# Tag Info

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When most people think of a non-parametric equivalent of ANOVA, they think of the Kruskal-Wallis test. The Kruskal-Wallis test cannot be applied to a factorial structure, however. The first workaround to this is to run all of your conditions as a one-way analysis. This does not let you test your factors individually, but you may be able to get what you ...

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In the normal distribution the median = the mean. If you have reason to suspect that your data is not normally distributed then you should use a non-parametric test to compare them (i.e., the Mann-Whitney U-test).

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measurements that are close to each other will tend to be more correlated than others If you have some kind of adjacency you could try looking for "clumps" (groups of adjacent interesting results) then comparing to what you get by either permuting the locations or group membership. It wouldn't necessarily have to be physical adjacency. Rather than ...

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Yes it is possible to use the Kaplan method to estimate left-censored data. The Wiki article is actually pretty decent Check it out

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I think your permutation idea is solid. Assume there are only two measurements y1 and y2. Here's how I'd do it in R: set.seed(1245) group <- c(rep(0, 50), rep(1, 50)) z <- rnorm(100) # Direct common cause of y1 and y2 y1 <- .7 * group + 1.5 * z + rnorm(100) # explanitory variable y2 <- 0 * group + 1.5 * z + rnorm(100) t1.stat <- ...

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Like Ziggystar said. The answer is yes. If you know the CDF (i.e., $F$) of $X$, then $E[I(x)] = F(x+h)-F(x-h)$, which I'm sure you know.

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If you know the range of your data, you can use the inverse probit transformation. On a couple of examples, the fit looked very satisfying visually. This approach is explained in more detail in a clear paper[1]. I think there should be an R implementation but I couldn't find it (perhaps you can contact the author). The approach can also be adapted to ...

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I found a very easy to understand guide for constructing Mann-Whitney confidence bounds here: https://onlinecourses.science.psu.edu/stat464/node/39 The article gives very good instructions, but references a table in a textbook (which luckily my uni has). Once I look at the table I will try to find an equivalent one online (unless anyone knows of one ...

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Yes, if your alternative is a shift alternative, the corresponding estimate is the Hodges-Lehmann. The estimate is easy enough. The Hodges Lehmann estimate is the sample median of the cross-sample pairwise differences. For really large samples, there are efficient methods (for example, ones that sort the two samples and then use the information in that ...

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The first issue that appears to me is that of multiplicity. If our objective is to consider each of the 80 places on the body where these individuals are measured and test whether there's a difference between diseased and healthy individuals, then having a 0.05 level test at each of the 80 places will result in an average of 4 type I errors. That's no good. ...

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Broadly, there are two possible approaches to your problem: one which is well-justified from a theoretical perspective, but potentially impossible to implement in practice, while the other is more heuristic. The theoretically optimal approach (which you probably won't actually be able to use, unfortunately) is to calculate a regression by reverting to ...

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Rather than relying on a test for normality of the residuals try assessing the normality with rational judgment. Normality tests do not tell you that your data is normal, only that it's not. They also get very sensitive at large N's and you may have approximately normal residuals. Your N is in that range where sensitivity starts getting high. If you run the ...

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