Transcript Lecture 9

First LHC Collisions seen by CMS, today!
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Last two Lectures - Overview
Last week’s Lecture:
•
Introduction of Experiment #4
•
Reminders from last lecture:

c2 Test of a distribution
Today (and last) Lecture:
– Issues in Experiment #4
– Covariance and Correlation
– Problems and examples before final
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End of quarter Logistics
• Review Session will be held on:
This Tue, (Tomorrow) Nov 24th, 1900 at Ledden Auditorium.
• Final:
Next week same time/place
Monday, Nov 30th, 1500. York 2622 (this room).
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Review Session
• Will be held tomorrow:
This Tue, Nov 24th, 1900 at Ledden Auditorium.
• The TAs will go with you over:
– questions you may have
– example problems
>> Please try to attend
>> Come prepared
• Make sure you have done all the homework assignments
from Taylor!
– Some of the final questions will be very similar to them
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Final Exam
You can use:
Your calculator
Your textbook
Your notes
Principal Formulas and Tables
Don’t forget to bring
No Laptop computers
On our beloved web site - 2bl.ucsd.edu
- Lecture notes
Bring your own calculator!!
- Experiment Guidelines
and make sure its charged…
- Homework solutions
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ThanksGiving Related Logistics
•
Thursday is ThanksGiving Holiday
•
If your session is on Thu or Wed evening, you have a few
choices:
–
–
The lab was open today for make-ups.
You may still join another section on Tue or Wed (NOTE: last
session has been canceled!)
– Please:
-
Tell TAs which section you belong to, when you join another section.
Turn in your reports by Wednesday to the lab (in the proper location
for your section). BEFORE 5PM !!!
First,
Visit from CAPE…
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Instructor’s Questions
A. Course goals and objectives were made clear from the beginning.
B. Lecture (Professor) and lab (TA) instruction are connected and consistent.
C. Material on the web (lecture notes, experiment guidelines, rubric, etc.) is
helpful.
D. Course content is aligned with credit and effort level.
E. There is enough time to cover the material during lectures.
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QuickTime™ and a
decompressor
are needed to see this picture.
Stolen from Dr. P. Riley
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Introduction Correlated Errors :
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Correlation is Quantifiable
Correlation Coefficient
 xy
r
 x y
1 N
 x   ( xi  x) 2
N i 1
N
1
 y 2   ( yi  y ) 2
N i 1
2
Magnitude and direction
of a linear relationship
between two variables
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Covariance
Variation of two random
variables together
 xy
1

N
N
 ( x  x)( y
i 1
i
i
 y)
Magnitude of covariance dependent
upon size and sign of the errors!
 xy  f (x, y )
Both variables are greater than
expected.
One variables is greater than expected,
while the other is less than expected.
 xy  0
Negative correlation:  xy  0
No correlation:  xy  0
Positive correlation:
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Review of Error propagation: uncorrelated variables
q  q ( x, y )
Summation in quadrature:
2
 q  2  q  2
     x     y
 x 
 y 
2
2
q
Assuming errors in x and y to be independent
and normally distributed
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Error Propagation – Product (re-visited)
q  xy
• Take x, y and q as xbest, ybest, qbest
From 1st lecture
• Measure (x + dx) and (y + dy)
• Compute (q + dq)
q  d q  ( x  d x )( y  d y )  xy  xd y  yd x
q
q
d q  xd y  yd x  d y  d x
y
x
(neglect dxdy)
Subtract q from both sides of
the above equation
What if we cannot neglect dxdy ?
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Error propagation: correlated variables
q  q ( x, y )
2
q q
 q  2  q  2
     x     y  2
 xy
x y
 x 
 y 
2
2
q
“interaction” term
Regardless of whether or not errors in x and y
can be assumed independent and normally
distributed
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Significance of Correlation Coefficient
•
Assess whether or not a given event belongs to the sampled data set
•
Rigorous error propagation in the case that the measurements are NOT
independent and normally distributed
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Deducing (unexpected) conclusions from data
•
Probability that N measurements of uncorrelated variables would give r
as large as observed ro
QuickTime™ and a
decompressor
are needed to see this picture.
Can use:
Prob N ( r  ro )
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Does
belong to this sample?
Significant
Prob N ( r  ro )  5%
Highly Significant
Prob N ( r  ro )  1%
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Example: Correlation Coefficient
Calculate the covariance and the correlation coefficient r for the following
six pairs of measurements of two sides x and y of a rectangle. Would you
say these data show a significant linear correlation coefficient? Highly
significant?
A B C D E F
y
x = 71 72 73 75 76 77 mm
y = 95 96 96 98 98 99 mm
x  74
x
y  97
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covariance  xy   (xi  x )( yi  y)  (3)  (2)  ...  3  2  3
N
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 xy
(xi  x )( yi  y)


