Drexel dragonThe Math ForumDonate to the Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.stat.math.independent

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Luis A. Afonso

Posts: 4,743
From: LIsbon (Portugal)
Registered: 2/16/05
A way to test normality from Skewness/Kurtosis Coefficients
Posted: Aug 14, 2012 8:39 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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
___k= 0.8_______0.824___0.050___0.099___0.027__
___k= 1.2_______0.851___0.023___0.111___0.015__
___k= 1.6_______0.864___0.011___0.117___0.008__

___S= 0.8
___k= 0.4_______0.839___0.145___0.002___0.014__n=40
___k= 0.8_______0.919___0.066___0.004___0.011__
___k= 1.2_______0.955___0.030___0.007___0.008__
___k= 1.6_______0.970___0.014___0.010___0.006__

___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___


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

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum 1994-2015. All Rights Reserved.