# Tag Info

15

If you check the properties and relations sections of the examples for both CorrelationTest and SpearmanRankTest you will see that these are not identical tests. In the first case we are checking whether the correlation coefficient is a particular value and the p-value is computed using a Fisher Z transformation. This transformation allows us to test ...

3

Pearson's correlation does assume Normality, while Spearman's correlation is a rank based correlation measure and does not assume Normality. So, instead of using PearsonCorrelationTest[A, B, "TestDataTable"] you should use SpearmanRankTest[A, B, "TestDataTable"] EDITED Pearson's correlation does not assume Normality, however the sampling distribution ...

3

Using SeedRandom[1] you get 3 observations higher than 1. When you use Histogram[data,{{-2,0,1}}] you're excluding those 3 observations... If you exclude those 3 observations, now your probability (i.e., "PDF") should be based on 12 observations, and not 15...

2

The command ARProcess requires that the input term $e_t$ be a zero mean white noise with a specified variance. What you are trying to do is to have your input term be $10+e_t$ which is clearly not zero mean. What you can do is to change variables. In your case, if you define a new process $z_t=y_t - 25$ then it will be the same as the $y_t$ process. To see ...

2

I would say there is something wrong with your partitioning. I think the second argument of StandardDeviationFilter is the number of number of elements to take both to the right and to the left -- i.e., if the second argument is 1, the standard deviation will be computed with 3 elements. Reducing the data set and simplifying sd1 and sd2 tends to confirm my ...

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