Central Limit Theorem• Example:
(NOTE THAT THE ANSWER IS CORRECTED COMPARED TO NOTES5.PPT)
– 5 chemists independently synthesize a compound 1 time each.– Each reaction should produce 10ml of a substance.– Historically, the amount produced by each reaction has been normally
distributed with std dev 0.5ml.1. What’s the probability that less than 49.8mls of the substance are made in
total?2. What’s the probability that the average amount produced is more than
10.1ml?3. Suppose the average amount produced is more than 11.0ml. Is that a rare
event? Why or why not? If more than 11.0ml are made, what might that suggest?
Answer:• Central limit theorem:
If E(Xi)= and Var(Xi)=2 for all i (and independent) then:X1+…+Xn ~ N(n,n2)
(X1+…+Xn)/n ~ N(,2/n)
Lab:
1. Let Y = total amount made. Y~N(5*10,5*0.52) (by CLT)Pr(Y<49.8) = Pr[(Y-50)/1.12 < (49.8-50)/1.12]=Pr(Z < -0.18) = 0.43
2. Let W = average amount made.W~N(10,0.52/5) (by CLT)Pr(W > 10.1) = Pr[Z > (10.1 – 10)/0.22]=Pr(Z > 0.45) = 0.33
Lab (continued)
3. One definition of rare:It’s a rare event if Pr(W > 11.0) is small(i.e. if “Seeing probability of 11.0 or something more extreme is small”)Pr(W>11) = Pr[Z > (11-10)/0.22] = Pr(Z>4.55) = approximately zero.
This suggests that perhaps either the true mean is not 10 or true std dev is not 0.5 (or not normally distributed…)
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
Sample size: 1006(source: gallup.com)
• Let Xi = 1 if person i thinks the Presidentis hiding something and 0 otherwise.
• Suppose E(Xi) = p and Var(Xi) = p(1-p) and each person’s opinion is independent.
• Let Y = total number of “yesses”= X1+…+ X1006
• Y ~ Bin(1006,p)• Suppose p = 0.36 (this is the estimate…)• What is Pr(Y < 352)?
Note that this definitionturns three outcomes intotwo outcomes
Normal Approximation to the binomial CDF
– Even with computers, as n gets large, computing things like this can become difficult. (1006 is OK, but how about 1,000,000?)
– Idea: Use the central limit theorem approximate this probability– Y is approximately
N[1006*0.36, (0.36)*(0.64)*1006]= N(362.16,231.8) (by central limit theorem)
Pr[ (Y-362.16)/15.2 < (352-362.16)/15.2]= Pr(Z < -0.67) = 0.25
Pr(Y<352) = Pr(Y=0)+…+Pr(Y=351), where Pr(Y=k) = (1006 choose k)0.36k0.641006-k
Normal Approximation to the binomial CDF
Black “step function” is plots of bin(1006,0.36) pdf versus Y (integers)
Blue line is plot of Normal(362.16,231.8) pdf
Normal Approximation to the binomial CDF
Area under blue curve toleft of 352
is approximately equal to the
sum of areas ofrectangles (blackStepfunction) to the left of 352
Comments about normal approximation of the binomial :
Rule of thumb is that it’s OK if np>5 and n(1-p)>5.
“Continuity correction”
Y is binomial.
If we use the normal approximation to the probability that Y<k, we should calculate Pr(Y<k+.5)
If we use the normal approximation to the probability that Y>k, we should calculate Pr(Y<k-.5)
(see picture on board)
Probability meaning of 6 sigma
• Even if you shift the process mean for the center of the specifications to 1.5 standard deviations toward one of the specifications, then you will expect no more than 3.4 out of a million defects outside of the specification toward which you shifted.
• (I know it’s convoluted, but that’s the definition…)
What does 6 sigma mean?(example)
• Suppose a product has a quantitative specification:ex: “Make the gap between the car door and the car body between 3.4 and 4.6mm.”
