We all learn from L1 that Positive skewness: mode < median < mean; Negative skewness: mode > median > mean. But in the following example, mode < median < mean while its skewness < 0. Or more specifically, mode(73) < median(74.5) < mean(74.63333) and skewness(-0.3495061)<0! Why?? Can anyone tell what’s going wrong? I shall have posted it in L1 forum, but I’m L2 candidate.
“Source: Little SAS Book 4th Edition, Section 8.7”. ============================================= The UNIVARIATE Procedure Variable: Score Moments N 30 Sum Weights 30 Mean 74.6333333 Sum Observations 2239 Std Deviation 12.5848385 Variance 158.378161 Skewness -0.3495061 Kurtosis 0.10385765 Uncorrected SS 171697 Corrected SS 4592.96667 Coeff Variation 16.8622222 Std Error Mean 2.29766665 Basic Statistical Measures Location Variability Mean 74.63333 Std Deviation 12.58484 Median 74.50000 Variance 158.37816 Mode 73.00000 Range 56.00000 =========================================== /*data file*/ Scores.dat 56 78 84 73 90 44 76 87 92 75 85 67 90 84 74 64 73 78 69 56 87 73 100 54 81 78 69 64 73 65 /*SAS code*/ DATA class; INFILE ‘c:\MyRawData\Scores.dat’; INPUT Score @@; RUN; PROC UNIVARIATE DATA = class; VAR Score; TITLE; RUN;
What kind of dataset are you using? If it is not normally (or at least close to normally) distributed then that relationship need not hold as far as I know.
Just googled this. I stopped reading pretty much after the first section, but for anyone who’s interested here’s more detail: http://www.amstat.org/publications/jse/v13n2/vonhippel.html
I got positively screwed in the exams
Cool, Carson. It’s “a rule of thumb”, but it fails with extremely high frequency. The SAS scores example is only one of them…