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PHYS2150 Lecture 2
� The Gaussian distribution – Chapter 5 in Taylor
� What it looks like
� Where and why it shows up
� Mean, sigma, and all that
� Statistical and systematic error and error propagation
� Example : e/m measurement
Assignments
• E-Class survey ends today at 11pm
• Read Taylor Ch 6
• This week: first lab starts.
• Radiation safety CAPA must be done before
your first lab this week! You’ll need to bring
print-out of completed CAPA to your first lab.
• First lab write-up due on Friday 9/16, 4pm.
Two types of uncertainty:
Statistical Systematic
sometimes high, sometimes low
centered on the true value
� ‘random’ distribution
always high, or always low
“systematically” off-center
We try to find “true value” X
This Lecture: How to deal with random
uncertainties
Defined as: Measure an infinite number of
times and take the average.
(can’t do in practice, but a very large number of
measurements get’s us pretty close to the same value)
1. Repeat the measurement many times
2. Calculate the average:
3. Estimate uncertainty (how far is this
average value off from the real value)
In practice we can and do:
Warning: Systematic uncertainty is NOT addressed
by statistical methods! Dealing with it requires
knowledge of the experimental setup and its
limitations.
∆xN
µ
Calculating uncertainty using statistical
analysis
Step 1: Make a histogram.
a) Split up the x-axis into n intervals. (n<<N)
b) Record the number of times a value falls
into a given range when you repeat the
measurement over and over.
c) Normalize: Divide the number in each bin
by the total number of measurements
Example: Exam score of a large class
160
120
80
40
0
Num
ber
of
Stu
de
tns
10080604020
Exam Score
76 students got a
score between 44
and 53 points
The class average is
µ = 68.7
(not normalized)
N = 608
What is the sum of the histogram
entries (ni) after Nmeasurements?
As recorded (i.e. before normalization):
After normalization:
bin i in timesofnumber : ; th
i
bins
i nNn =∑
1/ N ni
bins
∑ =1
160
120
80
40
0
Num
ber
of S
tude
tns
10080604020
Exam Score
ni
ni+
1
N=608 students
Example: Measure the circumference of 150 cows
3 4 5 6Bovine Circumference (m)
# o
f co
ws/
0.2
m
σ
✷The values appear to be
centered on the average value µ(‘mean’) .
✷The values appear to occur
more frequently near the average
and less frequently further away.
✷The mean (µ) and width
(“standard deviation” σ) describe the
relevant features of the distribution.
µ
Gaussian Distribution
• Shows up just about everywhere
• Synonyms: Normal Distribution, Bell
Curve
• Most basic form is “Unit Gaussian”:
centered at zero, unit integral, unit
σ:
• This is the probability density for a
continuous, normally distributed
random variable with mean zero
and standard deviation of 1.
x
Probability density function:
F(x) =1
2πexp −
x 2
2
F(x)dx =1−∞
+∞
∫
x
• What can you do to the Unit Gaussian, and
still keep it a Gaussian?
– Change its mean from zero to μ
– Change its width from 1 to σ (while
increasing height by 1/σ)
– You cannot change its integral, or else it
is not a probability distribution anymore.
As with any probability density
function, this still integrates to 1.
F(x) =1
σ 2πexp −
(x − µ)2
2σ 2
Gaussian Distribution
Gaussian probability distribution:
Mean, Sigma, and all that
• Let’s go back to our Cow histogram :
Circumference of 150 cows
• Calculate the mean circumference
• Standard deviation is calculated by:
3 4 5 6Bovine Circumference (m)
Co
ws/
0.2
m
Read up in Taylor, Chapter 5, on definitions and uses of variance,
standard deviation and standard deviation on mean!
μ=4.39 m,
σ=0.70 m
� = �̅ = 1� � ��
�= 4.39 m
�� = ∑ (�� − �)������ − 1
= 0.7 m
Circumference of Cow data
3 4 5 6Bovine Circumference (m)
Co
ws/
0.2
m
Note that we know the mean to much better than σ of individual measurements.
So
• Mean: 4.39 m, STD: 0.70 m
• 68% of cows have circumference between
(4.39−0.70) and (4.39+0.70) m. This is
because the integral of the normalized
Gaussian from μ−σ to μ +σ is 0.68. (68% of
confidence level)
• How well do we know the mean
circumference? Need std. dev. on the mean:
�� = ��� = 0.06 m
µ = (4.39 ± 0.06) m
Where it shows up
• If you don’t have a clue what the probability distribution of a
random quantity is (say, the circumferences of cows at the hbar
Ranch), it’s highly likely to be approximately Gaussian!
