Chapter 11. Hypothesis Testing (II)

Transcription

1 Chapter 11. Hypothesis Testing (II) 11.1 Likelihood Ratio Tests one of the most popular ways of constructing tests when both null and alternative hypotheses are composite (i.e. not a single point). Let X f (, θ). Consider hypotheses H 0 : θ Θ 0 vs H 1 : θ Θ Θ 0. The likelihood ratio test will reject H 0 for the large values of the statistic LR = LR(X) sup θ Θ f (X, θ) sup θ Θ0 f (X, θ) = f (X, θ) / f (X, θ), where θ the (unconstrained) MLE, and θ is the constrained MLE under hypothesis H 0.

2 Remark. (i) It is easy to see that LR 1. (ii) The exact sampling distributions of LR are usually unknown, except in a few special cases. Example 1. (One-sample t -test) Let X = (X 1,, X n ) τ be a random sample from N (µ, σ 2 ). We are interested in testing hypotheses H 0 : µ = µ 0 against H 1 : µ µ 0, where µ 0 is given, and σ 2 is unknown and is a nuisance parameter. Now both H 0 and H 1 are composite. The likelihood function is L(µ, σ 2 ) = C σ n exp { 1 2σ 2 n (X i µ) 2}.

3 The unconstrained MLEs are and the constrained MLE is The LR-ratio statistic is then Since µ = X, σ 2 = 1 n σ 2 = 1 n n (X i X ) 2, n (X i µ 0 ) 2. LR = L( µ, σ2 ) L(µ 0, σ 2 ) = ( σ2 / σ 2 ) n/2. n σ 2 = n σ 2 + n( X µ 0 ) 2,

4 it holds that σ 2 / σ 2 = 1 + T 2 /(n 1), where T = n( X µ 0 ) /{ 1 n 1 n (X i X ) 2} 1/2. Note that T t n 1 under H 0. The LRT will reject H 0 iff T > t n 1,α/2, where t k,α is the upper α-point of the t -distribution with k degrees of freedom.

5 Asymptotic Distribution of Likelihood ratio test statistic Let X = (X 1,, X n ) τ, and assume certain regularity conditions. Then as n, the distribution of 2 log(lr) under H 0 converges to the χ 2 - distribution with d d 0 degrees of freedom, where d is the dimension of Θ and d 0 is the dimension of Θ 0. To make the computation of dimension easy, reparametrisation is often adopted. Suppose that the parameter θ may be written in two parts θ = (ψ, λ) where ψ is k 1 parameter of interest, and λ is of little interest and is called nuisance parameters. The hypotheses to be tested may be expressed as H 0 : ψ = ψ 0 vs H 1 : ψ ψ 0.

6 Now the LR-statistic is of the form LR = L( ψ, λ; X) L(ψ 0, λ; X), where ( ψ, λ) is unconstrained MLE while λ is the constrained MLE of λ subject to ψ = ψ 0. Then as n, 2 log(lr) D χ 2 k under H 0. Example 2. Let X 1,, X n be independent, and X j N (µ j, 1). Consider the null hypothesis H 0 : µ 1 = = µ n.

7 The likelihood function is L(µ 1,, µ n ) = C exp { 1 2 n (X j µ j ) 2}, where C > 0 is a constant independent of µ j. Then the unconstrained MLE are µ j = X j and the constrained MLE is µ = X. Hence Hence LR = L( µ 1,, µ n ) L( µ,, µ) 2 log(lr) = = exp { 1 2 n (X j X ) 2}. n (X j X ) 2 χn 1 2 under H 0, which is true for any finite n as well. How to calculate the degree of freedom?

8 Since d = n, d 0 = 1, the d.f. is d d 0 = n 1. Alternatively we may adopt the following reparametrisation: µ j = µ 1 + ψ j for 2 j n. Then the null hypothesis can be expressed as H 0 : ψ 2 = = ψ n = 0. Therefore ψ = (ψ 2,, ψ n ) τ has n 1 component, i.e. k = n The permutation test a nonparametric method for testing if two distributions are the same. It is particularly appealing when sample sizes are small, as it does not rely on any asymptotic theory.

