meta user
network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 4 months |
| seen | 7 mins ago | |
| stats | profile views | 37 |
|
May 15 |
comment |
Evaluating the different calculations You could also have two different Mathematica versions installed in your machine. Put MMA9 to run 2-3 days and use, for instance, MMA8 for simpler calculations... |
|
May 15 |
comment |
Evaluating the different calculations What have you done so far? Instead of a simple question, you could put any code to show what exactly you're trying to achieve... |
|
May 13 |
revised |
Computing Correlations and p-values added 230 characters in body |
|
May 13 |
comment |
Computing Correlations and p-values OK, let me correct myself: the sampling distribution for Pearson's correlation does assume Normality, while the measure itself does not assume Normality. That's why on should use Spearman correlation instead of Pearson. |
|
May 13 |
comment |
Computing Correlations and p-values These assumptions must also be true for the Pearson correlation test... |
|
May 13 |
revised |
Computing Correlations and p-values deleted 2 characters in body |
|
May 13 |
answered | Computing Correlations and p-values |
|
May 13 |
comment |
Computing Correlations and p-values Or even SpearmanRankTest[], why not? |
|
May 13 |
comment |
Safe values of $\mu$ and $\sigma$ when randomly sampling from a Log-Normal Distribution? What definition of "outlier" are you using? A simple "visual test" usually does not work here. Try to generate your data, compute the IQR (interquartile range) and then you can effectively say if the observation is an outler... |
|
May 13 |
comment |
Computing Correlations and p-values Pearson's correlation does assume Normality, while Spearman's correlation is a rank based correlation measure and does not assume Normality. |
|
May 13 |
comment |
Computing Correlations and p-values Take a look at this post. It might help. |
|
May 3 |
awarded | Enthusiast |
|
May 2 |
awarded | Citizen Patrol |
|
May 1 |
comment |
Override Equation structure change in mathematica - CopyToLaTeX Try this link. |
|
Apr 30 |
comment |
Probability density histogram with unequal bin widths When you add 10^-13 to one block of the histogram, you are, at the same time, "removing" slightly the probability of the other blocks... That's why you get the difference... Anyway I think, too, that the difference is considerable... |
|
Apr 30 |
revised |
Probability density histogram with unequal bin widths added 14 characters in body |
|
Apr 30 |
answered | Probability density histogram with unequal bin widths |
|
Apr 30 |
answered | Probability density histogram with unequal bin widths |
|
Apr 30 |
comment |
Probability density histogram with unequal bin widths It does matter when you're computing PDF's. If you exclude those 3 observations, now your probability (i.e., "PDF") should be based on 12 observations, and not 15... |
|
Apr 30 |
comment |
Probability density histogram with unequal bin widths Using SeedRandom[1] you get 3 observations higher than 1. When you use Histogram[data,{{-2,0,1}}] you're excluding those 3 observations... |