• Central Limit Theorem: a sum of a large enough number of
random numbers has a Gaussian distribution, no matter what the
initial distribution might have been.
• Aside: Distribution of counts of a process with a uniform rate in a
finite amount of time is Poisson-distributed (see a later lecture)
but is approximately Gaussian in the high-number limit. This is
another example of the Central Limit Theorem.
• If there is no systematic bias, then the mean of measurements of a
quantity is the best estimate of its true value.
Using the Gaussian• Can use the same mathematics to describe the
results of repeated measurements of the same
quantity, where there is random error in the
instrument.
• The distribution will be centered on a mean
(assume for now that this is the correct value)
• The distribution will have a standard deviation
• Can fit this to a Gaussian (Lecture 4,5) or just
calculate mean and sigma directly as discussed.
• Uncertainty on the mean is now
• Note: More measurements → smaller error on
the mean!
μ
σ
σ µ =σN
What does sigma mean?
• First, if the measurements are from a Gaussian Fμ,σ(x), then the
probability of measuring a value in the range (a,b) is:
• For a normalized Gaussian,
• The general integral can’t be expressed analytically. Use error
function (erf(x)) tables for values other than 1σ. (Taylor Appx. A&B)
• So, saying “J=5.4±0.9” means one can say the true value of J is
between 4.5 and 6.3 with 68% confidence level.
� ��,! � "�#
$
So far we discuss “statistical uncertainty”.
Now what about uncertainty involved with
measurement itself ?
(Systematic Uncertainty)
Analyzing the error on a quantity
• You measure the voltage of a Josephson voltage standard with a $4
volt meter off e-bay
• You measure it 2983 times.
• The histogram of your results is at right.
• You calculate the mean to be 0.89 V, and the standard deviation to be
0.21 V.
• The 1σ uncertainty on the mean is
0.0038 V.
• If we assume the underlying
distribution is Gaussian and
centered on the true value, we can
turn this into a confidence level:
The true voltage of the Josephson
device is in the rage of 0.886 V to
0.894 V with a 68% confidence. 0 0.3 0.6 0.9 1.2 [V]
Analyzing the error on a quantity
0 0.3 0.6 0.9 1.2 [V]
• You measure the same Josephson voltage standard with a NIST
calibrated voltmeter.
• You find the true value is 1.200 V
• We are off by 75 sigma, which is extremely unlikely to be
correct!
• No matter how many times you measure, the systematic error
will not go away!
Statistical (random) vs. Systematic Uncertainties
• Keep statistical, systematic errors separate. Report results as something like:
g = [965 ± 3(stat) ± 2(syst)] cm/s2
• Add in quadrature (note that this assumes Gaussian distribution) to compare
with known values: g = [965 ± 4(total)] cm/s2
STATISTICAL SYSTEMATIC
NO PREFERRED DIRECTIONBIAS ON THE MEASUREMENT: ONLY ONE
DIRECTION (THOUGH OFTEN DON’T KNOW WHICH)
CHANGES WITH EACH DATA POINT: TAKNG MORE DATA REDUCES ERROR ON THE MEAN
STAYS THE SAME FOR EACH MEASUREMENT: MORE DATA WON’T HELP YOU!
GAUSSIAN MODEL IS USUALLY GOOD (EXCEPT COUNTING EXPERIMENTS WITH FEW EVENTS)
GAUSSIAN MODEL IS USUALLY TERRIBLE. BUT WE USE IT ANYWAY IF DON’T HAVE A BETTER
MODEL.
Propagation of Errors
• Often, we aren’t directly measuring the quantity our
experiment is after: we measure some lab quantities and
our final “physics result” is a function of them.
– Kaon experiment: we measure curvature of tracks, and
from them calculate the momentum of the pions, and
then calculate the mass of the kaon from that.
• We know the errors on the ‘lab quantities.’ How do we find
the error on the final physical quantity?
• This is a specific case of the general problem of finding the
error on a quantity that is a function of random (uncertain)
variables.