9 Let X 1,, X m be sample from distribution F x and Y 1,,Y n be a sample from distribution F y. We are interested in testing H 0 : F x = F y versus H 1 : F x F y. Key idea: under H 0, {X 1,, X m,y 1,,Y n } form a sample of size m + n from a single distribution. Choose a test statistic T = T (X 1,, X m,y 1,,Y n ) which is capable to tell the difference between the two distribution, e.g. T = X Y, or T = X Y 2 + Sx 2 Sy 2. Consider all (m + n)! permutations of (X 1,, X m,y 1,,Y n ), compute the test statistic T for each permutation, yielding the values T 1,,T (m+n)!.

10 The p-value of the test is defined as p = 1 (m + n)! (m+n)! I (T j > t obs ), where t obs = T (X 1,, X m,y 1,,Y n ). We reject H 0 at the significance level α if p α. Note. When H 0 holds, all those (m + n)! T j s are on the equal footing, and t obs = T (X 1,, X m,y 1,,Y n ) is one of them. Therefore t obs is unlikely to be an extreme value among T j s.

11 Algorithm for Permutation Tests: 1. Compute t obs = T (X 1,, X m,y 1,,Y n ). 2. Randomly permute the data. Compute T again using the permuted date. 3. Repeat Step 2 B times, and let T 1,,T B denote the resulting values. 4. The approximate p-value is B 1 1 j B I (T j > t obs ). Remark. Let Z = (X 1,, X m,y 1,,Y n ) (Z <- c(x,y)). A permutation of Z may be obtained in R as Zp <- sample(z, n+m) You may also use the R-function sample.int: k <- sample.int(n+m, n+m) Now k is a permutation of {1, 2,, n + m}.

12 Example 3. Class A was taught using detailed PowerPoint slides. The marks in the final exam are 45, 55, 39, 60, 64, 85, 80, 64, 48, 62, 75, 77, 50. Students in Class B were required to read books and answer questions in class discussions. The marks in the final exam are 45, 59, 48, 74, 73, 78, 66, 69, 79, 81, 60, 52. Can we infer that the marks from the two classes are significantly different? We conduct the permutation test using the test statistic T = X Y in R: > x <- c(45, 55, 39, 60, 64, 85, 80, 64, 48, 62, 75, 77, 50) > y <- c(45, 59, 48, 74, 73, 78, 66, 69, 79, 81, 60, 52) > length(x); length(y) [1] 13

13 [1] 12 > summary(x) Min. 1st Qu. Median Mean 3rd Qu. Max > summary(y) Min. 1st Qu. Median Mean 3rd Qu. Max > Tobs <- abs(mean(x)-mean(y)) > z <- c(x,y) > k <- 0 > for(i in 1:5000) { + zp <- sample(z, 25) # zp is a permutation of z + T <- abs(mean(zp[1:13])-mean(zp[14:25])) + if(t>tobs) k <- k+1 + } cat("p-value:", k/5000, "\n") p-value: Since p-value is , we cannot reject the null-hypothesis that the mark distributions of the two classes are the same.

14 We also apply the t -sample, obtaining the similar results: > t.test(x, y, var.equal=t) # mu=0 is the default Two Sample t-test data: x and y t = , df = 23, p-value = alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval:

15 11.3 χ 2 -tests Goodness-of-fit tests: to test if a given distribution fits the data. Let {X 1,, X n } be a random sample from a discrete distribution of k categories denoted by 1,, k. Denote the probability function Then p j 0 and k p j = 1. p j = P (X i = j ), j = 1,, k. Typically n >> k. Therefore the data are often compressed into a table: Category 1 2 k Frequency Z 1 Z 2 Z k

16 where Obviously k Z j = n. Z j = No. of X i s equal to j, j = 1,, k. To test the null hypothesis H 0 : p i = p i (θ), i = 1,, k, where the function forms of p i (θ) are known but the parameter θ is unknown. For example, p i (θ) = θ i 1 e θ /(i 1)! (i.e. Poisson distribution). We first estimate θ by, for example, its MLE θ. The expected frequencies under H 0 are E i = np i ( θ), i = 1,, k.

