PHYSICS 2150 EXPERIMENTAL MODERN PHYSICS. Lecture 6 Chi Square Test, Poisson Distribution

Transcription

1 PHYSICS 2150 EXPERIMENTAL MODERN PHYSICS Lecture 6 Chi Square Test, Poisson Distribution

2 LAB REPORT FORMAT Experiment title Objective : Several sentence description of scientific goal (not to learn about. ) Idea/Technique: One paragraph or two to give background and physics to be tested. What is being measured and how? Apparatus: Explanation of equipment. Drawing diagrams strongly encouraged! (Do not copy the manual!)

3 OBJECTIVE (EXAMPLE) When hydrogen gas is placed in an electrical discharge, the exited atoms may radiate photons at specific wavelength. Hydrogen gas has four wavelengths it may emit in the visible. These are the first four emission lines in the Balmer series. Our objective in this experiment is to find the wavelength of these emission lines. Planck s constant reflects the size of discrete energy levels in quantum mechanics. In performing this experiment, we sought to accurately determine Planck s constant by examining the quantized emission spectrum of the Hydrogen atom.

4 YOUR LAB REPORT: FORMAT V Conclusions: Short discussion summarizing results (give the final result again here) Further comment on uncertainties: explain basis of assigning them, possible hidden errors, etc. Compare results to accepted value: what is level of agreement based on your error estimate? Stick to scientific conclusions! No opinions, no personal comments.

5 CONCLUSIONS (EXAMPLE)... Our calculated value for Planck s constant is around a factor of 3 off from the known value. That is not as close as we would like to get but we are in the same power for the exponent which is not bad....the discrepancies from the accepted value of the de Broglie wavelength at these voltages are σ, 0.459σ, 1.46σ respectively. The error in the first set of measurements is considerably lower than the error in the two final sets. This could be due to our changed method of measurement taking throughout the lab...

6 COMPARE RESULTS TO ACCEPTED VALUE e/m example of lecture 2: Measured Value: e m =(1.95 ± 0.02(stat.) ± 0.06(sys.)) 1011 C/kg Accepted Value: e m = ( ± ) 1011 C/kg Discrepancy from Accepted Value: e m =( ) 1011 C/kg = C/kg

7 COMPARE RESULTS TO ACCEPTED VALUE Significance of Discrepancy: e m total = =2.8 Assess the probability for your result being off by 2.8 sigma: Probability for x inside 2.8σ: erf 2.8 = Probability for x outside 2.8σ: µ 2.8 µ µ µ+ µ erf 2.8 =0.0051

8 LINEAR FITS: EXAMPLE y =(0.53 ± 0.06) x + (0.3 ± 0.3) y =(0.47 ± 0.07) x + (0.6 ± 0.4) y =1.0 8 y = y =(0.4 ± 0.1)x + (1.1 ± 0.6) y =2.0 True Model? y =0.5 x

9 HOW GOOD IS THE FIT? Often want to know how good a fit is. 2 = 2 (f(x i p 1,...,p N ) y i ) 2 Minimising what the best fit to your function was told you The value of that minimum actually fit your function 2 can tell you how well data

10 2 CHI SQUARED TEST test is a particular type of goodness-of-fit test Generally for N measurements of y 1,...,y N : 2 = E i where is the expected value and error on. Need to know i and degree of freedom (dof) : N (number of measurements) minus the number of fitting parameters (e.g. 2 for line, slope and intercept) Now 2 = should be about 1. i y i 2 E i i y i 2 dof

11 CHI SQUARED TEST Look up the probabilities using Appendix D in Taylor: If Prob( 2 20) is high (>5%) then the fit is good Use of this test: determine if the fit is good if bad, either data is bad, or fit function is wrong

12 COUNTING EXPERIMENTS Start with a large of radioactive material with a very long halflife (compared to the experiment). Measure the decays with a detector at a fixed distance. Measure the number of decays in five 100 s intervals. Should the five intervals all have the same number of decays? Should there be a consistent trend?

13 COUNTING EXPERIMENTS Should there be a consistent trend? (in this case Should the decay rate be changing?) No! The long half-life assures that over the time of the experiment, the decay rate isn t changing significantly. Should the five intervals all have the same number of decays? No! The decay is a random process.

14 A POISSON PROCESS Assume a decay rate of 0.01 decays/sec. On average will get 1 decay in 100s. Sometimes we will get zero. Sometimes we will get two. Rarely will even get three or more. What is the probability of getting exactly n decays in 100s? Answer: The Poisson Distribution. Applies to any discrete process where events occur randomly at a constant rate.