 0.98
correlation coefficient r 
2
2
 x y
 (xi  x )  ( yi  y)


0.3 %
Table C Prob6 ( r  0.98)  0.3%
 evidence for linear correlation
is both significant and highly significant
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Conclusion on Correlations
• Unknown correlations in the observables may seriously
affect the statistical interpretation of the data.
• Calculating Covariance and Correlation Coefficients allows
one to check for possible correlations, and correct the
statistical analysis accordingly.
Real Life Example:
Grades of 2BL students
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What is the problem?
• Many (sometime 14) lab sections
– Possible large variations in TAs grading standards between
different sections (affect Lab reports quizzes)
• How can we ensure fair grading for all students, in
different sections?
• Could equalize average of all sections
– What is the problem with this solution?
• Need an objective, uniform view of students performance
– Final exam is graded in such a way
– Each problem is graded for the whole class by the same person
– No biases
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Solution we use
• Look at correlation between:
– Lab grades (reports & quizzes)
– Final exam grades
Compare average grades by section.
• Discuss a few possible example issues
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Example Problems
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A student measures the period of a pendulum. The measurement is repeated 5 times with the
same pendulum. The times measured are:
T = 1.42, 1.44, 1.47, 1.4, and 1.39 s.
What is the mean period?
What is the RMS error (the standard deviation) in the period?
What is the error in the mean period (the standard deviation of the mean)?
How many measurements are required to give an uncertainty in the mean period of 0.001 s?
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 (1.42  1.44  1.47  1.4  1.39)  1.424 s
5
1
1
2
T 
(
T

T
)

(0.004 2  0.0162  0.0462  0.024 2  0.034 2 )  0.0321s  0.03s

i
N 1
4

0.03
T  T 
 0.01342 s  0.013s
N
5
T 
1
N
 Ti 
2
    0.03 
N  T  
  900
  T   0.001 
2
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Taylor 3.14
A visitor to a castle measures the depth of a well by dropping a stone
and timing its fall. She finds the time to fall is: t  3.0  0.5sec
and calculates the depth as: d  1 gt 2
2
What is her conclusion, if she takes:
With negligible uncertainty?
g  9.80
m
s2
Solution:
d  0.5 * 9.80 * 3.02  44.1m
 d / d  2 * ( t / t)  2 * (0.5 / 3.0)  0.33
q(x)  x n
Based on: d q
 d  0.33* 44.1  14.7m
d  44.114.7m
q
 n
dx
x
d  44  15m
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Two students measure the radius of a planet and get final answers
RA=25,000±3,000 km and RB=19,000±2,500 km.
(a) Assuming all errors are independent and random, what is the discrepancy and what
is its uncertainty?
(b) Assuming all quantities are normally distributed as expected, what would be the
probability that the two measurements would disagree by more than this?
Do you consider the discrepancy in the measurements significant (at the 5% level)?
(a) RA  RB  25,000  19,000  6,000km
R
A  RB
  RA 2   RB 2  3,0002  2,5002  3,905km  4,000km
RA  RB  6,000  4,000km
(b) t 
6,000
 1.5
4,000
Table A: Probability to be within 1.5 is 86.64 % ≈ 87 %. Therefore, the
probability that the two measurements would disagree by more than this
is 100 – 87 = 13 %.
The discrepancy in the measurements is not significant (at the 5% level).
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Two students measure the radius of a planet and get final answers
RA=25,000±3,000 km and RB=19,000±2,500 km.
Find the best estimate of the true radius of a planet and the error in that estimate.
xwav 
wA x A  wB xB
wA  wB
RA
Rwav
wA 
1
wB 
 A2
1
 B2
 wav 
1
wA  wB
RB
25,000 19,000

