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First LHC Collisions seen by CMS, today!

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Last two Lectures - OverviewLast week’s Lecture:

• Introduction of Experiment #4

• Reminders from last lecture: 2 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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You can use: Your calculatorYour textbookYour notesPrincipal Formulas and Tables

On our beloved web site - 2bl.ucsd.edu- Lecture notes- Experiment Guidelines- Homework solutions

Don’t forget to bring

Final Exam

Bring your own calculator!!and make sure its charged…

No Laptop computers

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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

Magnitude and directionof a linear relationship between two variables

yx

xyrσσσ

=

Correlation Coefficient

∑−

−=N

iix xx

N 1

22 )(1

σ

∑−

−=N

iiy yy

N 1

22 )(1

σ

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Covariance

Magnitude of covariance dependent upon size and sign of the errors!

∑−

−−=N

iiixy yyxx

N 1

))((1

σ

Positive correlation:

Negative correlation:

No correlation:

0>xyσ0<xyσ

0=xyσ

),( yxfxy ΔΔ=σ

Variation of two random variables together

Both variables are greater than expected.

One variables is greater than expected, while the other is less than expected.

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Review of Error propagation: uncorrelated variables

2

2

22

2

),(

yxq y

q

x

q

yxqq

σσσ ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠

⎞⎜⎝

⎛∂∂

=

=Summation in quadrature:

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

• Measure (x + x) and (y + y)• Compute (q + q)

q qq x y y x y x

y x ∂ ∂= + = +

∂ ∂

Subtract q from both sides of the above equation

( )( )q q x x y y xy x y y x + = + + = + + (neglect xy)

From 1st lecture

What if we cannot neglect xy ?

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Error propagation: correlated variables

xyyxq y

q

x

q

y

q

x

q

yxqq

σσσσ∂∂

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠

⎞⎜⎝

⎛∂∂

=

=

2

),(

22

22

2

“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

• Deducing (unexpected) conclusions from data

• Probability that N measurements of uncorrelated variables would give r as large as observed ro

)(Prob oN rr ≥

QuickTime™ and a decompressor

are needed to see this picture.

Can use:

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Does belong to this sample?

%5)(Prob ≤≥ oN rr

%1)(Prob ≤≥ oN rr

Significant

Highly Significant

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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 Fx = 71 72 73 75 76 77 mmy = 95 96 96 98 98 99 mm

x =74y=97

σ xy =1N

(xi∑ −x)(yi −y) =16(−3)×(−2) + ...+ 3×2( ) =3

r =σ xy

σ xσ y

=(xi∑ −x)(yi −y)

(xi −x)2∑ (yi −y)2∑=0.98

Prob6( r ≥0.98) ≈0.3%

covariance

correlation coefficient

Table C

evidence for linear correlation is both significant and highly significant

0.3 %

y

x

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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 gradesCompare 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?

2 2 2 2 2 2

2 2

1 1(1.42 1.44 1.47 1.4 1.39) 1.424

5

1 1( ) (0.004 0.016 0.046 0.024 0.034 ) 0.0321 0.03

1 40.03

0.01342 0.0135

0.03900

0.001

i

T i

TT

T

T

T T sN

T T s sN

s sN

N

σ

σσ

σσ

= = + + + + =

= − = + + + + = →−

= = = →

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

∑ ∑

∑

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d =44.1±14.7m

Taylor 3.14A visitor to a castle measures the depth of a well by dropping a stoneand timing its fall. She finds the time to fall is: and calculates the depth as:

What is her conclusion, if she takes:With negligible uncertainty?

Solution:

t =3.0 ±0.5sec

d =12

gt2

g =9.80

ms2

Based on:

q(x)=xn

qq= n

xx

d =0.5* 9.80 * 3.02 =44.1mσ d / d=2* (σ t / t) =2 * (0.5 / 3.0) =0.33

σ d =0.33* 44.1=14.7m

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)?

2 2 2 2

25,000 19,000 6,000

3,000 2,500 3,905 4,000

6,000 4,000

6,0001.5

4,000

A B A B

A B

R R R R

A B

R R km

km km

R R km

t

σ σ σ−

− = − =

= + = + = →

− = ±

= =

Table A: Probability to be within 1.5σ 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).

(a)

(b)

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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.

2 2 2 2

2 2 2 2

2 2 2 2

25,000 19,0003,000 2,500 21,459 21,500

1 1 1 13,000 2,500

1 11,921 1,900

1 1 1 13,000 2,500

21,500 1,900

A B

A Bwav

A B

wav

A B

wav

R R

R km km

km km

R km

σ σ

σ σ

σ

σ σ

+ += = = →

+ +

= = = →+ +

= ±

2 2

1 1 1 A A B B

wav A B wavA B A B A B

w x w xx w w

w w w wσ

σ σ+

= = = =+ +

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Chauvenet’s CriterionA 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 2.3Should any of these measurements be dropped?

• Add up all the periods and divide by 14 to get the average, T=2.7 seconds.

• Compute the standard deviation from the data, σ=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).

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??

x =1n

xii=1

n

∑

σ x2 =

1n−1

xi −x( )i=1

n

∑2

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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) Between 75 and 85 points?b) Higher than 90 points?c) Between 70 and 90 points?d) Bellow 65 Points?

Solution:a) This grades range is: +/- 1σ therefore, 68% of the scores should be within this range

275*0.68 = 187 students.• Grades higher than 90 points are 2σ and above the average. Therefore:

• Scores between 70-90 points are within 2σ of the mean. Therefore 95.45% of scores are expected to be within this range.

d) Scores bellow 65 points are more than 3σ bellow the average score. We know that +/-3σ 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(2σ )) * 275 =0.5 * (1−0.9545)* 275 =6

N =0.9545 * 275 =262

N =0.5 * (1−prob(3σ )) * 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 σh. Then we bin the measurements as follows:

– Less than (<h>‐σh): 9

– Between (<h>‐σh) and <h>: 15

– Between <h> and (<h>+σh): 19

– More than (<h>+σh): 7

• 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>‐σh): 0.5 - 0.34 = 0.16 0.16*50= 8

– Between (<h>‐σh) and <h>: 0.34 0.34*50=17

– Between <h> and (<h>+σh): 0.34 0.34*50=17

– More than (<h>+σh): 0.5 - 0.34 = 0.16 0.16*50= 8

Ok 9 15 15 9Ek 8 17 17 8

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

Final Review:TA’s guided, problem solving Sessions:

• Tuesday, Nov 24, 7PM Ledden Auditorium

Enjoy the Holiday Good Luck in the Final!!