Sampling& meanandvariance 2 . Let Then n :: = (R 1 + R 2 + . + R - - PowerPoint PPT Presentation

Albert R Meyer, Ma y 13, 2013 confidence.1

Mathematics for Computer Science

MIT 6.042J/18.062J

Sampling & Confidence

Albert R Meyer, Ma y 13, 2013

Pairwise Independent Sampling Let R1,…,Rn be pairwise independent random vars with the same finite mean μ and variance σ2. Let Then

A

n ::= (R1 + R2 +.+ R n) /

n.

Pr[ |An - µ |> δ ] ≤ 1 n σ δ      

2

Theorem:

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coliform count in Charles River for swimming EPA requires average CMD < 200 (Coliform Microbial Density)

Sampling

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Sampling Questions

Make 32 measurements

• f CMD at random

times and locations

1

Then

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Sampling Questions

A few of the 32 counts turn out to be > 200 but their average is 180. Convince the EPA that avg in whole river is < 200?

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Sampling Questions

That is, convince EPA that the estimate based on 32 samples is within 20 of the actual average?

Albert R Meyer, Ma y 13, 2013

Sampling parameters

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c ::= actual average CMD in river

CMD sample ↔ ran var with μ = c n samples ↔ n mutually indep ran vars with μ = c

An ::= avg of the n CMD samples

Albert R Meyer, Ma y 13, 2013

Pr[|An - µ |> δ] ≤ 1 n σ δ      

2

Pairwise Independent Sampling

n = 32, µ = c, δ = 20

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Albert R Meyer, May 13, 2013

Pr[A32 -c|> 20 ] ≤ 1 32 σ 20      

2

Pairwise Independent Sampling

n = 32, µ = c, δ = 20

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?? don’t know (

Albert R Meyer, May 13, 2013

suppose L is max possible difference of samples Bound for (

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= 50 worst σ = L 2

n = 32, µ = c, δ = 20

Pr[A32 -c|> 20 ] ≤ 1 32 σ 20      

2

Albert R Meyer, May 13, 2013

Pairwise Independent Sampling

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1 32 25 20      

2

< 0.05

Pr[ |A32 -c| ≤ 20] > 0.95

Pr[A32 -c|> 20 ] ≤

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Confidence

tempting to say: “the probability that c = 180 ± 20 is at least 0.95”

• -technically wrong!

−not Probable Reality

3

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c is the actual average in the river.

c is unknown,

but not a random variable!

Confidence

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The possible outcomes of our sampling process is a random

• variable. We can say that the

“ probability that our sampling process will yield an average that is ± 20 of the true average at least 0.95”

Confidence

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Tell the EPA that with probability 0.95 our estimate method for avg CMD will be within 20 of the actual avg, c, in the river. Confidence

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For simplicity we say that

c = 180 ± 20 at the 95% confidence level

Confidence

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Confidence Confidence Moral: when you are told that some fact holds at a high confidence level, remember that a random experiment lies behind this claim. Ask yourself “what experiment?”

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Moral: Also ask “Why am I hearing about this particular experiment? How many

• thers were tried and not

reported?” See http://xkcd.com/882/

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