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Interpreting the results of Chi-Square test
We statisticians are used to make the strongest assertions with even more little samples. Related variables are not uncommon at all, they are the norm instead, so why you bother so much? If your sample is representative of your population of interest, even if it is little, it provides a strong evidence for those associations to exist. Those two sets of p-values are each very little, so, even if they are not precise, evidence is still very strong.
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Interpreting the results of Chi-Square test
@whuber is it correct to call "fisher exact test" a monte carlo version of chi square test?
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Interpreting the results of Chi-Square test
if I understand this correctly, that is a sample of your data. it's impossible to get p-values that low with only five observations. how many do you have?
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What are the branches of statistics?
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Possible reasons for AUC=1 (from a fitted glm model)?
As I just pointed out in the comment above, the number of predictors you have is your biggest problem. Anyway, binomial quasi-likelihood (called quasibinomial in R) only makes sense if you have grouped data, while in the sample you posted I only see individual observations. So, no, quasi-likelihood here is pointless
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Possible reasons for AUC=1 (from a fitted glm model)?
so you have how many observations? and 500.000 possible predictors? if your sample is little enaugh even completely random predictors could very well have a perfect association with your response variable, if you choose it from such a large poll
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Possible reasons for AUC=1 (from a fitted glm model)?
I see that you used LASSO to select variables, ok. what I was suggesting is to use LASSO to estimate the final model, directly or after a first selection, in general there is no preferred behaviour, even if you can of course try to see what gets you better out-of-fold performance.
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Showing t-distribution from multivariate standard normals
$a'X$ and $\sqrt{X'X}$ are not independent. $\hat \sigma^2$ is independent on $\sqrt{X'X}$, but total sum of squares isn't.
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Understanding formula for comparing players based on certain metrics
so you need a similarity measure to compare players, and you don't need the exact one used by that tool. You can compute your own similarity measure if you want, choosing among the most popular ones. Check here: en.wikipedia.org/wiki/Similarity_measure
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Statistically significant difference between distributions only knowing the mean and std dev?
I don't think this answer is the best one. ANOVA, even if it is weighted, estimates variances from data points, ignoring the information in s.d. column. using it to weight observations surlely improves the trustworthiness of group means estimates, but variance estimates are not improved as they could.
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R: two way ANOVA and ANCOVA functions return same result: is the code wrong?
consider ANCOVA is a generalization of ANOVA, if your model is simple, they are the same thing
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ANCOVA baseline as covariate
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