17 Listing them together with observed frequencies, we have Category 1 2 k Frequency Z 1 Z 2 Z k Expected frequency E 1 E 2 E k If H 0 is true, we expect Z j E j = np j ( θ) when n is large, as, by the LLN, it holds Z j n = 1 n I (X n i = j ) E {I (X i = j )} = P (X i = j ) = p j (θ). i =1 Test statistic: T = k (Z j E j ) 2/ E j. Theorem. Under H 0, T components in θ. D χ 2 k 1 d as n, where d is the number of

18 Remark. (i) It is important that E i 5 at least. This may be achieved by combining together the categories with smaller expected frequencies. (ii) When p j are completely specified (i.e. known) under H 0, d = 0. Example 4. A supermarket recorded the numbers of arrivals over 100 one-minute intervals. The data were summarized as follows No. of arrivals Frequency Do the data match a Poisson distribution? The null hypothesis is H 0 : p i = λ i e λ /i! for i = 0, 1,. We find the MLE for λ first.

19 The likelihood function: L(λ) = 100 i =1 λx i 100 X i! e λ λ i =1 X i e 100λ. The log-likelihood function: l (λ) = log(λ) 100 i =1 X i 100λ. Let d dλ l (λ) = 0, leading to λ = i =1 X i = X. Since we are only given the counts Z j instead of X i, we need to compute X from Z j. Recall Z j = no. of X i equal to j. Hence = X = 1 n n X i = 1 n i =1 k j Z j 1 ( ) = 1.81.

20 With λ = 1.81, the expected frequencies are E i = n p i ( λ) = 100 (1.81)i e 1.81, i! i = 0, 1,. We combine the last three categories to make sure E i 5. No. of arrivals Total Frequency Z i p i ( λ) = λ i e λ /i! Expected frequency E i Difference Z i E i (Z i E i ) 2 /E i Note under H 0, T = 4 i =0 (Z i E i ) 2 /E i χ = χ2. Since T = < 3 χ 0.10, 2 = 6.25, we cannot reject the assumption that the data follow a 3 Poisson distribution.

21 Remark. (i) The goodness-of-fit test has been widely used in practice. However we should bear in mind that when H 0 cannot be rejected, we are not in the position to conclude that the assumed distribution is true, as not reject accept" (ii) The above test may be used to test the goodness-of-fit of a continuous distribution via discretization. However there exist more appropriate methods such as Kolmogorov-Smirnov test and Cramér-von Mises test, which deal with the goodness-of-fit for continuous distributions directly.

22 Tests for contingency tables Tests for independence of two discrete random variables Let (X,Y ) be two discrete random variables, and X have r categories and Y have c categories. Let p ij = P (X = i, Y = j ), i = 1,, r, j = 1,, c. Then p ij 0 and i,j p ij r i =1 cj =1 p ij = 1. Let p i = P (X = i ) and p j = P (Y = j ). It is easy to see that p i = c P (X = i, Y = j ) = c p ij = j p ij Similarly, p j = i p ij

23 X and Y are independent iff p ij = p i p j for i = 1,, r and j = 1,, c. Suppose we have n pairs of observations from (X,Y ). The data are presented in a contingency table below Y 1 2 c 1 Z 11 Z 12 Z 1c X 2. Z 21. Z 22.. Z 2c. r Z r 1 Z r 2 Z r c where Z ij = no. of the pairs equal to (i, j ).

24 It is often useful to add the marginals into the table: c r r Z i = Z ij, Z j = Z ij, Z = Z i = i =1 i =1 c Z j = n Y 1 2 c 1 Z 11 Z 12 Z 1c Z 1 X 2. Z 21. Z 22.. Z 2c. Z 2. r Z r 1 Z r 2 Z r c Z r Z 1 Z 2 Z c Z = n

25 We are interested in testing the independence H 0 : p ij = p i p j, i = 1,, r, j = 1,, c. Under H 0, a natural estimator for p ij is p ij = p i p j = Z i n Hence the expected frequency at the (i, j )-th cell is E ij = n p ij = Z i Z j /n = Z i Z j /Z, i = 1,, r, j = 1,, c. If H 0 is true, we expect Z ij defined as T = Z j n E ij. The goodness-of-fit test statistic is r i =1 c (Z ij E ij ) 2/ E ij. We reject H 0 for large values of T.

