Lecture Outline. Hypothesis Testing. Simple vs. Composite Testing. Stat 111. Hypothesis Testing Framework

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1 Stat 111 Lecture Outline Lecture 14: Intro to Hypothesis Testing Sections in DeGroot 1 Hypothesis Testing Consider a statistical problem involving a parameter θ whose value is unknown but must lie in a certain parameter space Ω. Suppose now that Ω can be partitioned into two disjoint subsets Ω 0 and Ω 1, and the statistician is interested in whether θ lies in determining which of two hypotheses is true: H 0 : θ ϵ Ω 0 or H A : θ ϵ Ω 1. Since Ω has been partitioned into two disjoint subsets, either H 0 or H A is true (and not both). This is the formal set up of a problem of testing statistical hypotheses. * Note: the text writes H A as H 1. Either is fine. Simple vs. Composite Testing A hypothesis test problem always has a null hypothesis (H 0 ) and an alternative hypothesis (H A ), and each can be in one of two forms: Simple hypotheses are hypotheses where θ can only take on a single value. (ex: H 0 : θ = 0 or H A : θ = 1) Composite hypotheses are hypotheses where θ can take on multiple or a range of values. (ex: H 0 : θ 0 or H A : θ ϵ {-1,1}) In a particular problem, your null and alternative hypotheses could have any combination of simple and composite alternatives. Most common: H 0 is simple or composite, and H A is composite

2 One-sided and Two-sided Tests The most common set-up of hypotheses are called onesided tests and two-sided tests: A hypothesis test problem always has a null hypothesis (H 0 ) and an alternative hypothesis (H A ), and each can be in one of two forms: One-sided tests: both hypotheses are composite and represent one-side of the parameter space around some value θ 0 : H 0 : θ θ 0 vs. H A : θ < θ 0, or H 0 : θ θ 0 vs. H A : θ > θ 0 Two-sided tests: H 0 is a simple hypothesis, and H A represents the rest of the parameter space of θ: H 0 : θ = θ 0 vs. H A : θ θ 0 5 Test Statistic and Rejection Regions Let X 1, X n be a random sample from a distribution with parameter θ. Let T = r(x 1, X n ) be a statistic and let R be a subset of the real line. For a specific set of hypotheses (H 0 vs. H A ) in a testing problem, suppose we choose to reject H 0 if T ϵ R. Then T is called a test statistic and R is called the critical region or rejection region of the test. What in the world does that mean? Classically, a test statistic will be based on an estimator of the parameter (like using X to estimate μ for a normal, or using p to estimate p for a Bernoulli or binomial). A rejection region is usually in the form of T c, T c, or T c for some specifically chosen value c: the critical value. What value of c should we choose? Depends on how much (and what type) of error we want to allow for. 6 Lecture Outline Error and Power There are two types of error: Type I error: when we incorrectly reject a null hypothesis when it is actually true. Type II error: when we fail to reject the null hypothesis when the alternative hypothesis is true. A related idea is the power of a test: Power is the probability or rejecting the null hypothesis given a specific value of the parameter Power varies depending on the value of θ, and the book defines something called the power function, π(θ), which is the function describing this probability for varying values of the parameter we are testing, θ. 7 8

3 Choosing a critical value/region The most common method of choosing a critical value, c, is one that specifies the largest value of the power function for any value in the null hypothesis parameter space. The specified value we denote as α, the Type I error rate for the test (often called the level or size of the test). This produces an asymmetry in the testing procedure: we have fixed the Type I error rate for the test, but have not worried about the Type II error rate. In many situations this is quite natural: making a type I error can often be viewed as a worse error. So we will often set up our hypotheses in such a way that Type I error is the worse error. p-values The p-value of a test is the smallest level such that we would reject the null hypothesis for the data we have observed. If the critical region of the parameter space is based on critical value (of the form: reject H 0 if T c, T c, or T c), then the p-value is easy to calculate. It can be thought of as: p-value: the probability of observing our test statistic, or a more extreme one in the direction of the alternative, given the null hypothesis is true Lecture Outline A General Testing Example Let s suppose the following scenario: you go to a casino to play a dice game (based on one die with 6 sides), and you are told there are 3 dice: 1 fair die, and loaded dice that have probability 0.3 and 0.5 of showing up 6 on the top face. You are randomly given one die, and are asked to roll it 0 times. Let X = # rolls (out of 0) showing with a 6 face-up. Set-up a hypothesis test to determine whether you were given the fair die or a loaded die. What are your hypotheses? What is your test statistic? What is your critical region? What is your type I error? What is the power function for your test? You actually observe X = 5 rolls with a 6 showing face-up. What is the p-value for your test? What is your conclusion?

