Statistical Inference (Chapter 10) Statistical inference = learn about a population based on the information provided by a sample.

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1 Statistical Iferece (Chapter 10) Statistical iferece = lear about a populatio based o the iformatio provided by a sample. Populatio: The set of all values of a radom variable X of iterest. Characterized by the distributio (PDF or PMF) of X. The distributio of X depeds o some ukow parameter θ. Goal: lear about θ. Sample: Radomly select a sample of size from the populatio. Sample data: X 1, X 2,, X The sample observatios are idepedet, ad the distributio of each idividual observatio is same as the distributio of the populatio. I other words, the data are idepedetly ad idetically distributed (i.i.d.). Ex: What proportio of Americas thik Presidet Bush is doig a good job? Describe the populatio radom variable ad the parameter of iterest. Modes of statistical iferece: Parameter (poit) estimatio: Estimate θ by a sigle value. Cofidece itervals: Estimate θ by a iterval of values. Hypothesis testig: Test whether data are cosistet with a particular value of θ. Model fittig: Fit a model to the data. Parameter estimatio Statistic = ay fuctio W(X 1,, X ) of sample data. Estimator of θ = ay statistic used to estimate θ deoted as ˆ θ. Desirable properties of ˆ θ : 1. Ubiasedess ˆ θ is ubiased for θ if E( ˆ θ ) = θ. Estimator is correct o average. 1

2 2. Small variace Variace = ucertaity. Larger variace = less precise. We would like to have small variace or high precisio. 3. Cosistecy ˆ θ is cosistet for θ if P ( ˆ θ θ > ) 0 as. I other words, i the limit, ˆ θ = θ. Sufficiet coditio for cosistecy if ˆ θ is ubiased: Var[ ˆ θ ] 0 as. Ex 1: Estimatio of populatio mea µ = E(X) = θ. Suppose populatio variace = σ 2. Natural estimator: Ubiased? Cosistet? Does it have the smallest variace? Ex 2: Estimatio of populatio proportio of successes p = E(X) = θ. X is Beroulli (p). Populatio cosists of 0 s (failures) ad 1 s (successes). p is also a mea. Natural estimator: Similar properties as sample mea i Ex 1. Ex 3: Estimatio of populatio variace σ 2 = Var(X) = θ. Natural estimator: Also ubiased ad cosistet. 2

3 I geeral, the choice of a estimator may ot be obvious. Two geeral methods of estimatio: Method of momets ad method of maximum likelihood. - Both produce cosistet estimators. - Maximum likelihood is more popular the best if is large. - Oly the method of momet i this course as it is simpler (ad does ot require calculus!). - But remember that the method of momet estimator may ot be the best. Method of momets: k k-th populatio momet: µ = E( X ) k-th sample momet: M k 1 = k k X i i= 1 To estimate d parameters, solve the system of d equatios, M1 Md = L = µ µ 1 d Their solutio = MOME. The # equatios should be equal to the # parameters i the model. Ex: (Presidet Bush example) Populatio: X ~ Beroulli (p), where p = proportio of Americas who approve Presidet Bush s job performace. Parameter: p Sample: X 1, X 2,, X. What is the MOME of p? Is it ubiased? Is it cosistet? 3

4 Ex: Assume that the time to failure i years, X, of a telephoe switchig system is expoetially distributed with parameter λ (> 0). What is the MOME of λ based o a radom sample of times to failure, say X 1,, X? Is it ubiased? Is it cosistet? Ex: Derive the MOMEs of mea µ ad variace σ 2 of a N(µ,σ 2 ) distributio based o a radom sample X 1,, X from this populatio. 4

5 Cofidece itervals (sectio ) Estimator ˆ θ is a sigle umber that gives a plausible value of the ukow θ. Rarely the two will be equal. So, ofte it is preferable to give a iterval of plausible values a cofidece iterval (CI), which cotais the ukow θ with high probability. 100 (1 α) % cofidece iterval for θ = A iterval [L, U], where L = L(X 1,, X ) ad U = U(X 1,, X ) are statistics such that P L θ U = 1. ( ) α L ad U are radom, so the CI is radom. Parameter θ is ot radom it is ukow but fixed. (1 α) = cofidece coefficiet or cofidece level. I practice, (1 α) = 90% or 95% (most commo) or 99%. 100 (1 α) % CI for mea µ of a Normal (µ, σ 2 ) populatio (assumig σ 2 kow): X z σ σ / 2, X + zα /, α 2 where z α / 2 is such that area uder the stadard ormal curve to the right of z α / 2 = α/2. Derivatio: Step 1: Estimate θ = µ by its estimator X 1 = X i. Step 2: Recall that X ~ Normal 2 σ, X µ =. σ / µ. Therefore, Z ~ Normal( 0,1) Step 3: Fid critical values z α / 2 Step 4: The we have Step 5: Solve for µ to get: ± such that P ( z Z ) = 1 α. α / 2 zα / 2 X µ P zα / 2 zα / 2 = 1 α σ / P X zα / 2σ zα / 2σ µ X + = 1 α 5

6 Ex: Suppose that a observed sample of size 10 from a N(µ, 4) populatio gives x = Fid the 95% CI for µ. Notice that this iterval is fixed it s a umerical iterval. There is othig radom about it. Questio: Ca we say that this observed iterval cotais the true value of µ with 95% probability? So, how do we really iterpret a CI? Recall, σ σ P X zα / 2 µ X + zα / 2 = 1 α = 0.95, I other words, if we repeat a large umber of times the process of takig a radom sample of size from Normal (µ, σ 2 ) populatio ad costruct the CI usig the above formula, the roughly 95% of times the observed CI s will be correct, i.e., it will capture the true value of µ. This CI formula gives a icorrect iterval oly 5% (small) of the times. It is wrog to say that the observed iterval cotais the true value of µ with 95% probability. It either cotais the true value of µ or it does ot we do t kow what the case is. Thus, i a sese, we have 95% cofidece i the CI formula it gives the correct aswer 95% of the times. Ex: Let s use simulatio to verify this iterpretatio. I draw a radom sample of size 10 from N(5, 10) distributio ad use the observed sample to costruct a 95% CI for µ usig the above formula. I repeated this procedure 100 times. The figure below plots the costructed CI s. 6