Propagation of Errors
From PHYS 1140: The general formula for errors on a function f(x,y,z),
Addition : if (q=x+y) or (q=x-y),
The systematic error in q is
x,y,z.. variables with random
uncertainties δx, δy, δz…
Multiplication: if q=xyz,
Thus, fractional error is
Power:if q=x2y/z4, fractional error is
Examples:
Example: e/m Experiment
• Electrons accelerated with 40, 60, or 80 V
• Acted on by magnetic field perpendicular to velocity
• r F = e
r v ×
r B forces electrons into circular paths
• Measuring radius of circle give centripetal force by F =mv 2
r
• Energy of electrons given by accelerating potential 1
2mv 2 = eV
• Magnetic field from Helmholtz coils is B =8µ0NI
a 125
• After putting together, e
m= 3.906
Va2
µ0
2N 2I 2r2
e/m Experiment e
m= 3.906
Va2
µ0
2N 2I2r2
Variable Definition How Determined
V Accelerating Potential measured
a Helmholtz Coil Radius measured
μ0 Permeability of Free Space constant
N Number of turns in each coil given
r Electron beam radius given
I=IT-I0 Net Current calculated
IT Total Current measured
I0 Cancellation Current measured
✹Want an answer with systematic and statistical uncertainties !!
Data for e/m example
Entry I(A) r(m) Voltage (V) e/m (C/kg x 1011)
1 1.75 0.0572 40.0 2.08
2 1.99 0.0509 40.0 2.08
3 2.28 0.0447 40.0 2.00
4 2.67 0.0384 40.0 1.98
5 3.20 0.0321 40.0 1.97
6 2.20 0.0572 60.0 1.97
7 2.49 0.0509 60.0 1.94
8 2.85 0.0447 60.0 1.92
9 3.34 0.0384 60.0 1.90
10 4.03 0.0321 60.0 1.86
11 2.58 0.0572 80.0 1.91
12 2.90 0.0509 80.0 1.91
13 3.30 0.0447 80.0 1.88
14 3.39 0.0384 80.0 1.85
Measure I0 three times and find I0 = 0.17, 0.20, and 0.23;
giving a mean of 0.20 and an uncertainty of 0.03, i.e. I0 = 0.20±0.03 A
Compute e/m for 14 different voltages and pin radii
e
m= 3.906
Va2
µ0
2N 2I2r2e/m Compute Statistical Uncertainty
• Mean: x =xi
i=1
N
∑N
=1.946 × 1011C/kg
• Standard deviation: σ x =(xi
i=1
N
∑ − x )2
N −1= 0.072 ×1011C/kg
• Uncertainty on mean: σ x =σ x
N=
0.072
14= 0.019 ×1011C/kg
Result with statistical uncertainty only:
e
m= (1.946 ± 0.019)×1011C/kg
e/m : Compute Systematic Uncertainty
from statistical uncertainty: I0 = 0.17, 0.20, and 0.23;
I0 = 0.20±0.03 A
from meter (0.003x0.20+0.01) = 0.011A
from meter (0.003x2.85+0.01) = 0.019A
I: net current
IT: total current
I0: cancellation current
e/m Systematic Uncertainty
e/m measurement: Summary of Results
✸ Compare Measured value
[1.946 ± 0.019(stat.) ±0.064(sys.)] x 1011 C/kg
to accepted value 1.75882 x 1011 C/kg
✸ Discrepancy from the accepted value
[1.946-1.75882] x 1011 C/kg = 0.187 x 1011 C/kg
✸ Significance of Discrepancy : Ratio of discrepancy and σtotal
often stated as “off” by 2.8σ
Is this a good or bad measurement???
e/m measurement: Summary of Results
✸What is the probability of value being further than 2.8σ
from the mean for a gaussian distribution?
x
Probability of being inside of 2.8σ is
99.49% (look up the table p.287 Taylor)
Thus , probability of being outside of
2.8σ is 0.51%!!!
➙ Very Unlikely that a reasonable
measurement be that far from the
mean!
➙ Either mistakes or underestimates
systematic uncertainties.
✹If this happen to you, you will need to discuss more about
what sources of error might have caused this.
Assignments
• E-Class survey ends today at 11pm
• Read Taylor Ch 6
• This week: first lab starts.
• Radiation safety CAPA must be done before
your first lab this week! You’ll need to bring
print-out of completed CAPA to your first lab.
• First lab write-up due on Friday 9/16, 4pm.