Goodness of Fit Goodness of fit - 2 classes

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1 Goodness of Fit Goodness of fit - 2 classes A B Do these data correspond reasonably to the proportions 3:1? We previously discussed options for testing p A =0.75! Exact p-value Exact confidence interval Normal approximation

2 Goodness of fit - 3 classes AA AB BB Do these data correspond reasonably to the proportions 1:2:1? Multinomial distribution Imagine an urn with k types of balls. Let p i denote the proportion of type i. Draw n balls with replacement. Outcome: (n 1, n 2,...,n k ), with i n i = n, where n i is the no. balls drawn that were of type i. n! P(X 1 =n 1,...,X k =n k )= n 1! n k! pn 1 1 pn k k if 0 n i n, i n i =n Otherwise P(X 1 =n 1,...,X k =n k ) = 0.

3 Example Let (p 1, p 2, p 3 ) =(0.25,0.50,0.25)andn=100. P(X 1 =35, X 2 =43, X 3 =22) = 100! 35! 43! 22! Rather brutal, numerically speaking. Take logs (and use a computer). Goodness of fit test We observe (n 1, n 2, n 3 ) Multinomial(n,p={p 1, p 2, p 3 }). We seek to test H 0 : p 1 = 0.25, p 2 = 0.5, p 3 = versus H a : H 0 is false. We need two things: Ateststatistic. The null distribution of the test statistic.

4 The likelihood-ratio test (LRT) Back to the first example: A n A B n B Test H 0 :(p A, p B )=(π A, π B ) versus H a :(p A, p B ) (π A, π B ). MLE under H a : ˆp A = n A /n where n = n A + n B. Likelihood under H a : Likelihood under H 0 : L a =Pr(n A p A = ˆp A )= ( ) n n A ˆp n A A (1 ˆp A ) n n A L 0 =Pr(n A p A = π A )= ( ) n n A π n A A (1 π A) n n A Likelihood ratio test statistic: LRT = 2 ln (L a /L 0 ) Some clever people have shown that if H 0 is true, then LRT follows a χ 2 (df=1) distribution (approximately). Likelihood-ratio test for the example We observed n A =78andn B =22. H 0 :(p A, p B )=(0.75,0.25) H a :(p A, p B ) (0.75,0.25) L a =Pr(n A =78 p A =0.78) = ( ) = L 0 =Pr(n A =78 p A =0.75) = ( ) = LRT = 2 ln (L a /L 0 ) = Using a χ 2 (df=1) distribution, we get a p-value of We therefore have no evidence against the null hypothesis. In R: p-value = 1 - pchisq(0.49,1)

5 Null distribution Obs LRT A little math... n=n A +n B, n 0 A =E[n A H 0 ]=n π A, n 0 B =E[n B H 0 ]=n π B. ( ) na ( Then L a /L 0 = na n nb 0 n A 0 B ) nb ) Or equivalently LRT = 2 n A ln( na n 0 A ) +2 n B ln( nb. n 0 B Why do this?

6 Generalization to more than two groups If we have k groups, then the likelihood ratio test statistic is LRT = 2 k i=1 n i ln ( ni n 0 i ) If H 0 is true, LRT χ 2 (df=k-1) Null distributions 3 groups: χ 2 (df=2) 5 groups: χ 2 (df=4) groups: χ 2 (df=6) 9 groups: χ 2 (df=8)

7 Example In a dihybrid cross of tomatos we expect the ratio of the phenotypes to be 9:3:3:1. In 1611 tomatos, we observe the numbers 926, 288, 293, 104. Do these numbers support our hypothesis? Phenotype n i n 0 i n i /n 0 i n i ln ( ) n i /n 0 i Tall, cut-leaf Tall, potato-leaf Dwarf, cut-leaf Dwarf, potato-leaf Sum Results Obs LRT The test statistics LRT is Using a χ 2 (df=3) distribution, we get a p-value of We therefore have no evidence against the hypothesis that the ratio of the phenotypes is 9:3:3:1.