 A2  B 2 3,0002 2,5002


 21, 459km  21,500km
1
1
1
1
 2

2
2
A B
3,000 2,5002
 wav 

1
1
 A2

1
 B2

1
1
1

3,0002 2,5002
 1,921km  1,900km
Rwav  21,500  1,900km
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Chauvenet’s Criterion
A student makes 14 measurements of the period of a pendulum. She gets
the following measurements, all with the same estimated error.
T= 2.7, 2.3, 2.9, 2.3, 2.6, 2.9, 2.8, 2.7, 2.8, 3.2, 2.5, 2.9, 2.9, and 2.3
Should any of these measurements be dropped?
n
•
Add up all the periods and divide by 14 to get the average, T=2.7 seconds.
•
Compute the standard deviation from the data, =0.27 seconds.
•
The measurement furthest from the mean is 3.2 seconds giving t=0.5/0.27=1.85.
•
Look up the probability to be further off, P=6.43%.
•
Multiply by the number of trials to get the expected number of events
that far off, nexp=(14)(0.0643)=0.9
•
Do not drop this measurement (or any of the others).
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x   xi
n i1
 
1 n
2
x 
 x x
n  1 i1 i
2
Assume the student made a
15th measurement but her
partner bumped the
pendulum during the
measurement. She got a
period of 2.8 seconds.
Should she drop this
measurement?
Why?? 27
The Normal Distribution
The grades of students in a course where found to be normally distributed with a mean
of 80 points and standard deviation of 5 points. There were 275 students in the
course. How many students would you expect to have scores:
a)
b)
c)
d)
Between 75 and 85 points?
Higher than 90 points?
Between 70 and 90 points?
Bellow 65 Points?
Solution:
a) This grades range is: +/- 1 therefore, 68% of the scores should be within this range
275*0.68 = 187 students.
b) Grades higher than 90 points are 2 and above the average. Therefore:
N  0.5 * (1  prob(2 )) * 275  0.5 * (1  0.9545) * 275  6
c) Scores between 70-90 points are within 2 of the mean. Therefore 95.45% of scores are
expected to be within this range.
N  0.9545 * 275  262
d) Scores bellow 65 points are more than 3 bellow the average score. We know that +/-3 is
expected to contain 99.7% of the scores. The negative tail (outside this range) is expected to
contain half as many:
N  0.5 * (1  prob(3 )) * 275  0.5 * (1  0.997) * 275  0.4
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Example
• We measure the height of every student in the class (say of
50).We compute a mean height, <h> with a standard deviation
h. Then we bin the measurements as follows:
–
–
–
–
Less than (<h>‐h):
Between (<h>‐h) and <h>:
Between <h> and (<h>+h):
More than (<h>+h):
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15
19
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• Assuming the measurements are normally distributed:
– What is the expected number of measurements in each bin?
– What is the  2?
– Do you conclude that the students heights are distributed normally?
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Solution
From Table B:
– Less than (<h>‐h):
– Between (<h>‐h) and <h>:
– Between <h> and (<h>+h):
– More than (<h>+h):
Ok
Ek
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8
15
17
15
17
0.5 - 0.34 = 0.16
0.34
0.34
0.5 - 0.34 = 0.16
0.16*50= 8
0.34*50=17
0.34*50=17
0.16*50= 8
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Finish it yourselves…
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For Final Exam, Review…
•
•
•
•
•
•
•
•
•
RMS errors
Propagation of errors
Adding errors in quadrature
Averages
Probability distributions
Normal distribution
Confidence Levels
Chauvenet’s criterion
Principle of Maximum Likelihood
•
•
•
•
•
Weighted averages
Best fit straight line
Chi-square
Degrees of freedom
Probability of chi-square
Final Review:
TA’s guided, problem solving
Sessions:
• Tuesday, Nov 24, 7PM
Ledden Auditorium
Enjoy the Holiday
Good Luck in the Final!!
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