(2014-Nov-02, 19:21:44)Meng Hu Wrote: You know that I always hated significance test. S-W is useless because when N is large, the p-value will always be lower than 0.05, let alone the problem of arbitrary cut off values for significance.

It will only be so if the data is non-normal. Of course, if N is very large (e.g. 10k), then even a very slight deviation from normality will cause p to be lower than .05 or .01. Instead of looking at the p value, you can look at the W value, which is a measure of normality. It is only .93 for these data. Usually, it is close to .99 when the data looks normal.

Try:

Code:

`x = rnorm(5000)`

shapiro.test(x)

>Shapiro-Wilk normality test

data: x

W = 0.9997, p-value = 0.8392

You see that W is very close to 1, and p is high even though N=5000 (the function in R is limited to N=5k for some odd reason).

Quote:I don't understand why you keep using it. I will never approve a paper that uses S-W to examine normality, instead of histogram, P-P and Q-Q plots. As I said, look at histogram, P-P plot and Q-Q plot. That's all you need. If you don't want to display all the graphs but only one, I think you should probably select histogram.

Because it gives a numerical estimate of the normality of the data. Eye-balling cannot do that. So e.g. if one wants to compare 100 samples automatically for which one is the most normal, one will need a test.

Quote:By the way, If my memory is correct, the spatial transferability theory states that immigrant IQ will stay the same, and that the country of origin will predict test performance. Your paper says the result is consistent with ST theory, but it's just a correlational analysis, so it's only about the second prediction of the theory.

If the IQs changed, but stayed in the same relative order, then correlation analysis will not detect it, that's right. ST hypothesis says they will generally stay the same, which also implies the order will stay generally the same, and for this reason the usual correlates of IQ will be found.