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4

With some random data fhat = RandomReal[{-1, 1}, {1000, 4}]; One can get a BoxWhiskerChart with the value of the mean placed below each box and grid lines with BoxWhiskerChart[Transpose[fhat], "Mean", BarOrigin -> Left, LabelingFunction -> (Placed[Mean[#], Below] &), GridLines -> Automatic] Or with specified vertical grid lines, no ...


3

First, let's find a in functional form: Probability[Abs[x] < a, x \[Distributed] NormalDistribution[]] which shows that f[a_] := Erf[a/Sqrt[2]] Now you want to solve for the a that has some probability, say 0.1. Then Solve[f[a] == 0.1, a] {{a -> 0.125661}} A little more generally: Solve[f[a] == p, a] {{a -> Sqrt[2] InverseErf[p]}} ...


3

There's an example exactly like this in the documentation for BoxWhiskerChart. All you need to do is reshape your data into a list of doubles: xdata = {{"A", {1, 2, 5}}, {"B", {5, 7, 2, 2, 5}}, {"C", {3, 2, 5, 7}}}; ydata = {{"A", {7, 2}}, {"B", {7, 2, 5}}, {"C", {6, 7, 3}}}; labels = {xdata[[All, 1]], None} xdata = xdata[[All, 2]]; ydata = ydata[[All, ...


2

As this is simple enough, you could use definitions : dist = MultinormalDistribution[{0, 0, 0}, IdentityMatrix[3]] ; prob[a_] = Integrate[PDF[dist, {z1, z2, z3}], {z2, -Infinity, Infinity}, {z3, -Infinity, Infinity}, {z1, a Sqrt[z2^2 + z3^2], Infinity}] (* ConditionalExpression[1/4 (2 - ...


1

No error nv 10.0.2. On windows 7, 64 bit SeedRandom[0]; x = RandomReal[{-5, 5}, 100]; y = 2 x + 1 + RandomReal[{-0.1, 0.1}, 100]; X = Transpose[{x, y}]; ListPlot[X] Correlation[X] // MatrixForm CorrelationTest[X, 99995/100000, "PearsonCorrelation"] No error messages.


1

There is an add-on for Mathematica, called MathStatica, which does exactly what you want, if I understood you right. Find an example on www.mathstatica.com/examples and click on the left side on the item "Person Fitting." Ths is an excellent add-on which fits seamless into mma and extends its statistical capabilities. Best regards Volker



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