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Hi Pal --
There is one interpretation of the Z-Score that takes an observation from a
population and returns a Z-Score statistic, where the Z-Score is a
measurement of the number of standard deviations that that specific
observation deviates from the mean. If this is the interpretation you
intend, the following afl code returns the Z-Score of the Close of the most
recent 50 days of an end-of-day price series and plots it. Copy this code
and paste it into Indicator Builder.
//-------------------------
// ZScore of Close
//
ZLen = 50;
ZScore = (C-MA(C,ZLen))/StDev(C,ZLen);
Plot(C,"C",colorBlack,style=styleCandle);
Plot(ZScore,"ZScore",colorBlue,styleOwnScale|styleNoLabel,-3,3);
Plot(0,"",colorRed,styleOwnScale|styleNoLabel,-3,3);
Plot(-2.0,"", colorRed,styleOwnScale|styleNoLabel,-3,3 );
Plot(2.0,"", colorRed,styleOwnScale|styleNoLabel,-3,3);
//-------------------------
Note that most of the Closes (95 percent, on average) will have ZScore
values between -2.0 and +2.0.
My apologies if I misinterpreted your request.
Howard
/////-----------------------------
Date: Fri, 19 Sep 2003 19:01:08 -0000
From: "palsanand" <palsanand@xxxxxxxxx>
Subject: Z-Score
Hi,
I was looking for an algorithm that will tell me the z-score of a
given percentile. In other words, I was looking for an algorithm
that will replicate a reverse lookup in a normal distribution table.
Abramowitz & Stegun is my friend ("Handbook of Mathematical
Functions"). Formula 26.2.22 gives an approximation in z accurate to
within 3e-3, and formula 26.2.23 gives a more complicated but more
accurate formula.
26.2.22 says this (in pseudo-Fortran). p appears to be the
*upper*-tail probability (ie. p=P(Z>z)) and has to be less than 0.5
for this to work. (If not, call the function with argument 1-p and
put
a minus sign on the result.)
t=sqrt(ln(1/p**2)) [ p squared ]
a0=2.30753
a1=0.27061
b1=0.99229
b2=0.04481
z=t-(a0+a1*t)/(1-t*(b1+b2*t))
The problem is to implement this efficiently in AB. Ay ideas?
Thanks,
Pal
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