26 Under H 0, T χ 2 p d, where p = no. of cells -1 = r c 1 d = no. of estimated free parameters = r + c 2. Note. 1. The sum of r c counts Z ij is n fixed. So knowing r c 1 of them, the other one is also known. This is why p = r c The estimated parameters are p i and p j. But r i =1 p i = 1 and c p j = 1. Hence d = (r 1) + (c 1) = r + c For testing independence, it always holds that Z i E i = 0 and Z j E j = 0. Those are useful facts in checking for computational errors. The proofs are simple, as, for example, Z i E i = Z i E ij = Z i j j Z i Z j Z = Z i Z i Z Z = 0.

27 Theorem. Under H 0, the limiting distribution of T is χ 2 with (r 1)(c 1) degrees of freedom, as n.

28 Example. The table below lists the counts on the beer preference and gender of beer drinker from randomly selected 150 individuals. Test at the 5% significance level the hypothesis that the preference is independent of the gender. Beer preference Light ale Lager Bitter Total Gender Male Female Total The expected frequencies are: E 11 = E 21 = = 26.67, E 12 = = 23.33, E 22 = = 37.33, E 13 = = 32.67, E 33 = = 16, = 14.

29 Under the null hypothesis of independence, T = i,j (Z ij E ij ) 2 /E ij χ 2 2. Note the degree freedom is (2 1)(3 1) = 2. Since T = > χ 0.05, 2 = 5.991, we reject the null hypothesis, i.e. there 2 is significant evidence from the data indicating that the beer preference and the gender of beer drinker are not independent.

30 Tests for several binomial distributions Consider a real example: Three independent samples of sizes 80, 120 and 100 are taken respectively from single, married, and widowed or divorced persons. Each individual was asked to if friends and social life or job and primary activity contributes most to their general well-being. The counts from the three samples are summarized in the table below. Friends and social life Job or primary activity Single Married Widowed or divorced Total

31 Conditional Inference: Sometimes we conduct inference under the assumption that all the row (or column) margins are fixed. Different from the tables for independent tests, now Z 1j Bi n(z j, p 1j ), j = 1, 2, 3, where Z j are fixed constants sample sizes. Furthermore, p 2j = 1 p 1j. We are interested in testing hypothesis H 0 : p 11 = p 12 = p 13. Under H 0, the three independent samples may be seen from the same population. Furthermore, Z 11 + Z 12 + Z 13 Bi n(z 1 + Z 2 + Z 3, p),

32 where p denotes the common value of p 11, p 12 and p 13. Therefore the MLE is p = Z 11 + Z 12 + Z 13 Z 1 + Z 2 + Z 3 = The expected frequencies are = E 1j = pz j and E 2j = Z j E 1j, j = 1, 2, Total Total 0 0 0

33 Total

34 Under H 0, T = i,j (Z ij E ij ) 2 /E ij χ 2 p d = χ2 2, where p = no. of free counts Z ij = 3 d = no. of estimated free parameters = 1. Since T = < χ ,2 = 4.605, we cannot reject H 0, i.e. there is no significant difference among the three populations in terms of choosing between F&SL and J&PA as the more important factor towards their general well-being. Remark. Similar to the independence tests, it holds that Z i E i = 0 and Z j E j = 0.

35 Tests for r c tables a general description In general, we may test for different types of the structure in a r c table, for example, the symmetry (p ij = p j i ). The key is to compute expected frequencies E ij under null hypothesis H 0. Under H 0, the test statistic T = r i =1 c (Z ij E ij ) 2 p = no. of free counts among Z ij, E ij χ 2 p d, d = no. of the estimated free parameters. We reject H 0 if T > χ 2 α, p d. Remark. The R-function chisq.test performs both the goodness-of-fit test and the contingency table test.