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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 "...


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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 ...


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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.


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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 ...


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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. ...


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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 ...


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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 ...


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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 ...


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