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)

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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

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x=1N xi

i=1

N∑

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μ=limN→∞1N xi

i=1

N∑ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

Variance - > sta nda rdde via tion

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sx2= 1N−1 xi−x( )2

i=1

N∑

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σx2=limN→∞

1N−1 xi−μ( )2

i=1

N∑ ⎛ ⎝ ⎜ ⎞

⎠ ⎟Resu lt ± e rror

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x±sx/ N

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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:

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xn

⎛

⎝ ⎜ ⎞

⎠ ⎟=

n!

x!(n − x)!

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P(x : n, p) =n!

x!(n − x)!p x (1− p)n−x

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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

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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)

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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

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S =1

N −1(xi − x )2

i

∑

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= ≈ 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

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S(n,∞;μ)= P(x;μ)=e−μn

∞∑ μx

x!n

∞∑Ga uss ian:

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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:

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xi−x =(ui−u)∂x∂u+(vi−v)∂x∂v+.........

the n

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σx2=lim

N→∞

1N

(ui−u)∂x∂u

+(vi−v)∂x∂v

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

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=σu2 ∂x∂u ⎛ ⎝ ⎜ ⎞

⎠ ⎟2+σv

2 ∂x∂v ⎛ ⎝ ⎜ ⎞

⎠ ⎟2+2σuv

2 ∂x∂u ⎛ ⎝ ⎜ ⎞

⎠ ⎟∂x∂v ⎛ ⎝ ⎜ ⎞

⎠ ⎟

Sum&Diff: x = a u + bv ->

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σx2=a2σu

2+b2σv2±2abσuv

2

Prod.&Div: x = a. uv or a. u/v ->

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σx2

x2=σu2

u2+σv

2

v2+2σuv2

uv

Powers : x = a T4 ->

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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 ! ! !

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σx2=sx2= 1

N−1 xi−x( )2i=1

N∑

Be st e s t im a te= m ea n :

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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 :

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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

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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

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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 !!

Bottom line: 2/dof 1 not stringent(good) for dof>4