA can be non-significant while some of the pairwise t-tests erroneously report significant differences (i.e. those would be false positives).

2. My question is about standard, non-adjusted for multiple comparisons t-tests. If adjusted tests are used (like e.g. Tukey's HSD procedure), then it is possible that none of them turns out to be significant even though the overall ANOVA is. This is covered here in several questions, e.g. How can I get a significant overall ANOVA but no significant pairwise differences with Tukey's procedure? and Significant ANOVA interaction but non-significant pairwise comparisons.

3. Update. My question originally referred to the usual two-sample pairwise t-tests. However, as @whuber pointed out in the comments, in the ANOVA context, t-tests are usually understood as post hoc contrasts using the ANOVA estimate of the within-group variance, pooled across all groups (which is not what happens in a two-sample t-test). So there are actually two different versions of my question, and the answer to both of them turns out to be positive. See below.

Answer 1

Note: There was something wrong with my original example. I stupidly got caught by R's silent argument recycling. My new example is quite similar to my old one. Hopefully everything is right now.

Here's an example I made that has the ANOVA significant at the 5% level but none of the 6 pairwise comparisons are significant, even at the 5% level.

Here's the data:

g1:  10.71871  10.42931   9.46897   9.87644
g2:  10.64672   9.71863  10.04724  10.32505  10.22259  10.18082  10.76919  10.65447 
g3:  10.90556  10.94722  10.78947  10.96914  10.37724  10.81035  10.79333   9.94447 
g4:  10.81105  10.58746  10.96241  10.59571

Here's the ANOVA:

             Df Sum Sq Mean Sq F value Pr(>F)  
as.factor(g)  3  1.341  0.4469   3.191 0.0458 *
Residuals    20  2.800  0.1400        

Here's the two sample t-test p-values (equal variance assumption):

        g2     g3     g4
 g1   0.4680 0.0543 0.0809 
 g2          0.0550 0.0543 
 g3                 0.8108

With a little more fiddling with group means or individual points, the difference in significance could be made more striking (in that I could make the first p-value smaller and the lowest of the set of six p-values for the t-test higher).

--

Edit: Here's an additional example that was originally generated with noise about a trend, which shows how much better you can do if you move points around a little:

g1:  7.27374 10.31746 10.54047  9.76779
g2: 10.33672 11.33857 10.53057 11.13335 10.42108  9.97780 10.45676 10.16201
g3: 10.13160 10.79660  9.64026 10.74844 10.51241 11.08612 10.58339 10.86740
g4: 10.88055 13.47504 11.87896 10.11403

The F has a p-value below 3% and none of the t's has a p-value below 8%. (For a 3 group example - but with a somewhat larger p-value on the F - omit the second group)

And here's a really simple, if more artificial, example with 3 groups:

g1: 1.0  2.1
g2: 2.15 2.3 3.0 3.7 3.85
g3: 3.9  5.0

(In this case, the largest variance is on the middle group - but because of the larger sample size there, the standard error of the group mean is still smaller)

---

Multiple comparisons t-tests

whuber suggested I consider the multiple comparisons case. It proves to be quite interesting.

The case for multiple comparisons (all conducted at the original significance level - i.e. without adjusting alpha for multiple comparisons) is somewhat more difficult to achieve, as playing around with larger and smaller variances or more and fewer d.f. in the different groups don't help in the same way as they do with ordinary two-sample t-tests.

However, we do still have the tools of manipulating the number of groups and the significance level; if we choose more groups and smaller significance levels, it again becomes relatively straightforward to identify cases. Here's one:

Take eight groups with $n_i=2$. Define the values in the first four groups to be (2,2.5) and in the last four groups to be (3.5,4), and take $\alpha=0.0025$ (say). Then we have a significant F:

> summary(aov(values~ind,gs2))
            Df Sum Sq Mean Sq F value  Pr(>F)   
ind          7      9   1.286   10.29 0.00191 
Residuals    8      1   0.125                   

Yet the smallest p-value on the pairwise comparisons is not significant that that level:

