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



Luis A. Afonso
Posts:
4,758
From:
LIsbon (Portugal)
Registered:
2/16/05


A way to test normality from Skewness/Kurtosis Coefficients
Posted:
Aug 14, 2012 8:39 PM


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 40size 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 JarqueBera 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



