Weighted average and fits: Chapters 7-8
Announcements:
•  The requirement that you attend 10 lab sections has been
removed. You must take data during your normal lab section
(unless you have made prior arrangements). However, if you
only need to write up the lab, you do not need to attend your lab
section. I DO encourage you to attend in order to get help or to
check the lab for things you may have missed.
•  The deadline for Lab 2 has been extended:
–  Thursday section labs are due by 4pm on Monday, February 10
–  Tuesday section labs are due by 4pm on Friday, February 14
•  Remember, you need to do 6 labs. For at least 3 of them, you
need a full lab report. For the remainder, putting the analysis in
your lab notebook is fine. Should still make plots by computer
and past them in. Need to have same information as in report
but don’t need to strictly follow the format.
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Physics 2150 – Spring 2014
1
Combining data
•  First you should check that the values you are combining are
compatible. You can use the same test as you use to determine
if your measured value is compatible with the known value. If
the probability is small that the values are compatible, then it
probably doesn’t make sense to average them.
•  If they are compatible, it is a good idea to average as you can
get a more precise result.
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Physics 2150 – Spring 2014
2
Weighted average
•  Take N measurements x1, x2, …, xN with uncertainties σ1, σ2, …,
σN
N
∑w x
i i
•  Weighted average is xavg =
i=1
N
∑w
i
2
i
i=1
•  The weights are wi = 1 / σ
•  The uncertainty on xavg is σ avg =
1
N
∑w
i
i=1
•  If σ1 = σ2 = … = σN = σ then weighted average is the simple
average and the uncertainty is σ/√N.
•  Note that weights go as 1/σ2 so more precise measurements
count much more than less precise measurements.
•  This is the simplest example of least squares fitting (a fit to a
line with no slope)..
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Physics 2150 – Spring 2014
3
•  Suppose you make two
measurements of g and find
values of 9.802 ± 0.037 m/s2 and
9.774 ± 0.019 m/s2
•  Check the discrepancy between
the two numbers:
9.802 − 9.774
2
0.037 + 0.019
2
g (m/s2)
Example of weighted average
9.84
9.82
9.8
= 0.67σ ✔
•  The straight average would give
9.788 m/s2
•  The weighted average is 9.780 ±
0.017 m/s2
Meas 2
9.78
Weighted
Average
Average
9.76 Meas 1
•  The weighted average will have smaller uncertainty than any of
the measurements that go into the average.
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Physics 2150 – Spring 2014
4
Ways to use multiple measurements
•  Suppose you take N measurements of a quantity and you want
to get the best estimate of that quantity.
•  Two approaches:
A.  Calculate the mean, the standard deviation (which gives you the
uncertainty on each measurement), and the uncertainty on the
mean (SD/√N).
B.  Calculate the weighted average with corresponding uncertainty.
•  In A, you are determining the uncertainties from the spread of
data (good for large N), each of which is assumed to ~same.
•  In B, the mean uncertainty comes from the uncertainties from
each measurement.
•  How do you choose which method to use?
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Physics 2150 – Spring 2014
5
Choosing way to combine measurements
A.  Calculate the mean, the standard deviation (which gives you
the uncertainty on each measurement), and the uncertainty on
the mean (SD/√N).
•  Must use if uncertainty of each measurement is not known.
•  Best if uncertainties on each measurement are not well known,
the number of measurements is large (>10), and the uncertainties
on each measurement are expected to be the same.
B.  Calculate the weighted average with corresponding uncertainty.
•  Can only use if you know the individual measurement
uncertainties.
•  Should use if individual measurement uncertainties are well
known and/or are not all the same.
•  Also useful in cases where the number of measurements is small
(<10).
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Physics 2150 – Spring 2014
6
Example 1
•  10 measurements of a length (in cm): 9.0, 9.5, 9.8, 9.9, 10.0,
10.0, 10.1, 10.2, 10.5, 11.0
•  Method A: Average = 10.0cm, SD = 0.54cm, Uncertainty = SD/
√N = 0.54cm/√10 = 0.17cm
•  Method B (assuming each measurement has an uncertainty of
δ): Weighted average = 10.0cm, Uncertainty = δ/√N.
