On 2010-12-09 10:10:26 +0000, John Hall said:

In article 2010120908504475249-taharley@dundeeacuk,
Trevor Harley writes:
On 2010-12-08 20:10:13 +0000, Teignmouth said:

John,
If I use the SKEW function I get the following values:
Period Jan Feb Mar Apr May Jun Jul Aug
Sep Oct Nov Dec
1659-2009 (0.6) (0.5) (0.1) (0.1) 0.1 0.2 0.4
0.3 0.0 (0.1) 0.0
(0.2)
1971-2009 (0.8) (0.8) (0.2) 0.2 (0.1) (0.2) 0.8
0.4 (0.1) 0.0
(0.1) (0.8)
Now what do I do with the value and the Standard Deviation?
If I have a December mean for the period 1659-2009 of +4.1c, and the
STDEV is 1.7c, and the SKEW is -0.2c, do I get a revised STDEV of 1.5c
or 1.9c? Then do I use +4.1c and the revised STDEV x1 x2 x3 etc to
get the revised Standard Deviation thresholds?
Your help is much appreciated.
Thanks

No, just use the standard deviation. The skew is tiny. It's as good a
normal distribution as you'll ever get.

Surely a skew of -0.8, as for Decembers over the last 40 years, is
enough that it ought to be taken into account? It's interesting that the
winter month skew values have been larger over the last 40 years have
been larger that when one considers the whole CET record.


I did the descriptive stats for December since 1659 in SPSS.

1659-2009
Mean 4.08
Median 4.10
sd 1.72
Skewness -0.25
SE of skewness 0.13
Skewness Z = -1.92
Kurtosis -0.92
SE kurtosis .26

So there's no kurtosis (bumpiness) and the z-score for skewness isn't
(quite) significant either. (A negative skew means too long tail on the
left side.) But my understanding is that with large samples (like this)
you shouldn't worry too much about significant skew becaus the standard
error of the skew tends to be small. So when you've got a large sample
(over 200) you need a z score giving a p of at least 0.01 before you
need to start to worry.

So as you would expect the complete series is normally distributed.

From 1971-2009,

Mean = 4.97
Median = 5.3
SD = 1.5

Skew = -0.843
SE skew = 0.378
Z skewness = 2.23

kurtsosis = 1.52
se kurtosis = 0.741
Z kurtosis = 2.05

Both of which are significant 0.05p0.1.

Having said that, I wouldn't worry about it too much. If you're really
concerned you could transform the data, but the ones I'm familiar with
only work for positive skew. I've been told you can add a constant to
the mean and then apply a logarithmic transformation, but I'd be
surprised if it really makes a difference.

I'm no statistics expert; just a psychologist who uses them.

--

Trevor
Pedagogical in Lundie, near Dundee
http://www.personal.dundee.ac.uk/~taharley/