18.650 Statistics for Applications Chapter 5: Parametric hypothesis testing 1/37
Cherry Blossom run (1) ◮ The credit union Cherry Blossom Run is a 10 mile race that takes place every year in D.C. ◮ In 2009 there were 14974 participants ◮ Average running time was 103.5 minutes. Were runners faster in 2012? n runners To answer this question, select from the 2012 race at random and denote by X 1 , . . . , X n their running time. 2/37
Cherry Blossom run (2) We can see from past data that the running time has Gaussian distribution. The variance was 373. 3/37
Cherry Blossom run (3) ◮ We are given i.i.d r.v X 1 , . . . , X n and we want to know if X 1 ∼ N (103 . 5 , 373) ◮ This is a problem. hypothesis testing ◮ There are many ways this could be false: 1. I E[ X 1 ] = 103 . 5 2. var [ X 1 ] = 373 3. X 1 may not even be Gaussian. ◮ We are interested in a very specific question: is E[ X 1 ] < 103 . 5 ? I 4/37
Cherry Blossom run (4) ◮ We make the following assumptions : 1. var [ X 1 ] = 373 (variance is the same between 2009 and 2012) 2. X 1 is Gaussian. ◮ The only thing that we did not fix is I E[ X 1 ] = µ . µ = 103 . 5 or µ < 103 . 5 ”? ◮ Now we want to test (only): “Is is ◮ By making modeling assumptions , we have reduced the X 1 ∼ N (103 . 5 , 373) may number of ways the hypothesis be rejected. X 1 ∼ N ( µ, 373) for ◮ The only way it can be rejected is if some µ < 103 . 5 . ◮ We compare an expected value to a fixed reference number (103.5). 5/37
Cherry Blossom run (5) Simple heuristic: ¯ X n < 103 . 5 , µ < 103 . 5 ” “If then This could go wrong if I randomly pick only fast runners in my sample X 1 , . . . , X n . Better heuristic: ¯ X n < 103 . 5 − (something that then µ < 103 . 5 ” “If − − − → 0 ), n →∞ To make this intuition more precise, we need to take the size of the ¯ random fluctuations of X n into account! 6/37
Clinical trials (1) ◮ Pharmaceutical companies use hypothesis testing to test if a new drug is efficient. ◮ To do so, they administer a drug to a group of patients (test group) and a placebo to another group (control group). ◮ Assume that the drug is a cough syrup. ◮ Let µ control denote the expected number of expectorations per hour after a patient has used the placebo. ◮ Let µ drug denote the expected number of expectorations per hour after a patient has used the syrup. ◮ We want to know if µ drug < µ control ◮ We compare two expected values. No reference number. 7/37
Clinical trials (2) ◮ Let X 1 , . . . , X n drug denote n drug i.i.d r.v. with distribution Poiss ( µ drug ) ◮ Let Y 1 , . . . , Y n control denote n control i.i.d r.v. with distribution Poiss ( µ control ) ◮ We want to test if µ drug < µ control . Heuristic: ¯ ¯ “If X drug < X control − (something that − − − − − − − → 0 ), then n drug →∞ n control →∞ conclude that µ drug < µ control ” 8/37
Heuristics (1) Example 1: A coin is tossed 80 times, and Heads are obtained 54 times. Can we conclude that the coin is significantly unfair ? iid ◮ n = 80 , X 1 , . . . , X n ∼ Ber ( p ); ¯ ◮ X n = 54 / 80 = . 68 p = . 5 : By ◮ If it was true that CLT+Slutsky’s theorem, ¯ n − . 5 √ X n ≈ N (0 , 1) . J . 5(1 − . 5) ¯ n − . 5 √ X n ◮ J ≈ 3 . 22 ¯ . 5(1 − . 5) ◮ Conclusion: It seems quite reasonable to reject the p = . 5 . hypothesis 9/37
Heuristics (2) Example 2: A coin is tossed 30 times, and Heads are obtained 13 times. Can we conclude that the coin is significantly unfair ? iid ◮ n = 30 , X 1 , . . . , X n ∼ Ber ( p ); ¯ ◮ X n = 13 / 30 ≈ . 43 p = . 5 : By ◮ If it was true that CLT+Slutsky’s theorem, ¯ n − . 5 √ X n ≈ N (0 , 1) . J . 5(1 − . 5) ¯ √ X n − . 5 gives n ◮ Our data J ≈ − . 77 . 5(1 − . 5) ◮ The number . 77 is a plausible realization of a random variable Z ∼ N (0 , 1) . ◮ Conclusion: our data does not suggest that the coin is unfair. 10/37
Statistical formulation (1) ◮ Consider a sample X 1 , . . . , X n of i.i.d. random variables and a ( E, (I statistical model P θ ) θ ∈ Θ ) . ◮ Let Θ 0 and Θ 1 be disjoint subsets of Θ . � θ ∈ Θ 0 H 0 : ◮ Consider the two hypotheses: H 1 : θ ∈ Θ 1 null hypothesis , alternative hypothesis . ◮ H 0 is the H 1 is the θ is ◮ If we believe that the true either in Θ 0 or in Θ 1 , we may test H 0 against H 1 . want to reject H 0 (look ◮ We want to decide whether to for evidence against H 0 in the data). 11/37
