Your results illustrate the importance of interaction terms. This is no "illusion"; this phenomenon might be expected whenever the effect of one predictor on outcome depends on the value of another predictor.
The cookie-yield example in the Wikipedia entry on interactions in statistics provides an easy-to-grasp illustration. The outcome is the yield of "...
Joe Berkson coined the phrase. To my knowledge, it first shows up in 1963 in Edwards, Lindman, and Savage's "Bayesian Statistical Inference for Psychological Research" in Psychological Review, 70(3):
The preceding paragraph illustrates a procedure that statisticians of all schools find important but elusive. It has been called the interocular traumatic ...
If the data sample is large enough (I would say at least 10 times the number of categories), then you may apply a chi-square test of homogeneity for an uniform distribution.
Regarding the graphical methods, consider a bar chart showing a subset of categories, for instance, the top 5 and the bottom 5 categories.
Following my Comment, here is an example of finding powers of two
tests by means of simulation:
Suppose you have two samples of size $n_1 = n_2 = 10$ from
normal distributions with $\mu_1 = 1, \mu_2=3$ and $\sigma_1=\sigma_2 =1.$
A two sample t test is the natural test to see if data are likely to show a significant difference between the means. What is ...
I agree with EdM that this is not an illusion, but rather a demonstration of the importance of considering nonlinear models.
The negative sign on the interaction term slope may be important. The simple slopes implied by your interaction model results are
X: 1.3 - 0.68 W
W: 2.5 - 0.68 X
You don't say whether X and W are centered about their mean. ...
Let $C$ denote being a child and $C'=A$ being an adult. Similarly, let $L$ denote liking the cake. Then, based on the data, we can estimate the following probabilities for the population:
$$P(L|C)=0.9, \ \ P(L|C')=0.1$$
On your sample we have $P(C)=P(C')=0.5$ since number of adults and children are the same. But, you can also assign prior probabilities to ...
With such relatively small samples, I would not expect definitive results
from either the Shapiro-Wilk or the Kolmogorov-Smirnov tests. Usually, the
latter has poorer power than the former so I wonder why K-S (alone) finds group M
data non-normal. Even though all six of the P-values for normality tests
are about the same, I would want to see whether there ...
Ertxiem's answer is precisely what you want. Calculate a $\chi^2$ test with $1000-1=999$ degrees of freedom.
I personally am a big fan of simulating the null hypothesis a couple of times and plotting the results of such simulations, to get a feeling for the randomness that the null hypothesis would imply - and then comparing these plots to the actual data ...