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Lecture 20

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Parameters and statistics Example: A random sample of 1014 voters are asked if they think the President is too liberal, too conservative, or about right. 514 (51%) say ‘too liberal’. – The observed percentage in the sample, 51%, is a statistic. – The unknown percent of the population who would say yes is a parameter. Presumably it’s close to 51%, but we will never know. Question we’d like to ask: “How likely is the statistic to be how close to the parameter?”

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Sampling experiment em/

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Experiment Population (urn) – our class (x people) Question: Dogs vs Cats We will sample 5 – Assign numbers – Use R to sample Repeat 5 times ascertain truth and run R

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US Polls 2012 estimate of the number of eligible voters is 206,072,000. We sampled 1014 people at random and got 514 yes.

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Main idea We learned that given a model we can check if the data is consistent with it Idea: Find models that are consistent with the data.

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US Polls 2012 estimate of the number of eligible voters is 206,072,000. We sampled 1014 people at random and got 514 yes. We will consider various models: – True proportion is p = 0.05, 0.10, 0.15, … – Which of the models is the data in agreement with?

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Results

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“P-value” Consider proportion of fake samples < 514 Values close to 0 or 1 are not consistent with the model

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Cutoff Using better resolution of models A usual cutoff 0.05 split between both sides (0.975 and.025) Models selected: [.477,.538]

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Fake data 3 sub-groups – 42,343,562 (R) – 59,280,986 (D) – 104,447,452 (I) We sample roughly in proportion: – Sampled 220, 201 yes (R) – Sampled 284, 31 yes (D) – Sampled 510, 282 yes (I)

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Issue Too many levers to “fiddle” (three different Ks for each group) Cannot simply look how well the data fits. Needs more sophisticated statistics

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Naïve parametric bootstrap Compute a number/numbers estimated from the data (point estimate). Use this number to simulate a lot of fake data and see how the fake data can vary. Use this variability to estimate the uncertainty in our estimator

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Original problem Estimated K=206,072,000 * (514/1014)=104,458,588 Estimated p in (.476,.537)

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Stratified problem Estimate K for each group – combine to get joint p estimate pcombined=.20* * *.55=.50 – This is not that much different from ignoring stratification – There is a (small) gain in uncertainty – Bootstrap interval (.474,.525)

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Bootstrap Stratified

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Bootstrap SRS

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Homework Due this Thursday 11/19. Modify the bootstrap codes to get the bootstrap intervals for the population proportion of yes in the following two situations: 1.Population of 30000; SRS sample of 100, observe 32 yes. 2.Population of stratified to subpopulations of and Sample 30 observing 27 from fist subpopulation and sample 70 observing 5 from second subpopulation.

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