How To Understand The Correlation Between Weight And Height In Cui

Physics  2150  

Experimental  Physics  2  

Dmitry  Reznik  

Lecture  6:    

Correla@on  and  covariance  con@nued   Poisson  Distribu@on  

FCQs        

Homework  Due  Oct  12,  5:00  pm  

hLp://www.colorado.edu/physics/phys2150/ phys2150_fa15/Homework.pdf  

Last  Time  

Zero  when  no  correla@ons  

Posi@ve  when  posi@ve  correla@on:  if                          then  

σ _xy = 1 N _i=1

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Coefficient  of  linear  correla@on  

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r  indicates  how  well  pairs  x_i,y_i  sa@sfy  a  linear  rela@on.  

Example:  Is  there  a  correla@on  

between  weight  and  square  of  height  

of  CU  students?  

Weight  (x)   height    

squared  (y)   What  is  the  degree  of  correla@on?  

σ_weight =σ _x =15.48 σ _{height squared} = σ _y = 0.75 covariance   σ _xy = 1 N _i=1

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High  degree  of  correla@on  

3.619272 5.650122 14.025982 3.968822 5.028952 ,0.166658 2.032722 23.305522 21.991542 0.789802 x_i − x ( )₍y_i − y₎

Coun@ng  experiments  

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Examples:      

 coun@ng  radioac@ve  decay  events    birth  rates  

 rate  of  rare  diseases    car  accidents  

 failure  rate  of  oscilloscopes  in  the  Physics  1140  lab    traffic  flow  

Coun@ng  random  events  that  occur  at  a   well-­‐defined  average  rate  

Coun@ng  experiments  

Why  Gaussian  distribu@on  cannot  be  correct?  

a)  Because  randomness  in  the  case  of  coun@ng   experiments  is  not  really  random  

b)  Because  Gaussian  distribu@on  is  nonzero  for   nega@ve  values  

c)  Gaussian  distribu@on  does  not  work  for  large   numbers  of  counts.  

Poisson  Distribu@on  

P

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n

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Probability  that  we  will  have  n  counts  provided  that  the   average  number  of  counts  is  µ

Average  number  of  counts  per  seconds:  1  

What  is  the  probability  of  gecng  zero  counts  in  two  seconds?  

a)  0   b)  e-­‐1       c)  e-­‐2=0.135..   d)  2e-­‐2   n=0   µ=2  counts/2  seconds   P_{µ ( )}n = e−2 2 0 0! = e −2

What  is  the  probability  of  gecng  -­‐1  

counts??  

2e−2 e−2 2 2 2! = 2e −2 e−2 2 3 6 = 4 3 e −2 e−2

P

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Poisson  distribu@on  for  µ=2  

0" 0.05" 0.1" 0.15" 0.2" 0.25" 0.3" (10" (5" 0" 5" 10" 15" Series1"

Poisson  Distribu@on  

Very  Asymmetric  for  small  µ

If  m  is  an  integer:  two  most  

probable  points  are       Pµ µ

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If  m  is  not  an  integer:  the   most  probable  point  is     n=Int  (µ)  (round  down  µ)  

Approaches  Gaussian  distribu@on  for  large  µ

LARGE-MEAN POISSON

How  to  Use  Poisson  Distribu@on  in  

Coun@ng  

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For  large  µ  can  be  interpreted  similarly  to  Gaussian  σ

Also  add  in  quadrature  

Background  processes  may  be  contribu@ng  to  your   rate:  μ=μ_signal+μ_bkg.  Ohen,  can  measure  μ_bkg  by  

turning  off  the  signal,  so  when  you  then  measure   μ_total  you  can  subtract  the  background  to  measure   your  signal  rate.  

An  Example  

Claim  to  have  160  accidents  in  a  10  year  period  at  the   intersec@on  of  Broadway  and  Table    Mesa  Drive.  

How  many  accidents  per  year?   a) 16  +/-­‐  0.4   b) 16  +/-­‐  4   c) 16  +/-­‐  1.26   10 years :160 ± 160 =160±12.6 1 year: 1 10 ∗10 years= 1 10 ∗

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An  Example  

The  City  of  Boulder  decides  to  install  a  turn-­‐only  lane.  Over   the  next  two  years  they  record  26  accidents  at  the  

intersec@on    

The  city  makes  the  claim  that  the  accident  rate  has  gone   down.  Can  they  jus@fy  their  claim  sta@s@cally?  

(16  +/-­‐  1.26)  per  year  -­‐  (13.0+/-­‐  2.55)  per  year  =  (3  +/-­‐  2.8  )   accidents  per  year  –  Yes  it  is  significant!  (but  barely)  

A  Science  Example:  Ficng  

Radioisotope  Life@me  

•  Measure  count  rate  (background)  with  no   radioac@ve  material  present  

•  Introduce  radioac@ve  material  

•  Count  the  number  of  decays  in  a  1-­‐second   period;  remeasure  every  30  seconds  

•  Subtract  the  background  rate  

Measuring  Background  

•  Background  rate  should  be  constant,  so  each   trial  is  a  remeasurement  of  the  same  thing  

•  Get  the  background  rate  by  taking  mean:  

402/10=40.2  (same  as  measuring  for  10  sec.  and   divifing  by  10)    

•  Uncertainty  in  the  rate:  √402/10=20.0/10=2.0  

•  Background  rate  is  40.2±2.0  counts/second  

MEASURING BACKGROUND

• Background rate should be constant, so each trial is a remeasurement of the same thing

• Get the background rate by taking mean:

402/10=40.2 (note that this is exactly equivalent to simply measuring for 10 seconds and dividing by 10 to find the rate).

• Uncertainty in the rate: total counts is 402± 402 • So in 10 seconds, mean = 402±20

• Divide by 10 to get rate in 1 sec: 40.2±2.0

Signal  Data  

SIGNAL DATA

• Are all the trials measuring the same rate? • What are the uncertainties in the number of

counts?

• What is the measured signal rate?

• Subtract background (40.2 cts/sec)

• Statistical error remains the square root of the total number of counts

• Also there is a systematic error (not shown) due to uncertainty in the

background

What  is  the  measured  signal  rate?   What  are  the  uncertain@es  in  the   number  of  counts?  

•  Subtract  background  (40.2  cts/sec)   •  Sta@s@cal  error  remains  the  square   root  of  the  total  number  of  counts  

•  Also  there  is  a  systema@c  error  (not   shown)  due  to  uncertainty  in  the  

Ficng  for  the  mean  life@me  

FITTING FOR THE MEAN

LIFE

• Plot the rate vs. seconds • To fit to a line, take the log:

• N(t) = N0 exp( t/ ) ln(N) = ln(N0) t/ • Linear fit: y= ln(N); x=t.

• What’s the uncertainty on y if uncertainty on N is N?

Plot  the  rate  vs.  seconds    

Linear  fit:  y=  ln(N);  x=t.    

To  fit  to  a  line,  take  the  log:   N(t)  =  N₀exp(−t/τ)  ➠    

ln(N)  =  ln(N₀)  −  t/τ    

1/τ  is  the  slope  of  the  line