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Topic: A way to test normality from Skewness/Kurtosis Coefficients
Replies: 2   Last Post: Aug 15, 2012 8:01 PM

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Luis A. Afonso

Posts: 4,617
From: LIsbon (Portugal)
Registered: 2/16/05
A way to test normality from Skewness/Kurtosis Coefficients
Posted: Aug 14, 2012 8:39 PM
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A way to test normality from Skewness and Kurtosis Coefficients


The program ?JACARE? is mainly focused to show how normal data behaves towards the Skewness and Kurtosis (excess) Coefficients, respectively:
__S = m3/m2^3/2 and k = m4/m3^2 ? 3
where mj = (1/n) * Sum [x_i - m1]^j, j=2, 3, 4 , are the order j central moments and m1 the sample x_1, x_2, . . . , x_n mean value.

_____________

At the following Table the symbols [a, b]: a=0, a=1 respectively the S not exceeds, or does exceed the column 1 heads and similarly b for k.

____Data X ~ N(0, 1): n

______________ [0, 0]____[0, 1]___[1, 0]__ [1. 1]___
___S= 0.4
___k= 0.4_______0.763___0.111___0.078___0.048__n=40
_______________0.837___0.117___0.025___0.021__n=100
___k= 0.8_______0.824___0.050___0.099___0.027__
_______________0.918___0.036___0.036___0.010__
___k= 1.2_______0.851___0.023___0.111___0.015__
_______________0.944___0.011___0.041___0.004__
___k= 1.6_______0.864___0.011___0.117___0.008__
_______________0.951___0.004___0.044___0.002__


___S= 0.8
___k= 0.4_______0.839___0.145___0.002___0.014__n=40
_______________0.862___0.137___0.000___0.001__n=100
___k= 0.8_______0.919___0.066___0.004___0.011__
_______________0.954___0.045___0.000___0.001__
___k= 1.2_______0.955___0.030___0.007___0.008__
_______________0.984___0.015___0.000___0.001__
___k= 1.6_______0.970___0.014___0.010___0.006__
_______________0.994___0.005___0.000___0.000__


___S= 1.2
___k= 0.4_______0.841___0.158___0.000___0.001___
___k= 0.8_______0.923___0.076___0.000___0.001___
___k= 1.2_______0.961___0.037___0.000___0.001___
___k= 1.6_______0.980___0.018___0.000___0.001___

___S= 1.6
___k= 0.4_______0.841___0.158___0.000___0.000___
___k= 0.8_______0.923___0.077___0.000___0.000___
___k= 1.2_______0.962___0.038___0.000___0.000___
___k= 1.6_______0.980___0.019___0.000___0.000___



Application

Suppose a 40-size sample you got S= k= 1.6: the probability a normal sample to find such a result is less than 0.0005, so, you conclude that it is very unlike the sample was drawn from a normal Population (or the normal sample is very odd).
However it can be argued that the table above was built based on N(0,1) data, and at least, not the Population mean value of course, but the sigma, could invalidate the conclusion we did reach.
To confirmed the simulation data from N(0,100):40, 40´000 samples:
________S, k = (0.4, 1.6)______[1,1] = 0.008
_____________(0.8, 1.6)__________ = 0.005
_____________(1.2, 1.6)__________ = 0.001
_____________(1.6, 1.6)__________ = 0.000
which that is in accordance with the results from N(0, 1); 0.008, 0.006, 0.001, 0.000 showing that the [1, 1] frequencies do not depend from the Population variance.
Note: As I fully found the Jarque-Bera test, based on the same parameters estimators, has the drawback (in fact not at all supersede here) that one cannot to ascribe a significance level. The present method based on the output [a, b] leave freely to the researcher to make the YES - NOT NORMALITY decision following his/her common sense.



Luis A. Afonso



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