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RE: MDA vs LLD



This is something that was sent on the list last year. Hope it helps

Chris A. Marthaller RRPT
Phone (505) 234-8661
Sr. Training Coordinator - WIPP 
ChrisM@wipp.carlsbad.nm.us
Obviously, only my own views


-----Original Message-----
From: Rick Schoenig [mailto:raschoen@SoCA.com]
Sent: Monday, March 09, 1998 9:55 PM
To: Multiple recipients of list
Subject: Re: MDA vs LLD


ANTHONY F. ARMAGNO wrote:
 
> HELP!
     I am trying to revise a lesson plan to teach HPs about counting
statistics.
 I keep getting conflicting definitions for LLD and MDA.  The references
seem
 to agree on MDCR, when it is mentioned at all.  Someone, please shed
some
 light.
 
> Anthony F. (Tony) Armagno
> Northeast Utilities Millstone Station

> According to Cowboy Wisdom: There's more ways to skin a cat than stickin'
his
> head in a boot jack and jerkin' on his tail!

Tony,

Hope this is not a waste of bandwidth...but, this thread comes up every
year or so.  I'm also sorry, especially after a 12 hour day, that the
formulas have to be in ASCII format.

The following may help in understanding some of the published
calculations you may have seen concerning MDA and LLD determinations. 
Don't worry too much about the terms that I've used...it seems everyone
has their favorites...it's the formulas, and the assumptions
(qualifications) used in their derivation, that are important.  

The following is a brief simplification of most of the "quick and dirty"
formulas that are often quoted as gospel without qualifications.  To be
as inclusive as possible, I've included separate formulas to be used
when dealing with the standard deviations of total counts, and for use
when dealing in count rates and count times. Be careful, this is where a
lot of confusion normally comes into play.  

Note that the terms sd and Rd in the formulas are equivalent to the LLD
activity that you are looking for and can best determined only through
iteration.  This is why LLD is normally expressed as an approximation.
It is normally assumed that the LLD value is close enough to the
background level (by definition you're looking for the "lowest level
above background") that background can be used, negating the need for
iteration.  The k2 squared factor (or 2.71 at the 95% confidence
level),that you may see incorporated in LLD formulas, algebraically
compensates for most of the error this assumption introduces.  Also note
that all of these formulas are reasonably accurate only if the counting
times are sufficient for the total counts (both background and blank
sample) to be above ~30.  Finally note that the LLD value is valid only
if all samples above the MDA value are reported as being "positive" for
activity. 

Gosh...This isn't as simple as I thought it was going to be! 


TERMS USED:

MDA (Determistic or Decision Level, Ld) = Sample contains activity above
the false positive probability level.

LLD (Critical Level, Lc) = Pre-determined sample activity that could be
detected with both a false positive and a false negative probability
level.

k1 = Type I error (false positive) probability factor

k2 = Type II error (false negative) probability factor


FORMULAS FOR LLD AND MDA USING STANDARD DEVIATIONS, where;

sb = standard deviation of the background count,

sd = standard deviation of a blank sample count, and

sc = standard deviation of a sample count containing the "critical
level"


MDA = k1 x (sb + sd)
LLD = (k1 x (sb + sd)) + (k2 x (sb + sc))


If it is assumed that sd ~ sc, then;

MDA = k1 x (sb + sd)
LLD ~ (k1 + k2) x (sb + sd) + (k2 squared)

If k1 and k2 are both assumed to be at the 95% confidence level (1.645),
then;

MDA = 1.645 x (sb + sd)
LLD ~ 3.29 x (sb + sd) + 2.71


If it is assumed that sb and sc are equal (the background and sample
count times are equal), then;

MDA = 2.33 x sb
LLD ~ 4.66 x sb + 2.71

__________________________________________


GENERAL FORMULAS FOR LLD AND MDA USING COUNT RATES AND COUNT TIMES,
where;

Rb = background count rate,

Rd = LLD (Critical Level) count rate,

Tb = background count time, and

Ts = sample count time


MDA = k1 x sqrt (Rb/Tb + Rb/Ts)
LLD = (k1 x sqrt (Rb/Tb + Rb/Ts)) + (k2 x sqrt (Rb/Tb + Rd/Rs))


If it is assumed that Rd ~ Rb, then;

MDA = k1 x sqrt (Rb/Tb + Rb/Ts)
LLD ~ (k1 + k2) x sqrt (Rb/Tb + Rb/Ts) + (k2 squared)/Ts


If k1 and k2 are both assumed to be at the 95% confidence level (1.645),
then;

MDA = 1.645 x sqrt (Rb/Tb + Rb/Ts)
LLD ~ 3.29 x sqrt (Rb/Tb + Rb/Ts) + 2.71/Ts


If it is assumed that Tb and Ts are equal then;

MDA = 2.33 x sqrt (Rb/Tb)
LLD ~ 4.66 x sqrt (Rb/Tb) + 2.71/Ts


E-mail me privately if you have any specific comments or questions.   By
the way...my little kitty cat has waited paitently to be fed while I
prepared this for you.

-- 
Rick Schoenig .... raschoen@SoCA.com
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