I guess, because I'm lazy, I'll just repost exactly what I posted there here.

I don't know that looking at the resulting points from each match played is the best way to analyze the two conferences, but it's easy enough that I'm willing to do it before noon on a Sunday. The West has scored slightly more points on average and points per game than teams in the East. A KS test between the two sets of results based on individual results for each match (i.e. a set of 0's, 1's and 3's) comes back with a P value of 1.000, roughly translated as there is a 0.0% chance that the two data sets are statistically different.

Running a KS test with data sets based on each teams culmulative point total (i.e. 30, 28, 26, 19, etc...) comes back basically the same. P value of 0.975, meaning roughly only a 2.5% chance that the West is statistically better than the East.

Here's some numbers, if that's your thing.

**East Stats**Total Points 183

Average Points 18.30

Average Points Per Game 1.39

Standard Deviation 1.37

Games Played 132

__West Stats__Total Points 179

Average Points 19.89

Average Points Per Game 1.42

Standard Deviation 1.36

Games Played 126

**KS Test: Results**

Kolmogorov-Smirnov Comparison of Two Data Sets

Kolmogorov-Smirnov Comparison of Two Data Sets

The results of a Kolmogorov-Smirnov test performed at 10:17 on 17-JUN-2012

**The maximum difference between the cumulative distributions, D, is: 0.0278 with a corresponding P of: 1.000**--------------------------------------------------------------------------------

**Data Set 1:**132 data points were entered

Mean = 1.386

95% confidence interval for actual Mean: 1.150 thru 1.623

Standard Deviation = 1.37

High = 3.00 Low = 0.00

Third Quartile = 3.00 First Quartile = 0.00

Median = 1.000

Average Absolute Deviation from Median = 1.22

KS says it's unlikely this data is normally distributed: P= 0.00 where the normal distribution has mean= 1.524 and sdev= 0.9963

Items in Data Set 1:

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00

**Data Set 2:**

126 data points were entered

Mean = 1.421

95% confidence interval for actual Mean: 1.181 thru 1.660

Standard Deviation = 1.36

High = 3.00 Low = 0.00

Third Quartile = 3.00 First Quartile = 0.00

Median = 1.000

Average Absolute Deviation from Median = 1.20

KS says it's unlikely this data is normally distributed: P= 0.00 where the normal distribution has mean= 1.535 and sdev= 0.9958

Items in Data Set 2:

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00 3.00

--------------------------------------------------------------------------------

**KS Test: Results**

Kolmogorov-Smirnov Comparison of Two Data Sets

Kolmogorov-Smirnov Comparison of Two Data Sets

The results of a Kolmogorov-Smirnov test performed at 10:24 on 17-JUN-2012

*The maximum difference between the cumulative distributions, D, is: 0.2000 with a corresponding P of: 0.975*--------------------------------------------------------------------------------

**Data Set 1:**10 data points were entered

Mean = 18.30

95% confidence interval for actual Mean: 12.25 thru 24.35

Standard Deviation = 8.46

High = 30.0 Low = 3.00

Third Quartile = 26.5 First Quartile = 13.2

Median = 18.50

Average Absolute Deviation from Median = 6.10

KS finds the data is consistent with a normal distribution: P= 0.84 where the normal distribution has mean= 18.10 and sdev= 10.69

KS finds the data is consistent with a log normal distribution: P= 0.28 where the log normal distribution has geometric mean= 15.05 and multiplicative sdev= 2.834

Items in Data Set 1:

3.00 8.00 15.0 17.0 18.0 19.0 19.0 26.0 28.0 30.0

**Data Set 2:**10 data points were entered

Mean = 19.89

95% confidence interval for actual Mean: 14.89 thru 24.89

Standard Deviation = 6.98

High = 32.0 Low = 11.0

Third Quartile = 25.5 First Quartile = 13.0

Median = 19.45

Average Absolute Deviation from Median = 5.69

KS finds the data is consistent with a normal distribution: P= 0.98 where the normal distribution has mean= 20.24 and sdev= 8.109

KS finds the data is consistent with a log normal distribution: P= 1.00 where the log normal distribution has geometric mean= 18.78 and multiplicative sdev= 1.528

Items in Data Set 2:

11.0 13.0 13.0 15.0 19.0 19.9 24.0 25.0 27.0 32.0

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