15 THE POISSON DISTRIBUTION Probability to observe n counts if the µ mean number of counts P µ (n) = µn n! eµ Probability to observe n counts (%) µ = Number of counts n Distribution is asymmetric Distribution is discrete µ is a positive number

16 PROPERTIES OF THE POISSON Average number of counts: DISTRIBUTION n = = n=0 n=0 np µ (n) = µe µ n=0 n = µ n µn n! e µ = e µ n=1 µ n n! = µe µ e µ n µn n! = µe µ n=1 µ n 1 (n 1)! Average count is characterising parameter of the Poisson distribution

17 PROPERTIES OF THE POISSON DISTRIBUTION Standard deviation: 2 = (n n ) 2 = n 2 2n n + n 2 = n 2 2 n n + n 2 = n 2 n 2 d 2 P µ (n) =1 d µ 2 n=0 0=µ 2 2µ P µ (n)+ (n 2 n)p µ (n) = n 2 µ 2 µ = µ Standard deviation is square root of the mean

18 SMALL MEAN Consider Poisson distribution with µ =1 0.4 Gaussian: µ =1, = µ For small, Gaussian and Poisson is very different 0.1 Poisson: p(0 n 2) = P 1 (0) + P 1 (1) + P 1 (2) = 73.6% Gaussian: p(0 x 2) = 68%

19 LARGE MEAN: GAUSSIAN LIMIT Now let s consider a large 0.15 mean, e.g. µ = Well matched by a Gaussian with and µ =7.5 = Poisson: p(7.5 ± 2.7) = P 1 (0) P 1 (10) = 64% % vs. 64%: close enough Often use Gaussian approximation for large µ, generally for µ>5

20 GAUSS VS. POISSON G µ, (x) = 1 2 Symmetric 0.4 e 1 2( x µ ) 2 P µ (n) = µn n! eµ Asymmetric Continuous (x is real) Continuous (n is integer) Mean : is the most probable value µ µ Mean : most probable value is µ Standard deviation: Distribution describes results of measurements with accuracy 5 Standard deviation: Distribution describes results from counting experiment

21 HOW TO USE POISSON DISTRIBUTIONS IN COUNTING Uncertainty: = µ (for large µ, can be interpreted similarly to Gaussian ) Statistical uncertainty on the number of counts is thus the square root of the number of counts. Background processes may be contributing to your rate: µ = µ sig + µ bkg µ bkg µ. Often, can measure by turning off the signal, so when you then measure, you can subtract the background to measure your signal rate. How do you handle the error in this situation?

22 AN EXAMPLE Claim to have 160 accidents in a 10 year period at the intersection of Broadway and Tablemesa Drive. The City of Boulder decides to install an additional pedestrian light. Over the next two years they record 26 accidents at the intersection. The city makes the claim that the accident rate has gone down. Can they justify their claim statistically??

23 AN EXAMPLE: ANALYSIS Rate before installation of pedestrian light: 160 ± 160 rate b = =16.0 ± Rate after installation of pedestrian light: rate a = Change: 26 ± 26 2 = 13.0 ± 2.6 (3.0 ± 2.8) per year Yes, the city s claim is backed up by statistics!

24 SCIENCE EXAMPLE: RADIOISOTOPE LIFETIME Measure count rate (background) with no radioactive material present Introduce radioactive material Count the number of decays in a 1s period; remeasure every 30 seconds Subtract the background rate Fit the rate vs. time to an exponential lifetime

25 I: MEASURING THE BACKGROUND Trial Counts (in 1 s) Background rate should be constant, so each trial is a re-measurement of the same thing Total 402 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: (402± 402) counts/10s Background rate is 40.2±2.0 counts/s

26 DATA FROM RADIOACTIVE MATERIAL Seconds Counts Corrected ± ± ± ± ±31 933± ±26 658± ±23 486± ±19 313± ±17 245± ±15 177± ±12 110± ±11 82± ±11 72± ±9 46± ±9 38± ±7 13± ±7 11±7 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 counts/s) 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

27 FITTING FOR THE MEAN HALF LIFE 2000 Plot counts vs. time Counts Fit the decay: N(t) =N(0) e t/ Half Life: = (83.8 ± 1.1)s Time (seconds)

28 WHAT YOU SHOULD KNOW FROM LECTURES Error propagation for an arbitrary equation of which variables error are known Understand the difference between statistical and systematic uncertainties. Calculate mean, standard deviation, std on mean and know what they mean. Determine if measurements are compatible and how to compute weighted average.

29 WHAT YOU SHOULD KNOW FROM LECTURES Perform least-squares fit to data and extract the slope and intercepts and their uncertainties. How to determine if two variables are correlated. Understand what a counting experiment is and its distribution can be different from Gaussian.