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I still would like you to try to prove it, but I’ll give some intuition about those two identities I gave. Remember that variance (and standard deviation) have something to do with how spread out your data are. $var(X+a)=var(X)$ means that if you take data with some amount of spread and slide them up or down the number line, you do not change the spread. $... 3 You can't - you can have different samples with different means, yet having the same variance. Variance is not a function of the mean. 2 Your calculation of the test statistic is wrong. You're dividing the difference in means by the pooled standard deviation, rather than the standard error of the difference in means. The formula for the (equal-variance) two sample t-test statistic would be: $$\frac {{\bar {x}}_{1}-{\bar {x}}_{2}}{s_{p}\cdot {\sqrt {{\frac {1}{n_{1}}}+{\frac {1}{n_{2}}}}}}$$ ... 2 The term central tendency appears to have originated in psychology. It is so vague as to be nearly meaningless. It conveys very little about what it intended except in such sufficiently nice cases where almost any measure of it would agree with almost any other. It's a term that I think serves to obfuscate more than it enlightens and I've yet to see a book ... 2 For one, distributions more readily have a finite mean than a median or mode. Primary school analyses of these concepts for samples can obscure this issue with random variables. For the median to exist, you either require some value to have a CDF of exactly$\frac12$(which is far from guaranteed for discrete distributions), or define how we fudge a median ... 2 I believe the answer to this question can be expressed in more elemental terms. Beginning with the formula for the sample standard deviation $$s_{x} = \sqrt{\frac{ \sum_{i=1}^{n}{(x_i - \overline{x})^2} }{n - 1}},$$ I want to draw your attention to the expression$(x_i - \overline{x})$in the numerator. Ignore the summation notation for the purposes of ... 2 To supplement @Dave's answer, take a look at the following histograms which have the same x-axis. Data2 is just a shifted version of Data1 and therefore the standard deviation, which is a measure of spread, shouldn't change. R code to generate histograms: x <- rgamma(1000, 6, 2) #Simulate some data hist(x, xlim=c(0,100)) hist(100-x, xlim=c(0,100)) 2 You can resort to the bootstrap version of Student's t-test. It works as follows: Compute the sample mean and standard deviation for each group and label the results$X_1$and$s_1$for group 1, and$X_2$and$s_2$for group 2. Set$d_1=\frac{s_1^2}{n_1}$and$d_2=\frac{s_2^2}{n_2}$, where$n_1$and$n_2\$ are the sample sizes. Generate a bootstrap sample ...

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Utilizing your reproducible example, I have worked up what appears to be the idea behind VIM using nls. library(matrixStats) set.seed(101) df <- data.frame( "ID" = paste("ID", seq(1,100,1), sep = ""), "x1" = sample(90:220, size = 100, replace = T), "x2" = sample(90:220, size = 100, replace = T), "x3"...

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I believe it is because standard deviation is the root of variance, variance is the second moment of a random variable, and the second moment is minimized when taken about the first moment (i.e. the mean) of the random variable. As mentioned in the comments, there are dispersion metrics that are based on medians etc., but SD is necessarily based about the ...

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Whether the arithmetic mean is a good measure for a strongly skewed distribution depends on what you are using it for -- what does "good" mean? One variable that is usually very right skew is income and, usually, you see median income rather than mean income. But why? Because, usually, when we look at some measure of central tendency, we are looking for ...

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For that many subjects and that range of scores, I'd compare the raw scores, most probably using the paired t-test. Converting to percentiles, if the transformation is linear, would lead to equally valid results, but not that obvious to interpret. Categorizing the scores is, in addition, somewhat arbitrary and involves loss of information. I'd avoid it if ...

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