> with(gs2,pairwise.t.test(values,ind,p.adjust.method="none"))

        Pairwise comparisons using t tests with pooled SD 

data:  values and ind 

   g1     g2     g3     g4     g5     g6     g7    
g2 1.0000 -      -      -      -      -      -     
g3 1.0000 1.0000 -      -      -      -      -     
g4 1.0000 1.0000 1.0000 -      -      -      -     
g5 0.0028 0.0028 0.0028 0.0028 -      -      -     
g6 0.0028 0.0028 0.0028 0.0028 1.0000 -      -     
g7 0.0028 0.0028 0.0028 0.0028 1.0000 1.0000 -     
g8 0.0028 0.0028 0.0028 0.0028 1.0000 1.0000 1.0000

P value adjustment method: none 

Answer 2

Summary: I believe that this is possible, but very, very unlikely. The difference will be small, and if it happens, it's because an assumption has been violated (such as homoscedasticity of variance).

Here's some code that seeks out such a possibility. Note that it increments the seed by 1 each time it runs, so that the seed is stored (and the search through seeds is systematic).

stopNow <- FALSE
counter <- 0
while(stopNow == FALSE) {
  counter <- counter + 1
  print(counter)
  set.seed(counter)
  x <- rep(c(0:5), 100)
  y <- rnorm(600) + x * 0.01
  df  <-as.data.frame( cbind(x, y))
  df$x <- as.factor(df$x)
  fit <- (lm(y ~ x, data=df))
  anovaP <- anova(fit)$"Pr(>F)"[[1]]
       minTtestP <- 1
      for(loop1 in c(0:5)){
        for(loop2 in c(0:5)) {
          newTtestP <- t.test(df[x==loop1,]$y, df[x==loop2,]$y)$p.value
      minTtestP <- min(minTtestP, newTtestP )    
      }
   }

  if(minTtestP > 0.05 & anovaP < 0.05) stopNow <- TRUE 
  cat("\nminTtestP = ", minTtestP )
  cat("\nanovaP = ", anovaP )
  cat("\nCounter = ", counter, "\n\n" )
}

Searching for a significant R2 and no non-significant t-tests I have found nothing up to a seed of 18,000. Searching for a lower p-value from R2 than from the t-tests, I get a result at seed = 323, but the difference is very, very small. It's possible that tweaking the parameters (increasing the number of groups?) might help. The reason that the R2 p-value can be smaller is that when the standard error is calculated for the parameters in the regression, all groups are combined, so the standard error of the difference is potentially smaller than in the t-test.

I wondered if violating heteroscedasticity might help (as it were). It does. If I use

y <- (rnorm(600) + x * 0.01) * x * 5

To generate the y, then I find a suitable result at seed = 1889, where the minimum p-value from the t-tests is 0.061 and the p-value associated with R-squared is 0.046.

If I vary the group sizes (which increases the effect of violation of heteroscedasticity), by replacing the x sampling with:

x <- sample(c(0:5), 100, replace=TRUE)

I get a significant result at seed = 531, with the minimum t-test p-value at 0.063 and the p-value for R2 at 0.046.

If I stop correcting for heteroscedasticity in the t-test, by using:

newTtestP <- t.test(df[x==loop1,]$y, df[x==loop2,]$y, var.equal = TRUE)$p.value

My conclusion is that this is very unlikely to occur, and the difference is likely to be very small, unless you have violated the homoscedasticity assumption in regression. Try running your analysis with a robust/sandwich/whatever you want to call it correction.

Answer 3

It's entirely possible:

• One or more pairwise t-test is signfiicant but the overall F-test isn't
• The overall F-test is significant but none of the pairwise t-test is

The overall F test tests all contrasts simultaneously. As such, it must be less sensitive (less statistical power) to individual contrasts (eg: a pairwise test). The two tests are closely related to each other but they are not reporting exactly the same thing.

As you can see, the textbook recommendation of not doing planned comparisons unless the overall F-test is significant is not always correct. In fact, the recommendation may prevent us from finding significant differences because the overall F test has less power than planned comparisons for testing the specific differences.