Chapter 8: Hypothesis Testing1 HypothesesA hypothesis is a statement about a population parameter. Often, there aretwo complementary statements/hypotheses about θ, respectively called thenull hypothesis and alternative hypothesis.Let Θ be the parameter space. Let Θ0be a subset of the parameterspace, called null region. A pair of hypotheses are denoted by H0and H1,H0: θ ∈ Θ0vs H1: θ ∈ Θc0.Note Θ = Θ0∪ Θc0.Different Types of Hypotheses(1) simple hypotheses: both H0and H1consist of only one probabilitydistribution.H0: θ = θ0vs H1: θ = θ1.(2) composite hypotheses: either H0or H1has more than one distribution:• one-sided hypotheses: H0: θ ≥ θ0vs H1: θ<θ0.• one-sided hypotheses: H0: θ ≤ θ0vs H1: θ>θ0.• two-sided hypotheses: H0: θ = θ0vs H1: θ = θ0.Example 1:An ideal manufacturing process requires that all products are non-defective.This is very seldom. The goal is to keep the proportion of defective itemsas low as possible. Let p be the proportion of defective items, and 0.01 bethe maximum acceptable proportion of defective items.statement 1: p ≥ 0.01 (the proportion of defectives is unacceptably high)statement 2: p<0.01 (acceptable quality)Example 2:Let θ be the average change in a patient’s blood pressure after taking adrug. An experimenter might be interested in testingH0: θ = 0 (the drug has no effect on blood pressure)H1: θ = 0 (there is some effect)91Rejection regionA hypothesis testing procedure or hypothesis test is a rule that specifies:i. for which sample values H0is accepted as trueii. for which sample values H0is rejected and H1is accepted as true.The subset of the sample space for which H0will be rejected is denoted asR and called the rejection or critical region.The complement set Rcis called the acceptance region.The rejection region R of a hypothesis test is usually defined through a teststatistic W (X), a function of the sample. For example,R = {X : W (X) >b}.If X ∈ R, one rejects H0(or accepts H1); otherwise if X ∈ Rc, one acceptsH0(or rejects H1).Example 2: Let¯X be the average change of blood pressure for the sampledpatients. We are interested in testingH0: θ =0, vs H1: θ =0.The rejection region may look like R = {¯XS/√n> 3}.922 Two Types of ErrorsWhen deciding to accept or reject H0, we might make a mistake no matterwhatever the decision is:Type I error:ifH0is true (θ ∈ Θ0), but the test incorrectly rejects H0.Type II error:ifH0is false (θ ∈ Θc0), but the test incorrectly accepts H0DecisionTruth Accept H0Reject H0H0Correct decision Type I errorH1Type II error Correct decisionPower Function:The power function β(θ) of a hypothesis test with rejection region R isthe function of θ defined byβ(θ)=Pθ(X ∈ R)=Pθ(reject H0)=Prob(committing Type I error) if θ ∈ Θ01 − Prob(committing Type II error) if θ ∈ Θc0Therefore,Prob(committing Type I error) = β(θ), for θ ∈ Θ0Prob(committing Type II error) = 1 − β(θ), for θ ∈ Θc0.Remarks:• An ideal test should have the power function satisfying:β(θ) = 0 for all θ ∈ Θ0; β(θ) = 1 for all θ ∈ Θc0.• But such a test is almost impossible unless the truth is known. Inpractice, a good test should have the power function satisfyingβ(θ)isnear0(small)formostθ ∈ Θ0;β(θ) is near 1 (large) for most θ ∈ Θc0.93Example: (Binomial power function)Let X ∼Bin(5,θ). Consider testingH0: θ ≤12versus H1: θ>12.Test 1: reject H0if and only if all “successes” are observed, i.e R = {5}(1) Compute and plot the power function.(2) What is the maximal probability of making Type I error?(3) What is the probability of making Type II error if θ =23?94Test 2: reject H0if X =3, 4, or 5.(1) Compute and plot the power function.(2) What is the maximum Type I error probability?(3) What is the probability of making Type II error if θ =0.8?953 Likelihood Ratio Tests (LRT)Let L(θ|x) be the likelihood function of θ. The likelihood ratio test statisticfor testing H0: θ ∈ Θ0vs H1: θ ∈ Θc0isλ(x)=supθ∈Θ0L(θ|x)supθ∈ΘL(θ|x)=L(ˆθ0|x)L(ˆθ|x),whereˆθ0is the MLE of θ in Θ0(“restricted” maximization);ˆθ is the MLEof θ in the full set Θ = Θ0∪ Θc0(“unrestricted” maximization).A likelihood ratio test (LRT) is a test that has a rejection regionR : {x : λ(x) ≤ c},where c is any number satisfying 0 ≤ c ≤ 1.Comments:• The numerator in λ(x) is the maximum probability of the observedsample x computed over parameters in H0. The denominator is themaximum probability of the observed x over all possible parameters.•ˆθ0is the value in Θ0which makes the observation of data most likely;ˆθ is the value in Θ which makes the observation of data most likely.• If λ(x) is small, it implies that there are some parameter points inH1for which the observed sample is much more likely than for anyparameter in H0. So the LRT suggests we reject H0and accept H1.• The LRT statistic λ(x) is a function of x not a function of θ.• 0 ≤ λ(x) ≤ 1.About the cut-off value c• Different choices of c ∈ [0, 1] give different tests and rejection regions.• The smaller c, the smaller Type I error; The larges c, the smaller TypeII error.• We will discuss the ideal choice of c later.After finding an expression for λ(x), we should get the simplestexpression for R.96Example: (Normal One-sided LRT) X1,...,Xn∼ iid N(θ,σ2)withθ un-known and σ2known. Consider testingH0: θ ≤ θ0versus H1: θ>θ0.(i) Find the LRT and its power function.(ii) Comment on the decision rules given by different c’s.97Example: Let X1,...,Xnbe a random sample from a location-exponentialfamilyf(x|θ)=e−(x−θ)if x ≥ θ, −∞ <θ<∞.Test H0: θ ≤ θ0versus H1: θ>θ0. Find the LRT and its power function.98LRT based on Sufficient StatisticsIf T is sufficient for θ, then we can construct the LRT based on T and thelikelihood function L∗(θ|t)=g(t|θ). Since T(x) contains all the informationabout θ in x, the test based on T should be as good as the test based on x.In fact, the tests are equivalent.TheoremIf T (X) is a sufficient statistic for θ, λ∗(t) is the LRT statistic based onT ,andλ(x) is the LRT statistic based on x. Thenλ∗(T (x)) = λ(x)for every x in the sample space.Proof:Comment: The simplified expression for λ(x) should depend on x onlythrough T (x)ifT (X) is a sufficient statistic for θ.99Example: (Normal Two-sided LRT) X1,...,Xn∼ iid N (θ, σ2)withσ2known. TestH0: θ = θ0versus