Chi-squared (χ 2 ) (1.10.5) and F-tests (9.5.2) for the variance of a normal distribution ( )

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1 Chi-squared (χ ) (1.10.5) and F-tests (9.5.) for the variance of a normal distribution χ tests for goodness of fit and indepdendence ( ) Prof. Tesler Math 83 Fall 016 Prof. Tesler χ and F tests Math 83 / Fall / 41

2 Tests of means vs. tests of variances Data x 1,..., x n, sample mean x, sample var. s X Data y 1,..., y m, sample mean ȳ, sample var. s Y Tests for mean One-sample tests: H 0 : µ = µ 0 vs. H 1 : µ µ 0 test statistic: z = x µ 0 σ/ n or t = x µ 0 s/ n (df =n 1) Tests for variance One-sample test: H 0 : σ = σ 0 vs. H 1 : σ σ 0 test statistic: chi-squared χ = (n 1)s /σ 0 (df =n 1) Two-sample tests: H 0 : µ X = µ Y vs. H 1 : µ X µ Y test statistic: x ȳ z = or t = σ X n + σ Y m x ȳ s p 1n + 1 m (df = n + m ) Two-sample test: H 0 : σ X = σ Y vs. H 1 : σ X σ Y test statistic: F = s Y /s X (with m 1 and n 1 d.f.) Prof. Tesler χ and F tests Math 83 / Fall 016 / 41

3 Application: The fine print in the Z and t-tests One-sample z-test, H 0 : µ = µ 0 vs. H 1 : µ µ 0 This assumes that you know the value of σ, say σ = σ 0. A χ test could be used to verify that the data is consistent with H 0 : σ = σ 0 instead of H 1 : σ σ 0. Two-sample z-test, H 0 : µ X = µ Y vs. H 1 : µ X µ Y This assumes that you know the values of σ X and σ Y. Separate χ tests for σ X and σ Y could be performed to verify consistency with the assumed values. Two-sample t-test, H 0 : µ X = µ Y vs. H 1 : µ X µ Y This assumes σ X = σ Y (but doesn t assume that this common value is known to you). An F-test could be used to verify that the data is consistent with H 0 : σ X = σ Y instead of H 1 : σ X σ Y. If the variances are unequal, Welch s t-test can be used instead of the regular two-sample t-test (Ewens & Grant pp ). Prof. Tesler χ and F tests Math 83 / Fall / 41

4 The χ ( Chi-squared ) distribution Used for confidence intervals and hypothesis tests on the unknown parameter σ of the normal distribution, based on the test statistic s (sample variance). It has the same degrees of freedom as for the t distribution. Point these out on the graphs: The chi-squared distribution with k degrees of freedom has Range [0, ) Mean µ = k Mode χ = k (for k, the pdf is maximum for χ = k ) χ = 1 (for k = 1) Median k(1 9k )3 Between k and k 3. Asymptotically decreases k 3 as k. Variance σ = k x PDF (k/) 1 e x/ k/ Γ (k/) : Γ distrib. with shape r = k, rate λ = 1 Unlike z and t, the pdf for χ is NOT symmetric. Prof. Tesler χ and F tests Math 83 / Fall / 41

5 The graphs for 1 and degrees of freedom are decreasing: pdf 3 pdf µ = 1 µ = χ = 0 χ = 0 χ = chiinv(.5,1) = qchisq(.5,1) = χ = chiinv(.5,) = qchisq(.5,) = The rest are hump shaped and skewed to the right: pdf pdf µ = 3 µ = 8 χ = 1 χ = 6 χ = chiinv(.5,3) = qchisq(.5,3) =.3660 χ = chiinv(.5,8) = qchisq(.5,8) = Prof. Tesler χ and F tests Math 83 / Fall / 41 8

6 χ ( Chi-squared ) distribution Cutoffs leftsided critical region sided acceptance region: df=5, " = 0.05 pdf ",df pdf ,5 = ,5 = Define χ α,df as the number where the cdf (area left of it) is α: P(χ df χ α,df ) = α Different notation than z α and t α,df (area α on right) since pdf isn t symmetric. Matlab R χ 0.05,5 = chiinv(.05,5) = qchisq(.05,5) = χ 0.975,5 = chiinv(.975,5) = qchisq(.975,5) = chicdf(0.831,5) = pchisq(0.831,5) = 0.05 chicdf(1.835,5) = pchisq(1.835,5) = chipdf(0.831,5) = dchisq(0.831,5) = chipdf(1.835,5) = dchisq(1.835,5) = Prof. Tesler χ and F tests Math 83 / Fall / 41

7 Two-sided cutoff sided acceptance region: df=5, " = 0.05 pdf ,5 = ,5 = The mean, median, and mode are different, so it may not be obvious what values of χ are more consistent with the null H 0 : σ = vs. the alternative σ Closer to the median of χ is "more consistent" with H 0. For -sided hypothesis tests or confidence intervals with α = 5%, we still put 95% of the area in the middle and.5% at each end, but the pdf is not symmetric, so the lower and upper cutoffs are determined separately instead of ± each other. Prof. Tesler χ and F tests Math 83 / Fall / 41

