primer on errors 8.13 u.becker 9,2007 random & systematic errors distribution of random errors...
Post on 20-Dec-2015
234 views
TRANSCRIPT
PRIMER ON ERRORS8.13 U.Becker 9,2007
11 1Random & Systematic Errors11 1Distribution of random errors11 1Binomial, Poisson, Gaussian11 1Poisson <-> Gaussian relation11 1Propagation of Errors11 1Fits - Read Bevington ch. 1-4
• No measurement has infinite accuracy or precision”• K.F.Gauss: • “Experimental physics numbers must have errors and dimension.”
• G = (6.67310 0.00010)[m3 kg-1 s-2]• • Errors = deviations from Truth (unknown, but approached)
Large error: Result is insignificant. Small error: Good measurement, it will test theories.
Blunders : no,no, – repeat on Fridays correctly. Random Errors or “statistical” - independent, repeated
measurements give (slightly) different resultsSystematic Errors “in the system”, DVM 2% low,…
correct if you can, otherwise quote separate
Systematic error: inherent to the system or environment
Example:
Measure g with a pendulum: neglect m of thread)equipment (accuracy of scale, watch)environment (wind, big mass in basement)thermal expansion t 1.correct by l=l0(1+t) if t known or
2.give sys.error t from est. range of t.
large M below earth
limits ACCURACY (=closeness to truth)
€
T =2πl
g
Random (statistic) error
example: measure period T of the pendulum several times:
measurements of T jitter around “truth”.T improves with N measurements, if they are independent.
Statistics N:
limits PRECISION , but lim(N->∞)= truth
# Distribution
T [sec]
Evaluate: g = (l l) {2 /(TT)}2 = 9,809 ??
To find the statistical and systematic errors of g, we need error propagation (later) and do it separately- for statistical and for systematic errors. Let’s say we found:
g = (9.80913 .007stat .o11sys)m/s2
nonsense digits
Compare to the “accepted value”: (Physics.nist.gov/physRefData/contents.html) gives‘global value’
g = 9.80665 ( 0 ) by definition !!
conclusion: OK within (our) errors
90 95 100 105 110 x
N->N= 20
Distribution of Measurements with Random Error:Limit N-> Parent Distribution.
Measure resistors N times:manufacture 100 5%DVM accurate to 1%
contact resistance 0.2
INDEPENDENT !!!
# of R
N measured samples Parent distributionGraph histogram smooth curveAverage -> Mean
€
x=1N xi
i=1
N∑
€
μ=limN→∞1N xi
i=1
N∑ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟
Variance - > sta nda rdde via tion
€
sx2= 1N−1 xi−x( )2
i=1
N∑
€
σx2=limN→∞
1N−1 xi−μ( )2
i=1
N∑ ⎛ ⎝ ⎜ ⎞
⎠ ⎟Resu lt ± e rror
€
€
x±sx/ N
€
μ±0
Famous Probability Distibutions:
Binomial: Yes(Head): p No(Tail): q = 1-p x heads in n trials: px qn-x for ppppppp qqqqq seq., There are permutations, so that:
€
xn
⎛
⎝ ⎜ ⎞
⎠ ⎟=
n!
x!(n − x)!
€
P(x : n, p) =n!
x!(n − x)!p x (1− p)n−x
€
mean = n•p =: μvariance = n⋅p(1-p)
x
P
Seldom used, but grandfather of the following:
Poisson: follows for low p << 1, yet n• p = Obser ve r events i n ti meinterval t
€
P(r.μ)=μr
r!e−μ= (λ⋅t)rr! e−λt ⎛
⎝ ⎜ ⎞ ⎠ ⎟
mean = λ⋅t = μst. deviation = μ
Applicat ion for low rate s r
Inte rvals betwee n two ev e nts: P (0,λ• t) = e -λ• t -> t = 0 m ost like ly!!!!
Ga uss ian .. is th e case f or n• p >>1 (app lica ble to 1%, if np > 17)
€
P(x;μ,σ)= 12πσe
(x−μ)22σ2
mean = μSt. deviation σ = μ for counting experiments
Cen tral Limit The ore m:“Folding many different distribut ions - > a lways a Ga ussi a n”. This is the case in most o f your meas urements.
μ x
P
μ x
P
Γ = 2.354σ
Distribution of 2200 resistors
= 2.17 .01 k
= 0.023 k
Distribution of 100 5 resistors
= 104 ????
No good fit !!!
Still √s = 7.5
Someone selected out!!
Given a histogram of measurements:
How do you know it is
a) A Poisson distribution?
b) Since you do not know the true mean, check the
c) Sample Variance1/2
d) Against the standard deviation
€
S =1
N −1(xi − x )2
i
∑
€
= ≈ x
What doe this mean? Poisson Gaussianp <<1 n•p >>1
Source n=1010 atomsFe55-⇒Mn55γ =p ln2/τ1/2 =0.8 10
-8/ s n• p= 80/secτ1/2=2.7 = y 8.5 107sec
Expecte d averageDecays/ Tim e == n•p 0.8 counts/.01s 80 count /s s
As μ increases Gaussian
Triva ques tion :”If I make 100 of the left hand mea sure ments a nd lump toge the r in onedistribution, w ill I s ee the right han d shape ?” YES !!
