Statistics Lectures: Questions for discussion

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Statistics Lectures: Questions for discussion. Louis Lyons Imperial College and Oxford CERN Summer Students July 2014. Combining errors. z = x - y δ z = δ x – δ y [1] Why σ z 2 = σ x 2 + σ y 2 ? [2]. Combining errors. z = x - y - PowerPoint PPT Presentation

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PHYSTAT2011

Statistics Lectures: Questions for discussionLouis Lyons Imperial College and Oxford

CERN Summer Students July 20141Combining errors z = x - y z = x y [1]Why z2 = x2 + y2 ? [2]

3Combining errors z = x - y z = x y [1]Why z2 = x2 + y2 ? [2]

[1] is for specific x, yCould be so on average ? N.B. Mneumonic, not proof

2) z2 = z2 = x2 + y2 2 x y = x2 + y2 provided..43) Averaging is good for you: N measurements xi [1] xi or [2] xi /N ?

4) Tossing a coin:Score 0 for tails, 2 for heads (1 1)After 100 tosses, [1] 100 100 or [2] 100 10 ?

0 100 200Prob(0 or 200) = (1/2)99 ~ 10-30Compare age of Universe ~ 1018 seconds

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N.B. Better to combine data!BEWARE 10010 21? 11 or 50.55?

Combining Experimental Results6For your thought Poisson Pn = e - n/n!P0 = e P1 = e P2 = 2 e- /2

For small , P1 ~ , P2 ~ 2/2 If probability of 1 rare event ~ , why isnt probability of 2 events ~ 2 ?

7P2 = 2 e- or P2 = 2 /2 e- ?1) Pn = e - n/n! sums to unity2) n! comes from corresponding Binomial factor N!/{s!(N-s)!} 3) If first event occurs at t1 with prob , average prob of second event in t-t1 is /2. (Identical events)4) Cow kicks and horse kicks, each producing scream.Find prob of 2 screams in time interval t, by P2 and P2

2c, 0h (c2e-c) (e-h) ( c2e-c) (e-h)1c,1h (ce-c) (he-h) (ce-c) (he-h)0c,2h (e-c) (h2e-h) (e-c) ( h2e-h)Sum (c2 + hc + h2) e-(c+h) (c2 + 2hc + h2) e-(c+h)2 screams (c+h)2 e-(c+h) (c+h)2 e-(c+h) Wrong OK78PARADOXHistogram with 100 binsFit with 1 parameterSmin: 2 with NDF = 99 (Expected 2 = 99 14)

For our data, Smin(p0) = 90Is p2 acceptable if S(p2) = 115?

YES. Very acceptable 2 probability

NO. p from S(p0 +p) = Smin +1 = 91 But S(p2) S(p0) = 25 So p2 is 5 away from best value

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Comparing data with different hypothesesPeasant and Dog

Dog d has 50% probability of being 100 m. of Peasant pPeasant p has 50% probability of being within 100m of Dog d ?14dpxRiver x =0River x =1 km

15Given that: a) Dog d has 50% probability of being 100 m. of Peasant, is it true that: b) Peasant p has 50% probability of being within 100m of Dog d ?

Additional information Rivers at zero & 1 km. Peasant cannot cross them.

Dog can swim across river - Statement a) still trueIf dog at 101 m, Peasant cannot be within 100m of dogStatement b) untrue

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