4 Frequency A Reasonable Solution: Let X = # rolls (out of 0) showing with a 6 face-up. H 0 : fair die (p = 1/6) vs. H A : loaded die (p ϵ {0.3,0.5}) X is the test statistic. Critical region: we will choose to reject H 0 if X 6. Type I error: P(X 6 p = 1/6) = In R: > 1-pbinom(5,size=0,p=1/6) What is the power function for your test? P(X 6 p = 1/6) = P(X 6 p = 0.3) = [> 1-pbinom(5,size=0,p=0.3)] P(X 6 p = 0.5) = [> 1-pbinom(5,size=0,p=0.5)] A Reasonable Solution (cont.): Let X = 5. What is the p-value for your test? P(X 5 p = 1/6) = In R: > 1-pbinom(4,size=0,p=1/6) What is your conclusion? Straightforward: since X = 5 does not fall in our rejection region, we fail to reject H 0. What is the probability you were handed the fair die? (Note: this probability is usually not calculable in real-life since we do not know the prior probability of H 0 and H A to begin with) A Reasonable Solution (cont.): What is the probability you were handed the fair die? P( X 5 fair) fair) P( fair X 5) P( X 5 fair) fair) X 5 L1) L1) X 5 L) L) (0.0) (1/ 3) (0.0) (1/ 3) ) (1/ 3) ) (1/ 3) > dbinom(4,size=0,p=0.5) [1] > dbinom(4,size=0,p=0.3) [1] > dbinom(4,size=0,p=1/6) [1] Single Mean Testing Example On HW#1, we used are first day survey to determine whether or not Harvard students, on average, get the recommended amount of sleep (8 or more hours per night). Let s set-up this problem from scratch, with the new terminology and details we now have. Histogram of sleep Let X i be the self-reported typical amount of sleep for a randomly chosen Harvard student (for this problem, let s assume this class is a good random sample, n = 91). What distribution is it reasonable to assume the X i follow? sleep

5 power Single Mean Testing Example Set-up a hypothesis test to determine whether Harvard students get at least the recommended amount of sleep. What are the hypotheses? What is the test statistic? What is the critical region? What is the type I error? What is the power function for the test? * For now, assume σ = 1.0. Note: 91 = You actually observe the following sample statistics: n 91 x s 1.09 What is the p-value for your test? What is your conclusion? What test should you actually have run instead? 17 Solutions: H 0 : enough sleep (μ 8) vs. H A : lack of sleep (μ < 8) T = X is the test statistic, or better: Z = (X μ 0 )/(σ/ n). X ~ N(, / n) Type I error: let s set this at Therefore, our critical region will be: reject H 0 if Z So this means for X : X 0 z( / n) (1/ 91) 7.88 What is the power function for the test? This is tricky. We want π(μ) = P(T c μ), for various values of the population mean, μ. Key: subtracting off μ 0 = 8 is not the correct standardizing if μ 8. This then becomes: 18 Solutions: Calculating Power: X 0 X 0 ( ) P P / n / n / n X 0 P / n / n 0 P Z / n / n / 91 Why does this power curve make sense? mu=750:850/100 power=pnorm( (8-mu)/(1/sqrt(91))) plot(power~mu,type= l ) Power Curve (aka, Power Function) mu 19 Solutions (cont.): n 91 x You actually observe the following sample statistics: s 1.09 What is the p-value for your test? What is your conclusion? p-value = P(X H 0 : μ 8). Uh-oh, what value for μ should we choose? Choose the value for which this probability would be highest: μ = 8: p-value = P(X H 0 : μ = 8) = P((X μ 0 )/(σ/ n) ( )/(1/ 91)) = P(Z ) = 4.11 x We can reject the null hypothesis since p-value < α (or note our test statistic, Z = , is in the rejection/critical region). What test should you actually have run instead? We should have performed a t-test. No reason to assume σ =

6 Lecture Outline A look ahead: Likelihood Ratio Test Another way to form a test statistic is to compare the value of the likelihood functions for the two hypotheses: H 0 : θ ϵ Ω 0 vs. H A : θ ϵ Ω 1. Let the likelihood ratio statistic then be: max f ( x ) 0 ( x) LR max f ( x ) Where Ω = Ω 0 Ω 1 is the entire parameter space of θ. What is the range of values for this statistic? We will then reject H 0 if Λ(x) c for some critical value c. We ll cover this further in the next lecture. 1 Take Home Message There is a very formal set of procedures when performing a statistical hypothesis test. We are always comparing a null hypothesis to an alternative hypothesis, and either one can be a simple or composite hypothesis. A test statistic is calculated from the observations as the evidence for the competing hypotheses. We set up a critical region: the values for the test statistic for which we will decide to reject the null hypothesis. The p-value: the probability of observing our test statistic or a more extreme one, given the null hypothesis is true. Type I error is rejecting H 0 when it is actually true. Type II error is not rejecting H 0 when it is actually not true. The power function/curve is the probability of the test statistic falling in the critical region for values of θ Another way to perform a hypothesis test is through the likelihood ratio for θ ϵ Ω 0 vs. θ ϵ Ω. 3 6

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