–  If we estimated our measurement uncertainty to be 0.54cm, we
would get Uncertainty = 0.54cm/√10 = 0.17cm, same as before.
–  If we estimated our measurement uncertainty to be much smaller
(0.1cm), our uncertainty on the mean is 0.1cm/√10 = 0.03cm. But
then the spread of our measurements is not consistent with our
estimated uncertainty on each measurement.
•  For this case, Method A is better because it is a relatively large
number of measurements and the uncertainties on each
measurement are known to be the same.
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Physics 2150 – Spring 2014
7
Example 2
•  Two measurements, using different techniques, are made of g
and find values of 9.802 ± 0.037 m/s2 and 9.774 ± 0.019 m/s2
•  Method A: Average = 9.788 m/s2, SD = 0.020 m/s2, Uncertainty
= SD/√N = 0.020m/s2/√2 = 0.014 m/s2
•  Method B: Weighted average = 9.780, Uncertainty is 0.017 m/s2.
•  For Method A, we are trying to determine the standard deviation
and the uncertainty from just two measurements, which is
technically possible but is not a good idea. For just two
measurements, they may happen to lie close to each other
(giving a small uncertainty) or far apart (giving a large
uncertainty). Also, we know the two uncertainties are different
so Method A is disfavored.
•  Method B is the correct method in this case: low number of
measurements, well known measurement uncertainties, and
different uncertainties for each measurement.
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Physics 2150 – Spring 2014
8
Clicker question 1
Set frequency to AB
•  You take N measurements of a current in the e/m experiment
using an analog mete and use Method A to determine the mean
e/m, the standard deviation, and the uncertainty on the mean.
•  Now you start to think about systematic uncertainties.
•  You think of two systematic uncertainties related to the current:
A.  The meter is stated to have a potential bias of 0.001*I+0.01 A.
B.  Your reading of the meter could be off by 0.1A in either direction.
Which of these should be included as systematic uncertainty?
A)  A only
B)  B only
C)  Both A and B
D)  Neither A nor B
E)  Depends on how big the statistical uncertainty is
A is a potential systematic bias you should include. B is an
uncertainty that should be accounted for by the statistical uncertainty.
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Physics 2150 – Spring 2014
9
Clicker question 2
Set frequency to AB
•  You time the period of a pendulum 10 times with a stop watch.
•  The stop watch reports results to 1/100 of a second.
What is the best way to determine the period?
A)  Take the weighted average of the 10 times, with an uncertainty for
each time of 0.01 second, and use the calculated weighted average
uncertainty (0.01s/√10 in this case)
B)  Take the simple average of the 10 times and use the uncertainty of
the mean (SD/√10) as the uncertainty.
C)  As in A but also add 0.01s in quadrature as a systematic uncertainty.
D)  As in B but also add 0.01s in quadrature as a systematic uncertainty.
The stop watch accuracy of 0.01s is not the limit on the accuracy, it is
human error. Therefore, it is best to let the data determine the
uncertainty on each measurement (the standard deviation). The 0.01s
uncertainty from the stop watch is random so is included in the statistical
uncertainty and should not be added as a systematic uncertainty.
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Physics 2150 – Spring 2014
10
More general parameter estimation
Weighted average for single parameter
Sometimes data points lie on a line
Example is photoelectric effect
hf=Tmax+φ and eVs=Tmax
Vs = (h/e)f + (φ/e)
Strategy: plot Vs vs f
Fit the data points to a line; slope gives
(h/e) and intercept gives (φ/e).
•  If no uncertainties on measurements,
slope and intercept uncertainties are
obtained directly from data (linear fit).
•  If measurements have uncertainties,
slope and intercept uncertainties are
derived from them (weighted linear fit).