8 The chi-square test There is an alternative technique. The test is called the chi-square test, and has the greater tradition in the literature. For two groups, calculate the following: X 2 = (n A n 0 A) 2 n 0 A + (n B n 0 B) 2 n 0 B If H 0 is true, then X 2 is a draw from a χ 2 (df=1) distribution (approximately). Example In the first example we observed n A =78andn B = 22. Under the null hypothesis we have n 0 A =75andn 0 B =25.Wethereforeget X 2 = (78-75) (22-25)2 25 = =0.48. This corresponds to a p-value of We therefore have no evidence against the hypothesis (p A, p B )=(0.75,0.25). Note: using the likelihood ratio test we got a p-value of In R: chisq.test(c(78,22),p=c(0.75,0.25))

9 Generalization to more than two groups As with the likelihood ratio test, there is a generalization to more than just two groups. If we have k groups, the chi-square test statistic we use is X 2 = k i=1 (n i n 0 i ) 2 n 0 i χ 2 (df=k-1) Tomato example For the tomato example we get X 2 = ( ) ( ) ( ) ( ) = = 1.47 Using a χ 2 (df=3) distribution, we get a p-value of We therefore have no evidence against the hypothesis that the ratio of the phenotypes is 9:3:3:1. Using the likelihood ratio test we also got a p-value of In R: chisq.test(c(926,288,293,104),p=c(9,3,3,1)/16)

10 Test statistics Let n 0 i denote the expected count in group i if H 0 is true. LRT statistic LRT = 2 ln { } Pr(data p = MLE) Pr(data H 0 ) =...=2 i n i ln(n i /n 0 i ) χ 2 test statistic X 2 = (observed expected) 2 expected = i (n i n 0 i ) 2 n 0 i Null distribution of test statistic What values of LRT (or X 2 )shouldweexpect,ifh 0 were true? The null distributions of these statistics may be obtained by: Brute-force analytic calculations Computer simulations Asymptotic approximations If the sample size n is large, we have LRT χ 2 (k 1) and X 2 χ 2 (k 1)

11 The brute-force method Pr(LRT = g H 0 ) = n 1,n 2,n 3 giving LRT = g Pr(n 1, n 2, n 3 H 0 ) This is not feasible. Computer simulation 1. Simulate a table conforming to the null hypothesis. E.g., simulate (n 1, n 2, n 3 ) Multinomial(n=100, {1/4, 1/2, 1/4}) 2. Calculate your test statistic. 3. Repeat steps (1) and (2) many (e.g., 1000 or 10,000) times. Estimated critical value the 95th percentile of the results. Estimated P-value the prop n of results the observed value. In R, use rmultinom(n, size, prob) to do n simulations of a Multinomial(size, prob).

12 Example We observe the following data: AA AB BB We imagine that these are counts (n 1, n 2, n 3 ) Multinomial(n=100,{p 1, p 2, p 3 }). We seek to test H 0 : p 1 = 1/4, p 2 = 1/2, p 3 = 1/4. We calculate LRT 4.96 and X Referring to the asymptotic approximations (χ 2 dist n with 2 degrees of freedom), we obtain P 8.4% and P 6.9%. With 10,000 simulations under H 0,wegetP 8.9% and P 7.4%. Example Est d null dist n of LRT statistic Observed 95th %ile = LRT Est d null dist n of chi square statistic Observed 95th %ile = X 2

13 Summary and recommendation For either the LRT or the χ 2 test: The null distribution is approximately χ 2 (k 1) if the sample size is large. The null distribution can be approximated by simulating data under the null hypothesis. If the sample size is sufficiently large that the expected count in eachcell is 5, use the asymptotic approximation without worries. Otherwise, consider using computer simulations. Composite hypotheses Sometimes, we ask not p AA = 0.25, p AB = 0.5, p BB = 0.25 But rather something like: p AA = f 2, p AB = 2f(1 f), p BB =(1 f) 2 for some f. For example: Consider the genotypes, of a random sample of individuals, at a diallelic locus. Is the locus in Hardy-Weinberg equilibrium (as expected in the case of random mating)? Example data: AA AB BB