• When cars are actually made, the “std dev of car door gap is 0.1mm”. i.e. X1,…,Xn are gap widths. The sqrt(sample variance of X1,…,Xn)= 0.1mm
Lower specification
Upper specification
3.4mm 4.6mm4.6 – 3.4 = 1.2 = 12*0.1 = 12*sigma
Statistically, six sigma means that Upper Spec – Lower Spec > 12 sigma(i.e. Specs are fixed. Lower the manufactuing process variability.)
Center of spec = 4mm gap
Shifted mean= 3.85mm gap
Distribution of gap widths
Probability of beingout here is Pr( gap is less than 3.4 ) = Pr( (gap – 3.85)/0.1 < (3.4-3.85)/.1)
=Pr( Z < -4.5) = 3.4/1,000,000 Arbitrary “magic” number for 6
In general:Assume process mean is 1.5 standard deviations toward the lower spec: i.e. E(X)=4-1.5 and assume X has a normal distribution.When the process is in control enough so that the distance between the center of the specs and the lower spec is least 6, thenPr(X below lower spec) =Pr( X<4- 6)=Pr[(X- (4-1.5-6(4-1.5 ] =Pr(Z<-4.5) = 3.4/1,000,000
Probability meaning of 6 sigma
Control Charts
• Let X = an average of n measurements.• Each measurement has mean and
variance 2.• Fact:
– By the central limit theorem, almost all observations of X fall in the interval +/- 3/sqrt(n) (i.e. mean +/- 3 standard deviations)
– /sqrt(n) is also called x or standard error
Use the “fact” to detect changes in production quality
• Idea: let xi = average door gap from the n cars made by shift i at the car plant
+3 /sqrt(n)(Upper Control Limit)
-3 /sqrt(n)(Lower Control Limit)
shift
x1
x2
x3
x4
x5
x6
x7
x8
Points outside the +/- 3 std error bounds, are called “out of control”. They are evidence that and or are not the true mean and std dev any more, and the process needs to be readjusted. Calculate the “false alarm rate”… (= 26/10,000)
Assume 100 new people arepolled.
Assume true pr( a new person says yes) = 0.36.
Let P = “P hat”= number say yes/100
What’s an approximation tothe distribution of P-hat?
Use the approximation todetermine a number so thatthe Pr(p-hat> that number) = 0.95.
QuickTime™ and aTIFF (Uncompressed) decompressor
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EXAMPLE OF SAMPLING DISTRIBUTION OF P-HAT
Xk = 1 if person k says yes and 0 if not.
Note that E(Xk)=0.36=p and Var(Xk)=0.36*0.64=p(1-p) Note that Xk is binomial(1,0.36).
P-hat = (X1+…+X100)/100. By CLT, P-hat is approximately N(0.36,0.36*0.64/100).
(Rule of thumb is that this approximation is good if np>5 and n(1-p)>5.)
• Suppose true p is 0.36.• If survey is conducted again on 100 people, then
P-hat ~ N(.36,(.36)(.64)/100) = N(.36, 0.002304)
Want p0 so that Pr(P-hat<p0) = 0.95 Pr(P-hat<p0) = 0.95 means Pr(Z < (p0-.36)/0.048) = 0.95.Since Pr(Z<1.645) = 0.95,(p0-.36)/0.048 = 1.645(p0-.36) = 0.07896p0 = 0.43896
• Suppose true p is 0.40.
• If survey is conducted again on 49 people, what’s the probability of seeing 38% to 44% favorable responses?
Pr( 0.38 < P ”hat” < 0.44)
= Pr[(0.38-0.40)/sqrt(0.40*0.60/49) < Z < (0.44-0.40)/sqrt(0.40*0.60/49) ]
= Pr(-0.29 < Z < 0.57)= Pr(Z<0.57) – Pr(Z<-0.29)= 0.7157-0.3859=0.3298
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