Statistical formulation (2) ◮ H 0 and H 1 do not play a symmetric role: the data is is only used to try to disprove H 0 ◮ In particular lack of evidence, does not mean that H 0 is true (“innocent until proven guilty”) test is ψ ∈ { 0 , 1 } such ◮ A a statistic that: ψ = 0 , ◮ If H 0 is not rejected; ψ = 1 , ◮ If H 0 is rejected. example: H 0 : p = 1 / 2 vs. H 1 : p = 1 / 2 . ◮ Coin { } ¯ n − . 5 √ X > C ◮ ψ = 1 n C > 0 . J I , for some . 5(1 − . 5) threshold C ? ◮ How to choose the 12/37
Statistical formulation (3) ◮ Rejection region of a test ψ : R ψ = { x ∈ E n : ψ ( x ) = 1 } . ◮ Type 1 error of ψ (rejecting H 0 when a test it is actually true): α ψ : Θ 0 → I R θ P θ [ ψ = 1] . �→ I ◮ Type 2 error of ψ (not a test rejecting H 0 although H 1 is actually true): Θ 1 → β ψ : I R θ P θ [ ψ = 0] . �→ I ◮ Power of a test ψ : π ψ = inf (1 − β ψ ( θ )) . θ ∈ Θ 1 13/37
Statistical formulation (4) ψ has level α if ◮ A test α ψ ( θ ) ≤ α, ∀ θ ∈ Θ 0 . ψ has asymptotic level α if ◮ A test lim α ψ ( θ ) ≤ α, ∀ θ ∈ Θ 0 . n →∞ ◮ In general, a test has the form ψ = 1 I { T n > c } , c ∈ I for some statistic T n and threshold R . test statistic . The ◮ T n is called the rejection region is R ψ = { T n > c } . 14/37
Example (1) iid unknown p ∈ (0 , 1) . ◮ Let X 1 , . . . , X n ∼ Ber ( p ) , for some ◮ We want to test: H 0 : p = 1 / 2 vs. H 1 : p = 1 / 2 α ∈ (0 , 1) . with asymptotic level √ p ˆ n − 0 . 5 n ◮ Let T n = J , where p ˆ n is the MLE. . 5(1 − . 5) ◮ If H 0 is true, then by CLT and Slutsky’s theorem, I P[ T n > q α/ 2 ] − − − → 0 . 05 n →∞ I { T n > q α/ 2 } . ◮ Let ψ α = 1 15/37
Example (2) α = 5% , Coming back to the two previous coin examples: For q α/ 2 = 1 . 96 , so: ◮ In Example 1 , H 0 is rejected at the asymptotic level 5% by the test ψ 5% ; ◮ In Example 2 , H 0 is not rejected at the asymptotic level 5% by the test ψ 5% . Question: In α would Example 1 , for what level ψ α not reject H 0 α would ? And in Example 2 , at which level ψ α reject H 0 ? 16/37
p-value Definition p-value of The (asymptotic) a test ψ α is the smallest (asymptotic) α at level which ψ α rejects H 0 . It is random, it depends on the sample. Golden rule ≤ α ⇔ H 0 is p-value rejected by ψ α , at the (asymptotic) level α . The smaller the p-value, the more confidently one can reject H 0 . P[ | Z | > 3 . 21] ≪ . 01 . ◮ Example 1: p-value = I ◮ Example 2: p-value = I P[ | Z | > . 77] ≈ . 44 . 17/37
Neyman-Pearson’s paradigm Idea: For given hypotheses, among all tests of level/asymptotic level α , is it possible to find one that has maximal power ? ψ = 0 that Example: The trivial test never rejects H 0 has a level ( α = 0 ) but perfect poor power ( π ψ = 0 ). Neyman-Pearson’s theory provides (the most) powerful tests with given level. In 18.650, we only study several cases. 18/37
χ 2 distributions The Definition χ 2 (pronounced “Kai-squared”) For a positive integer d , the distribution with d degrees of freedom is the law of the random iid 2 + Z 2 . . . + Z 2 , Z 1 , . . . , Z d ∼ N (0 , 1) . variable Z 1 2 + where d Examples: Z ∼ N d ( 0 , I d ) , I Z I 2 2 ∼ χ 2 d . ◮ If then ◮ Recall that the sample variance is given by n n S n = 1 n ¯ n ) 2 = 1 n ¯ n ) 2 2 − ( X ( X i − X X i n n i =1 i =1 iid X 1 , . . . , X n ∼ N ( µ, σ 2 ) , ◮ Cochran’s theorem implies that for if S n is the sample variance, then nS n ∼ χ 2 n − 1 . σ 2 ◮ χ 2 2 = Exp (1 / 2) . 19/37
Student’s T distributions Definition Student’s T distribution with d For a positive integer d , the degrees of freedom (denoted by t d ) is the law of the random Z Z ∼ N (0 , 1) , V ∼ χ 2 and Z ⊥ ⊥ V ( Z is J variable , where d V/d V ). independent of Example: iid X 1 , . . . , X n ∼ N ( µ, σ 2 ) , ◮ Cochran’s theorem implies that for if S n is the sample variance, then √ ¯ n − µ X n − 1 ∼ t n − 1 . √ S n 20/37
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