STATISTICAL INFERENCE
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1 STATISTICAL INFERENCE POPULATION AND SAMPLE Populatio = all elemets of iterest Characterized by a distributio F with some parameter θ Sample = the data X 1,..., X, selected subset of the populatio = sample size Examples of F : Beroulli(p), Normal(µ, σ), Gamma(, λ), Poisso(λ), etc. 1
2 Statistical Iferece Statistical Iferece = iferece about the populatio based o a sample Parameter estimatio Cofidece itervals Hypothesis testig Model fittig 2
3 Statistical Iferece Parameter Estimatio Statistic = ay fuctio of data W (X 1,..., X ) Estimator of θ = ay statistic used to estimate parameter θ Estimator ˆθ is ubiased if E(ˆθ) = θ Stadard error of a estimator is its stadard deviatio Std(ˆθ) It is estimated by Ŝtd(ˆθ). It shows the accuracy, reliability of estimator ˆθ. 3
4 Parameter estimatio Estimatio of a mea Sample (X 1,..., X ) is collected from a populatio with E(X) = µ ad Var(X) = σ 2. Estimate the populatio mea θ = µ = E(X i ) by a sample mea Properties: E( X) = 1 Var( X) = 1 2 X = 1 EX i = 1 X i θ = θ VarX i = σ2 2 = σ2 So, X is ubiased, ad its stadard error is SE( X) = Std( X) = σ 4
5 Parameter Estimatio Estimatio of a variace Estimate the populatio variace by a sample variace θ = σ 2 = Var(X i ) S 2 = 1 1 (X i X) 2 It is also ubiased: E(S 2 ) = σ 2. The, the stadard error of X is estimated by Ŝtd( X) = S = (Xi X) 2 ( 1) 5
6 Parameter estimatio Estimatio of a proportio Sample (X 1,..., X ) is collected from Beroulli populatio with parameter p. Estimate the populatio proportio p = E(X i ) by a sample proportio ˆp = 1 X i = umber of X i = 1 Special case of a sample mea X E(ˆp) = p, Var(ˆp) = σ2 = p(1 p) So, ˆp is ubiased; p(1 p) its stadard error is SE(ˆp) = typically estimated by ŜE(ˆp) = ˆp(1 ˆp) 6
7 Parameter Estimatio Geeral Methods of Estimatio 1. Method of Momets k th populatio momet µ k = EX k k th sample momet M k = 1 Xi k To estimate d parameters, solve the system of d equatios M 1 = µ 1... M d = µ d M 1,..., M d are kow from the sample; µ 1,..., µ d are fuctios of ukow parameters 7
8 Method of momets Example: X 1,..., X are Expoetial(λ) Estimate λ. The umber of parameters is d = 1. eed 1 equatio. So, we M 1 = X; µ 1 = 1 λ Solve M 1 = µ 1 X = 1 λ ˆλ mom = 1 X 8
9 2. Method of Maximum Likelihood Maximize the probability (pmf, pdf) of seeig the really observed data Implemetatio Observe X 1,..., X from pdf or pmf f(x θ). Maximize f(x 1,..., X θ) = i θ. Simplificatio: maximize l f(x 1,..., X θ) = f(x i θ) l f(x i θ) Typically, compute θ l f(x 1,..., X θ) = equate to 0 ad solve i θ. θ l f(x i θ) 9
10 Method of maximum likelihood Example: X 1,..., X are Expoetial(λ) f(x 1,..., X θ) = λe λx i l f(x 1,..., X θ) = = l λ λ λ l f(x 1,..., X θ) = λ Solve for λ, ˆλ mle = Xi = 1 X l ( λe λx ) i X i X i =: 0 10
11 Maximum likelihood 2h f(x) This area = P {x h x x+h} (2h)f(x) x h x x+h Probability of observig almost X = x 11
12 Statistical Iferece Cofidece Itervals 100 (1 α) %-cofidece iterval is a iterval that cotais parameter θ with probability γ. That is, P {a θ b} = 1 α where a = a(x 1,..., X ) ad b = b(x 1,..., X ) are statistics. So, a ad b are radom, θ is ot. 12