#decaysper .01 s
Poisson
0 1 2 3 4
#decaysper s
20 40 60 80
Ga uss
For μ > 17 P oisson ≈ Gauss to 1%, Both h a ve va riance = √μ
Probability of getting more than n events from a distribution ofaverage ?”
Poi :sson
€
S(n,∞;μ)= P(x;μ)=e−μn
∞∑ μx
x!n
∞∑Ga uss ian:
€
S(n,∞;μ)=Erf(n,μ,σ) see tables for e rror funct ion
I am 2σ from the acc ep tedva lueis t he equipm ent broken??
The g raph te lls you 4.6 o f 100me asur e men ts may do th is.
If you do anot her meas urem e nt a ndst ill ar e > 2σ off:
-Your e rrors may h a ve b e en too sma ll-an und e tecte d syste mat ics is present-eva luat ion wa s flawed
equipme nt failure us ua lly let yo u notge t close t o 2σ !
PROPAGATION OF ERRORS Bevington ch.3
You measure u±u an d v±v , b ut evaluate x= u + v o r x= u/van d want theerro r of x.
Tayl orexpansion:
€
xi−x =(ui−u)∂x∂u+(vi−v)∂x∂v+.........
the n
€
σx2=lim
N→∞
1N
(ui−u)∂x∂u
+(vi−v)∂x∂v
⎛ ⎝ ⎜
⎞ ⎠ ⎟
2
€
=σu2 ∂x∂u ⎛ ⎝ ⎜ ⎞
⎠ ⎟2+σv
2 ∂x∂v ⎛ ⎝ ⎜ ⎞
⎠ ⎟2+2σuv
2 ∂x∂u ⎛ ⎝ ⎜ ⎞
⎠ ⎟∂x∂v ⎛ ⎝ ⎜ ⎞
⎠ ⎟
Sum&Diff: x = a u + bv ->
€
σx2=a2σu
2+b2σv2±2abσuv
2
Prod.&Div: x = a. uv or a. u/v ->
€
σx2
x2=σu2
u2+σv
2
v2+2σuv2
uv
Powers : x = a T4 ->
€
σxx=± 4 σT
T wa tch o ut!
= 0 if uncorrelated,otherwiseσμν covariance matrix -difficult
Make a measurement, get the mean
What is the err or ? NOT x ! ! !
€
σx2=sx2= 1
N−1 xi−x( )2i=1
N∑
Be st e s t im a te= m ea n :
€
x=1N xi
i=1
N∑ ±σxN err o r of th e me a n
Be vin gt o n e q 4 .1 2 m ak e m an y m e a s ur e m e nt s !
If m e a s ure m e n ts x i h av e d iffer e nt err o r s σ I
Co m b in e :
€
x=xiσi
2∑1σi
2∑ with σ = 11σi
2∑The y ar e ca lle d: w e ig hte d a vera g e ± co mb in e d s t . d ev .
x
Famous Confusion
N(t) = (300 20stat50syst) N(t) = (330 60stat10syst) exp{t/(2.05.03s} exp{t/(2.3.6s}
good not impressive
1 2 s
Deca ys/ s
Theo
1 2
Precise databut inaccurateT=0 value seems offBad fit. Find reasonDelete first point.
Accurate databut imprecise
More data, finer intervals needed
Theo
Muon lifetime
The Art of f itting or he χ2 - dist r ib ut io :n Bevin γτ on ,χh.4
Assum e wek now I the tru e function valu es in the i-t h bin.We measure n time s xi with rando m e rror i in eac h bin and evaluate
€
χ2 = xi −μiσi
⎛ ⎝ ⎜
⎞ ⎠ ⎟
2
i=1
n
∑Clear ly t h is q u ant it y h a s a d is tr ibut io n it s e lf , be cau s e th e n e xt rand om sa mp le o f n m e a su re m ent swill h a ve d iffer e nt χ 2 .
You us u a lly hav e t h e re ver s e cas e , t h e tr u e f u n cti on i s no t k n o wn , b u t ap pr o xim a te db y
a “ be s t fit” , bas ed on n m e a su reme n t s .
You wan t t o know : ” I s t h e f it accep t abl e ?”Th e χ 2 d is tri b ut ion g ive s you t h e p r o b a b ilit y t h at you r m ea su rem e nt h as t h isp art icu la r χ 2 . Defi n e χ 2=u a nd ν = n - #p ar a meter s = “ d e g ree s o f free do m ” .
(Wit h 2 data p o int s and 2 para m e t e rs th e cu rv e m us t go t hr ough th e m , th er e i s no “f r ee dom t o f it ”. )
The di str ibu t ion i s
€
P(u)=u2( )
ν2−1e2Γν2( )
u2
where Γ = Gamma funct.
Example:χ2 = 8 has stil l an acceptabl e probabilit y fo r n=5 ,but χ2 = 8 is very unlikely fo r n = 1 .I n th is cas e th e data o r th e hypothesi s (= th e theoreticalfit for mula) m ay b e wron .g Lo ok fo r a b etter f ormula ordata!!N :ote χ2 / do f < 1 f o r v > 4 is suspicious, f o r n > 2 obtaining χ2 = 0 has probabilit y = 0 !!