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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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A coin may be fair or it may be biased so that $p = P(\mathtt{Heads}) = 2/3.$
[Notice that I've gotten rid of the possibly confusing twist of letting $p$ be the probability of Tails.]
We want to test $H_0: p = 1/2$ against $H_1: p = 2/3.$ [This situation in which $H_0$ and $H_1$ each specify only one value is called 'simple vs. simple'.] Data for the test ...
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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.
answered Apr 1 at 22:11
Ertxiem - reinstate Monica
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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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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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The way to reconcile the opposing views is the following:
1) The t-test does assume normality.
2) The t-test is robust to deviations from said normality.
For #1, we do need normality in order for the test statistic to have the t distribution that we assume. However, #2 kicks in by slowing the data themselves to be non-normal but allowing for the test ...
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oh dear I went totally the wrong way there. The answer to your question is much simpler now you've clarified. It is still not very clear why you would want to do a test like this but here's how you do it:
this link might help you:
https://www.universalclass.com/articles/math/statistics/calculate-probabilities-normally-distributed-data.htm
What you might ...
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Your question is not completely clear and I can't comment because I don't have enough reputation but I'll do some assumptions here please correct if I'm wrong.
So your data (time series) looks like this:
Each timepoint is an hour with a timespan on one month.
For each timepoint you have a number of sad tweets and a number of other tweets.
So your data ...
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An interaction means that the effect of one variable depends on the level of the other. You can examine whether the difference between live_alone = YES and live_alone = NO differs between those with job=$j_1$ and those with job=$j_2$. If there is a difference between these differences, then you have an interaction. Of course, you need to statistically test ...
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Let me start pointwise
McNemar's test can't be used in this case. Had this been 2X2 and not 3X2, even then McNemar's test would have been wrong because it works on paired nominal data
Wilcox test is also referred for paired
One example of paired is pre-post i.e., earlier sample was not exposed to treatment (pre) and then sample was exposed to treatment (...
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Comparing two samples of such different sizes, when one sample is large, is actually quite similar to doing a one-sample comparison to a known distribution. To see this, imagine the more extreme case where you are comparing a sample of size $34$ to a sample of size $\infty$. In the latter case, the infinite sample is equivalent to knowledge of the ...
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You very likely have samples from the posterior. Each of these distributions likely belong to random variables of a model.
If you wish to know $P(\text{orange}>\text{purple})$ you can compare the samples directly. Here is an example
#Example
orange = rnorm(1000, 0, 1)
purple = rnorm(1000, 2, 1)
prob = mean(orange>purple)
This yields a ...
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