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0

There's nothing exactly "wrong" with this, you're showing the estimated mean absolute error and that the estimated standard deviation of the absolute errors is larger than the estimated mean absolute error. But, you might've been intending to construct a confidence interval for the MAE instead. A typical Stat 100 approach would be to construct a 95%...


0

In adition to Hassan's response, you need to be careful on interpreting standard deviation. Some people define it as the mean distance between every observation and its mean, but this is the definition of mean absolute deviation (MAD), thus wrong. For a better understanding of both concepts, variance and SD, I highly recommend Taleb's video series on ...


3

While the estimate of the variance is unbiased, the estimate of the standard deviation is not. An unbiased estimate $V_{est}$ is one such that the expected value of the estimate is the true value $V$ being estimated: $E[V_{est}] = V$. But applying a nonlinear operator like the square or square root destroys this property. In general, $E[\sqrt{V_{est}}] \neq \...


10

This is what's going on > summary(row_std) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.05212 0.69485 0.91762 0.94109 1.15915 2.56883 > summary(row_std^2) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.002717 0.482820 0.842034 1.001564 1.343628 6.598870 Because $s^2$ is unbiased for $\sigma^2$, it is not possible for $s$ to be ...


8

By Jensen’s inequality, the sample standard deviation is an underestimate (in expectation) of the true standard deviation, since the square root is concave and $S^2$ is unbiased for the second central moment.


1

The t statistic divides the estimate by the standard error, which dilates the result from the [-1,1] range to the [-$\infty$, $\infty$] range. You can prove the t distribution result mathematically, but I suspect a simple simulation will be as convincing and perhaps more illustrative: set.seed(12345) n = 100 b0 = 2 b1 = 0.5 x = rnorm(n) y = b0 + b1*x + rnorm(...


-2

First let me be humble and state that I am not a statistician but I think that the misunderstand may stem from the semantics used The population mean is well understood as it is simply the arithmetic average of the population values The sample mean is the arithmetic average of the means of each sample Thus it is more illustrative say the mean of samples as ...


1

This type of coefficient of variation is generally only useful if the mean is not 'arbitrary', that is, the data means something very different if the means is artificially raised. An example would be count data, but there are many practical situations where the coefficient of variation could be sensible with continuous data as well. For circular data, ...


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