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Events (or random variables) are independent when information on some of them tells you nothing about the probability of occurrence (/ distribution) of the others. Please DO NOT use this tag for independent variable use [predictor] instead.

2
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fair, or it must revert to tails soon. Indeed, this intuition has been formalized by statisticians into a test for randomness / independence (i.e., the runs test). One thing to realize is that with … least under independence; with dependent data, the prior result must be a part of, or stored somehow within, the data generating process). Likewise, $Pr(HHHHH)=.03125$, and $Pr(HHHHH|HHHH)=.03125 …
answered Dec 2 '12 by gung
1
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I don't know of a test for independence for ANOVAs that is analogous to, say, the Shapiro-Wilk test of normality or Levene's test of homogeneity of variance. Nor am I aware of a standard plot to … assess this. I think they aren't in textbooks because I don't think they exist. The typical case of non-independence in ANOVA would be something like having the same (or matched) study units used in …
answered Jul 26 '15 by gung
4
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+1 for a clearly laid-out question. Regarding terminology, people use terms in different ways--unfortunately, terms are not fully standardized. However, multivariate regression usually means regress …
answered Oct 28 '12 by gung
1
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If you only want to know if the proportion .2128 (40 / 188) differs from the value .154 (taken as set in stone), this is a simple binomial test. Here's an example in R: binom.test(x=40, n=188, p=. …
answered Dec 5 '14 by gung
2
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You don't want to use the chi-squared test to analyze this situation. It does not correspond to the question you want to answer. Instead, you should use logistic regression. To learn more about the …
answered Mar 4 '14 by gung
0
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Your setup doesn't quite match the idea of independence. Instead, you want to know if the estimated proportions are the same. You could rearrange your data to make that work with a chi-squared test …
answered Apr 12 '17 by gung
2
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The t-test is a special case of regression, so I'll discuss this in that context. The assumption of independence isn't that it's not possible to predict the observations, but that you can't predict …
answered Oct 28 '12 by gung
3
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nail down: If two variables are not independent, there are potentially infinite ways they could be dependent. For the sake of simplicity, let's specify the type of non-independence as being linearly … (given means, variances and degree of correlation)? Likewise, you could test for (this type of) non-independence with a simple product-moment correlation test. Of course, if you were interested in …
answered Sep 15 '13 by gung
0
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I don't really see any problem with this. In general, you could use a t-test here. The t-test does assume that the prices are normally distributed and the the standard deviations are the same for th …
answered Jul 31 '17 by gung
4
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When errors are dependent, for example with multi-level models, the matrix solution is: $\beta=(X'\Sigma^{-1}X)^{-1}X'\Sigma^{-1}Y$ Where $\Sigma$ is the covariance matrix. For more information, se …
answered Jan 7 '12 by gung
1
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Your data won't be independent. That means your confidence intervals will appear narrower than the correct width and your p-values will be incorrect. If the radient power is conditionally normal, yo …
answered Jan 31 '16 by gung
2
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combination, although it's actually more complicated than that. Then you would simply run a standard chi-squared test for independence. For the record, I don't see any value in pursuing this course. …
answered Mar 11 '12 by gung
5
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Instead of asking why the chi-squared test assesses independence, think about how you would test the association of two categorical variables against the null hypothesis of independence. For the … to test these variables for independence; what do we mean by "independence"? We aren't interested in testing if any of the proportions are any particular value. We want to know if being in a …
answered Feb 14 '16 by gung
2
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This is mostly just to support and explicate the points @Glen_b has made. The fact that variables are highly correlated does not necessarily mean they are redundant and that you want to throw one o …
answered Jul 17 '14 by gung
9
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explicit questions: Your plot shows serial autocorrelations / non-independence of your residuals. It means that your model is not appropriate in its current form. …
answered Feb 20 '15 by gung

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