E=h
e
+
Vs = stopping voltage
V
•
•
•
•
•
•
•
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Physics 2150 – Spring 2014
11
Clicker question 3
Set frequency to AB
The true value of the speed of light is c. If we make a
measurement of the speed of light (x), with uncertainty σx, 2then
"
%
the probability distribution is proportional to 1 exp $ − (x − c) '
2
σx
2
σ
#
&
x
If we are measuring the stopping voltage as a function of light
frequency, the measurements will not all be the same but will be
of the form Vi = φ/e + (h/e)f or yi = A + Bxi. We make
measurements of y with uncertainty of σy.
What does the probability distribution look like for yi in this case?
" (A + Bx)2 %
1
''
A)
exp $$ −
2
σy
2σ y &
#
" (y + (A + Bx))2 %
1
''
C)
exp $$ −
2
σy
2σ y
#
&
http://www.colorado.edu/physics/phys2150/
" (y − (A + Bx))2 %
1
''
B)
exp $$ −
2
σy
2σ y
#
&
! (A + Bx))2 $
! y2 $
1
&& exp ## 2 &&
D)
exp ##
2
σy
" 2σ y %
" 2σ y %
Physics 2150 – Spring 2014
12
Fitting to a straight line
•  Measure N points: (x1,y1), (x2,y2),…,(xN,yN) with negligible
uncertainty on x and σy uncertainty on all y measurements.
•  Assume measurements lie on line with y = A + Bx
•  For a given xi, fitted value of yi is A + Bxi
# −(y − (A + Bx ))2 &
1
•  Probability of obtaining yi is
i
∝
exp % i
(
σy
2σ y
%$
('
•  Probability of obtaining the set of measurements is the product
of all the individual probabilities is
2
N
2&
#
yi − (A + Bxi ))
1
−χ
(
2
∝ N exp %
( where χ = ∑
2
σy
σ
$ 2 '
y
i=1
•  In general, find the values of A and B that maximize the
probability (principle of maximum likelihood).
•  In this case, maximum probability is when χ2 is a minimum.
•  Here, can differentiate χ2 with respect to A and B and set to 0 to
obtain A and B.
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Physics 2150 – Spring 2014
13
Equations
χ2 fit to straight
line for straight line fit
Pages 183-184 of Taylor
N
N
2
2
!
!
∂χ
−2
−2
∂χ
(yi − A − Bxi ) = 0
xi (yi − A − Bxi ) = 0
= 2
= 2
∂A
σy i=1
∂B
σy i=1
can be rewritten as:
!
!
AN + B
xi =
yi
A
which can be solved for:
" 2"
" "
xi
y i − xi xi y i
A=
∆
" 2
" 2
where ∆ = N xi − ( xi )
!
xi + B
B=
N
!
"
x2i
=
xi y i −
∆
!
"
xi y i
xi
"
yi
Problem 8.9, page 201 gives results for cases when uncertainties on yi are
not all equal (weighted linear fit)
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January 31, 2006
Physics 2150 – Spring 2014
14
Physics 2150 Lecture 3 – p. 16
! σ2 , B = 2
A = Estimating
, 2∆ = N 2 xi −
2
y
∆
∆
+ 0.2 + 0.4
Estimating
σy . . . 2.0 = 15.4 V
!
! xi = 0.0
xi = 0.0
+
0.2
+
0.4
.
.
.
2.0
=
11.0
V
y
=
0.0
+
1.1
+
2.3
. . . 12.1 =was:
65.8 cm
i
!Uncertainties
Our
standard
deviation
from
many
measurements
!
2 linear
from
unweighted
fit
2
2
2
2
!
Our
standard
deviation
from
many
measuremen
x
=
0.0
+
0.2
+
0.4
.
.
.
2.0
=
15.4
V
"
x
y
=
0.0
·
0.0
+
0.2
·
1.1
+
0.4
·
2.3
.
.
.
2.0
·
!
i
N i i
!
2
!"N 2
(xiwas:
−!
x) 2
•  The
standard
deviation
i=1
2
2
y
=
0.0
+
1.1
+
2.3
.
.
.
12.1
=
65.8
cm
(x
−
x)
σ
=
i
i
x
∆
=
N
x
−
(
x
)
=
11
×
15.4
−
11.0
=
i=1
i
!
iσx =
N −1
1. 2.0
xi y i =
0.0
·
0.0
+
0.2
+
0.4
.