7 Roughly, 7% of these 100 itervals do ot cover the true value of µ, which is 5 i this simulatio. But 100 is ot cosidered a large # of repetitios i simulatio. So, if I repeat this process times, the oly 5.2% of the itervals do ot cover the true value 5. I other words, roughly 94.8% of the itervals cover the true value of µ. Questio: Give a radom sample, which CI would you prefer: a 95% CI or a 99% CI? Ex: Suppose that a observed sample of size 10 from a N(µ, 4) populatio gives x = Fid the 95% level ad 99% level CI s forµ. 7

8 Note the tradeoff: The precisio of a CI is give by its width. The reliability of a CI is give its cofidece level. Higher cofidece = lower precisio (wider). The width of a 100% CI is:. It is a useless iterval: extremely reliable but extremely imprecise! Questio: What ca we do to get a arrower CI without lowerig the cofidece? σ Width = 2 zα / 2 Icrease to make CI more precise. Choosig the sample size : Let w = desired CI width for 1 α cofidece. Margi of error = 2w σ Set 2 z α / 2 = w 2 σ Solvig for gives: = 2zα / 2 w Ex: Suppose that we wish to estimate the mea CPU service time of a job ad we wish to assert with 99% cofidece that the estimated value is withi less tha 0.5 s of the true value. Suppose that the past experiece suggests that CPU service time is ormally distributed with stadard deviatio σ = 1.5 s. How may observatios should we take? 8

9 100 (1 α) % CI for mea µ of a Normal (µ, σ 2 ) populatio (assumig σ 2 ukow): Ukow σ 2 is more realistic. Estimate σ by sample variace S = ( X i X ) 1 i X µ Use the fact that Z = ow follows a t distributio with ( 1) degrees of freedom, istead S / of the Normal (0, 1) distributio. A t 1 -distributio looks like a Normal (0, 1) but it has heavier tails. A heavier tail accouts for the fact that there is there is more ucertaity i Z whe S is used i place of σ. Whe is large, a t 1 -distributio Normal (0, 1). 2 Result: CI for µ = X ± t α / 2, 1 S The t critical poits are tabulated i the Appedix. Sample size calculatio ow becomes complicated tha before because S eeds to be kow before data are collected. Oe optio is to make a itelliget guess about S ad be coservative (guess a larger value of S so that larger tha ecessary is chose). Ex: Some studies of Alzheimer s disease (AD) have show a icrease of 14 C0 2 productio i patiets with the disease. I oe such study the followig 14 C0 2 values were obtaied from 16 eocortical biopsy samples from AD patiets Fid the 95% CI for mea 14 C0 2 value assumig that 14 C0 2 values are ormally distributed. 9

10 Large Sample Cofidece Itervals Populatio mea µ: So far we assumed that the populatio was ormal. Questio: Ca we use this CI eve whe the populatio is ot ormal? Yes, provided is large. S 100(1 α)% CI for mea µ of ay populatio: X ± zα / 2 Works because of the cetral limit theorem, which says X is ormal whe is large. Rule of thumb: > 30. May ot be valid for smaller sample sizes. Ex: We wish to estimate the mea executio time of a program. The program was ru 35 times o radomly selected iputs, ad the sample mea ad the sample stadard deviatio of the executio times were evaluated as x = 230 ms ad s = 14 ms, respectively. Fid a 95% CI for the true mea executio time µ. Populatio proportio of successes p: Populatio: X ~ Beroulli (p), where p = proportio of successes i populatio. Sample data: X 1,., X. (Note: they are 0 s ad 1 s). Recall: Estimator for p = p ˆ = X = proportio of successes i the sample Questio: What is the approximate distributio of pˆ for large? p(1 p) CLT whe is large, pˆ ~ Normal p, p ˆ(1 pˆ ) Estimate it Var( pˆ ) as. 10

11 pˆ p For large, Z = pˆ(1 pˆ) ~ Normal(0,1). Result: 100(1-α)% CI for p = pˆ(1 pˆ) pˆ ± z α / 2 where p ˆ = X. Rule of thumb: pˆ 10 ad ( 1 pˆ) 10. Ex: From a large populatio of RAM chips, a radom sample of 50 is take ad a test carried out o each to see whether they perform correctly. I the test, oly 20 chips are foud to perform correctly. Fid a 95% CI for p, the true proportio of chips that perform correctly. Choosig the sample size : Width of CI = 2 z α / 2 pˆ(1 pˆ) Let w = desired CI width for 1 α cofidece. Margi of error = 2w pˆ(1 pˆ ) Set 2 z α / 2 = w 2 4zα Solvig for gives: / 2 pˆ(1 pˆ ) = 2 w This formula ivolves pˆ, which is ot kow before the experimet. Oe alterative: take pˆ = 0.5 because pˆ (1 pˆ ) is maximum whe p ˆ = This strategy will yield a coservative values of. (The sample size will be larger tha ecessary.) 11

12 Ex: Suppose we are plaig a survey to estimate the proportio of America who approve of Presidet Bush s job. We would like our estimate to be withi 3% of the true proportio with 95% cofidece. How much sample size should we take? 12

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