14 Another example ABO blood groups 3 alleles A, B, O. Phenotype A genotype AA or AO B genotype BB or BO AB genotype AB O genotype O Allele frequencies: f A, f B, f O (Note that f A + f B + f O = 1) Under Hardy-Weinberg equilibrium, we expect p A = f 2 A + 2f A f O p B = f 2 B + 2f B f O p AB = 2f A f B p O = f 2 O Example data: O A B AB LRT for example 1 Data: (n AA, n AB, n BB ) Multinomial(n,{p AA, p AB, p BB }) We seek to test whether the data conform reasonably to H 0 : p AA = f 2, p AB = 2f(1 f), p BB =(1 f) 2 for some f. General MLEs: ˆp AA = n AA /n, ˆp AB = n AB /n, ˆp BB = n BB /n MLE under H 0 : ˆf =(naa + n AB /2)/n p AA = ˆf 2, p AB = 2ˆf (1 ˆf), p BB =(1 ˆf) 2 LRT statistic: LRT = 2 ln { } Pr(nAA, n AB, n BB ˆp AA, ˆp AB, ˆp BB ) Pr(n AA, n AB, n BB p AA, p AB, p BB )

15 LRT for example 2 Data: (n O, n A, n B, n AB ) Multinomial(n,{p O, p A, p B, p AB }) We seek to test whether the data conform reasonably to H 0 : p A = f 2 A + 2f A f O,p B = f 2 B + 2f B f O,p AB = 2f A f B,p O = f 2 O for some f O, f A, f B,wheref O + f A + f B = 1. General MLEs: ˆp O, ˆp A, ˆp B, ˆp AB, like before. MLE under H 0 : Requires numerical optimization Call them (ˆf O,ˆf A,ˆf B ) ( p O, p A, p B, p AB ) LRT statistic: LRT = 2 ln { } Pr(nO, n A, n B, n AB ˆp O, ˆp A, ˆp B, ˆp AB ) Pr(n O, n A, n B, n AB p O, p A, p B, p AB ) χ 2 test for these examples Obtain the MLE(s) under H 0. Calculate the corresponding cell probabilities. Turn these into (estimated) expected counts under H 0. Calculate X 2 = (observed expected) 2 expected

16 Null distribution for these cases Computer simulation (with one wrinkle) Simulate data under H 0 (plug in the MLEs for the observed data) Calculate the MLE with the simulated data Calculate the test statistic with the simulated data Repeat many times Asymptotic approximation Under H 0,ifthesamplesize,n,islarge,boththeLRTstatistic and the χ 2 statistic follow, approximately, a χ 2 distribution with k s 1degrees of freedom, where s is the number of parameters estimated under H 0. Note that s = 1 for example 1, and s = 2 for example 2, and so df = 1 for both examples. Example 1 Example data: AA AB BB MLE: ˆf =(5+20/2)/100=15% Expected counts: Test statistics: LRT statistic = 3.87 X 2 =4.65 Asymptotic χ 2 (df = 1) approx n: P 4.9% P 3.1% 10,000 computer simulations: P 8.2% P 2.4%

17 Example 1 Est d null dist n of LRT statistic Observed 95th %ile = LRT Est d null dist n of chi square statistic 95th %ile = 3.36 Observed X 2 Example 2 Example data: O A B AB MLE: ˆfO 62.8%, ˆf A 25.0%, ˆf B 12.2%. Expected counts: Test statistics: LRT statistic = 1.99 X 2 =2.10 Asymptotic χ 2 (df = 1) approx n: P 16% P 15% 10,000 computer simulations: P 17% P 15%

18 Example 2 Est d null dist n of LRT statistic Observed 95th %ile = LRT Est d null dist n of chi square statistic Observed 95th %ile = X 2 Example 3 Data on number of sperm bound to an egg: count Do these follow a Poisson distribution? MLE: ˆλ =sampleaverage=( ) / Expected counts n 0 i = n e ˆλ ˆλi / i!

19 Example observed expected X 2 = (obs exp) 2 exp =...=42.8 LRT = 2 obs log(obs/exp) =...=18.8 Compare to χ 2 (df = = 4) P-value = (χ 2 ) and (LRT). By simulation: p-value = 16/10,000 (χ 2 ) and 7/10,000 (LRT) Null simulation results Observed Simulated χ 2 statistic Observed Simulated LRT statistic

20 A final note With these sorts of goodness-of-fit tests, we are often happy when our model does fit. In other words, we often prefer to fail to reject H 0. Such a conclusion, that the data fit the model reasonably well, should be phrased and considered with caution. We should think: how much power do I have to detect, with these limited data, a reasonable deviation from H 0?

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