.
·
12.1
=
92.48
V
x2i· 1.1
yi −
xi · 2.3
xiN
yi −
15.4×65.8−11.0×92.48
!The
!
of=N −2 of
1 instead
of Nfrom
cameusing
from
x48.4
instead
=1 instead
=
•  The factor
of2factor
N-1A
instead
N The
comes
the average
2 of
2 using
∆
factor
of
N
−
of
N
came
from
us
∆
=
N
x
−
(
x
)
=
11
×
15.4
−
11.0
=
48.4
V
i
(unknown)
µ true
ithan the
N value.
xi yi(unknown)
− xi
yµ
11×92.48−11.0×65.8
i
value rather
B
=
=
= 6.0
2
xi
yi − xi
xi yi
∆
48.4
15.4×65.8−11.0×92.48
Similar
result
for estimating
theresult
uncertainy
in =
y: −0.08
•  ASimilar
for
estimating
the
uncertainty
in
measured
yuncertainy
=
cm
= result
!
Similar
for
estimating
the
in yo
∆ "
48.4
!
From
spread
of
measurements,
find
uncertainty
values:
N
xi yi − xi N y(y
11×92.48−11.0×65.8
i"
−A
− Bxi )2 "2N (yi −=A6.06
2
B=
cm/V
= kC
i=1 i=
−
Bx
)
i
i=1
σy =∆
(yiσ
−A−Bx
)
48.4
i
=
σy = N − 2 Ny −2
= 0.14
cm = σD
N
−
2
spread
measurements,
find (A,B)
uncertainty
ondata,
y, A,N and
B:
•  From
Since"
we Here
nowof
derive
two
parameters
from
the
gets
" Here
the expected value
for yithe
is expected
A + Bxi"
instead
value
forof
yi xisand
A +we
Bxhave
i inste
ting replaced
a line by
(2)
2
2
N-2.
(y
−A−Bx
)
x
i 2 instead
i of N due to iguessing on A and
15.4
N−
Bto(for
N = 2on
we
are B
N
−
2
instead
of
N
due
guessing
A and
σy =
=
0.14
cm
=
σ
σ
=
σ
=
0.14
×
=
0.08
cm
D
A
y
Non
−2A and B are found
∆ line
48.4
•
Uncertainties
to
be:
guaranteed
to
have
a
perfect
and
therefore
on
guaranteed
to
have
a
perfect
line and therefore
easurements of"x and y can be fit
to line of form"y = AnoCalibration
+information
Bx:
"
"
cons
!
!
uncertainties)
x2i
yi − xi uncertainties)
xi yix2
2
x
y
−
x
y
N
N
11
i i 15.4where
i
i
2
i B
σ
=
σ
=
0.14
×
=
0.07
cm/V
,
=
,
∆
=
N
x
−(
x
)
y
i
σA =
=B0.14 ×
∆48.4 = 0.08 cm48.4 kCi = (6.06±0.0
∆ σy
∆
"0.4 . . . 2.0 = 11.0
"
=
0.0
+
0.2
+
V7, can
February11
2006be found in Problem 8.19 (p.
P
i
•  Uncertainties
fit
N from weighted
σ204).
σy2 + ∆
=
=2 0.07
cm/V
2
2 0.14 ×
2
B+=
48.4
=
0.0
0.2
0.4
.
.
.
2.0
=
15.4
V
January
31,
2006
January 31, 2006
Physics 2150 Lecture 3 – p.Physi
17
i
•  Follow
same
fit asPhysics
for weighted/
= 0.0
+ 1.1
+ 2.3
.2006
. . 12.1for=weighted/unweighted
65.8 cm
February
7, criteria
2150 Lecture
unweighted
average.
0.0
·
0.0
+
0.2
. . . 2.0 · 12.1 = 92.48 V · cm15
i yi =http://www.colorado.edu/physics/phys2150/
Physics 2150 – Spring 2014
! 2
! 2 · 1.1 + 0.4 · 2.3
N xi − ( xi ) = 11 × 15.4 − 11.02 = 48.4 V2