13 Cofidece itervals Cofidece itervals for the same parameter θ obtaied from differet samples of data θ Cofidece itervals ad coverage of parameter θ. 13
14 Cofidece itervals Example: X 1,..., X from Normal(µ, σ) with ukow µ, kow σ 1. Estimate θ = µ by its estimator X = 1 Xi. 2. Fid its distributio: Normal with E( X) = µ Var( X) = Therefore, Z = X µ σ/ Var(X i ) = σ2 2 is Normal(0,1) = σ2 3. Fid critical values ±z α/2 such that for Z Normal(0,1). P { z α/2 < Z < z α/2 } 14
15 Cofidece itervals 4. The we have P { z α/2 < X µ σ/ < z α/2 } = 1 α Solve for µ: P { X z α/2 σ < µ < X + z α/2 σ } = 1 α 5. Hece, X ± z α/2 σ = [ X z α/2 σ, X + z α/2 σ ] is a (1 α)100% cofidece iterval for µ. X is approximately Normal for large ad ay distributio of X 1,..., X. 15
16 Cofidece itervals Whe σ is ukow Data X 1,..., X from Normal(µ, σ) with ukow µ, ukow σ 1. Estimate σ by S = ( Xi X ) 2 2. Use t-distributio with ( 1) degrees of freedom istead of Normal. For large, use Normal approximatio. Result: X ± t α/2, 1 S 16
17 TESTING HYPOTHESES Hypothesis H 0 ad alterative H A = mutually exclusive statemets about the ukow parameter θ. Collect data Coduct a test State if there is sufficiet evidece to reject H 0 i favour of H A. Coclusio Reject H 0 Accept H 0 H 0 is true Type I error correct H 0 is false correct Type II error Cotrol the sigificace level α = P { Type I error } 17
18 Hypotheses testig Data: X 1,..., X from Normal(µ, σ) with ukow µ, kow σ Test H 0 : µ = µ 0 vs H A : µ µ Fid ±z α/2. Acceptace regio: [ z α/2, z α/2 ]. 2. Compute the test statistic Z = X µ 0 σ/. 3. If Z belogs to the acceptace regio, do ot reject H 0. Otherwise, reject H 0. If H 0 is true, Z has Normal(0,1) distributio, ad P { Type I error } = P { Z > z α/2 } = α 18
19 Hypotheses testig Oe-sided, right-tail tests Test H 0 : µ = µ 0 vs H A : µ > µ Fid z α. The acceptace regio is (, z α ]. 2. Compute the test statistic Z = X µ 0 σ/. 3. If Z belogs to the acceptace regio, do ot reject H 0. Otherwise, reject H 0. 19
20 Hypotheses testig Oe-sided, left-tail tests Test H 0 : µ = µ 0 vs H A : µ < µ Fid z α. The acceptace regio is [ z α, + ). 2. Compute the test statistic Z = X µ 0 σ/. 3. If Z belogs to the acceptace regio, do ot reject H 0. Otherwise, reject H 0. 20
21 Hypotheses testig Case of ukow variace 1. Estimate σ by S = ( Xi X ) 2 2. Use t-distributio with ( 1) degrees of freedom. For large, use Normal approximatio. 21
22 Hypotheses testig, Z-tests Null hypothesis Parameter, estimator If H 0 is true: Test statistic H 0 θ, ˆθ E(ˆθ) Var(ˆθ) Z = ˆθ θ 0 Var(ˆθ) Oe-sample Z-tests for meas ad proportios, based o a sample of size µ = µ 0 µ, X µ 0 σ 2 p = p 0 p, ˆp p 0 p 0 (1 p 0 ) X µ 0 σ/ ˆp p 0 ˆp(1 ˆp) Two-sample Z-tests comparig meas ad proportios of two populatios, based o idepedet samples of size ad m µ X µ Y = D µ X µ Y, X Ȳ D σ 2 X + σ2 Y m X Ȳ D σ 2 X + σ2 Y m p 1 p 2 = D p 1 p 2, ˆp 1 ˆp 2 D p 1 (1 p 1 ) + p 2(1 p 2 ) m ˆp 1 ˆp 2 D ˆp 1 (1 ˆp 1 ) + ˆp 2(1 ˆp 2 ) m 22
23 Hypothesis testig, t-tests Hypothesis H 0 Coditios Test statistic t Degrees of freedom µ = µ 0 Sample size ; ukow σ t = X µ 0 s/ 1 µ X µ Y = D Sample sizes, m; ukow but equal σ X = σ y t = X Ȳ D s p m + m 2 µ X µ Y = D Sample sizes, m; ukow, uequal σ X σ y t = X Ȳ D s 2 X + s2 Y m Special formula 23
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