8 Two-sided hypothesis test for variance Test H 0 : σ = vs. H 1 : σ at sig. level α =.05 (In general, replace by σ 0 ; here, σ 0 = 100) Decision procedure 1 Get a sample x 1,..., x n. 650, 510, 470, 570, 410, 370 with n = 6 Calculate m = x 1+ +x n n and s = 1 n 1 n i=1 (x i m). m = , s = , s = Calculate the test-statistic χ = (n 1)s σ 0 χ = (n 1)s σ 0 = n i=1 = (6 1)( ) = 5.33 (x i m) σ 0 4 Accept H 0 if χ is between χ α/,n 1 and χ 1 α/,n 1. Reject H 0 otherwise. χ.05,5 =.831, χ.975,5 = Since χ = 5.33 is between these, we accept H 0. (Or, there is insufficient evidence to reject σ = ) Prof. Tesler χ and F tests Math 83 / Fall / 41

9 Doing the same test with a P-value pdf % 4.61% 37.69% Supports H 0 better Supports H 1 better median= P(χ ) = is the area left of 5.33 for χ with 5 d.f.: Matlab: chicdf(5.33,5) R: pchisq(5.33,5) Values at least as extreme as this are those at the 6.31th percentile or higher, OR at the 37.69th percentile or lower, so P = (1.631) = (.3769) = P > α (0.75 > 0.05) so accept H 0. To turn a one-sided P-value p 1 into a two-sided P-value, use P = min(p 1, 1 p 1 ). χ 5 Prof. Tesler χ and F tests Math 83 / Fall / 41

10 Two-sided 95% confidence interval for the variance Continue with data 650, 510, 470, 570, 410, 370 which has n = 6, m = , s = , s = Get bounds on σ in terms of s for the two-sided test: 0.95 = P(χ.05,5 < χ < χ.975,5) = P(0.831 < χ < 1.835) ) = P (0.831 < (6 1)S < σ ( ) = P (6 1)S > σ > (6 1)S A two-sided 95% ( confidence interval ) for the variance σ is (6 1)S 1.835, (6 1)S = ( , ) A two-sided 95% confidence interval for σ is ( ) (6 1)S 1.835, (6 1)S = (64.47, 53.31) Prof. Tesler χ and F tests Math 83 / Fall / 41

11 Properties of Chi-squared distribution 1 Definition of Chi-squared distribution: Let Z 1,..., Z k be independent standard normal variables. Let χ k = Z Z k. The pdf of the random variable χ k with k degrees of freedom. is the chi-squared distribution Pooling property: If U and V are independent χ random variables with q and r degrees of freedom respectively, then U + V is a χ random variable with q + r degrees of freedom. 3 Sample variance: Pick X 1,..., X n from a normal distribution N(µ, σ ). It turns out that n (X i X) σ = SS (n 1)S = σ σ i=1 has a χ distribution with df = n 1, so we test on χ = (n 1)s σ 0. Prof. Tesler χ and F tests Math 83 / Fall / 41

12 Distributions with an additive/pooling property For certain families of distributions, if U, V are independent random variables of that type, then U + V is too, with certain parameters combining additively. Parameters of Distribution U V U + V Binomial (n, p) (m, p) (n + m, p) Negative binomial (r, p) (s, p) (r + s, p) Gamma (r, λ) (s, λ) (r + s, λ) Poisson µ ν µ + ν χ q d.f. r d.f. q + r d.f. Prof. Tesler χ and F tests Math 83 / Fall / 41

13 F distribution Let U and V be independent χ random variables with q and r degrees of freedom, respectively. Then the random variable F = F q,r = U/q V/r is called the F distribution with q and r degrees of freedom. Range [0, ) Mean Mode r r if r > r(q ) q(r+) if q > Variance PDF r (q+r ) q(r ) (r 4) if r > 4 messy Median 1 if q = r > 1 if q > r < 1 if q < r Prof. Tesler χ and F tests Math 83 / Fall / 41

14 F distribution.5 1 F 1,1 F 3,1 F 1, 0.8 F 3, F 1,5 F 3,5 1.5 F 1, F 3,10 F 1,30 F 3,30 1 F 1, F 3,100 F 1, F 3, F 10,1 F 30,1 0.8 F 10, F 10,5 1.5 F 30, F 30,5 0.6 F 10,10 F 30,10 F 10,30 1 F 30, F 10,100 F 30, F 10, 0.5 F 30, Prof. Tesler χ and F tests Math 83 / Fall / 41

15 F distribution Define F α,q,r as the number where P(F F α,q,r ) = α (left-hand area α) sided acceptance region for F 7,5, =5% pdf F F 0.05,7,5 = ,7,5 = Matlab R F 0.05,7,5 = finv(.05,7,5) = qf(.05,7,5) = F 0.975,7,5 = finv(.975,7,5) = qf(.975,7,5) = F 7,5 fcdf(0.189,7,5) = pf(0.189,7,5) = 0.05 fcdf(6.853,7,5) = pf(6.853,7,5) = fpdf(0.189,7,5) = df(0.189,7,5) = fpdf(6.853,7,5) = df(6.853,7,5) = R s command df is probability density of the F distribution, not degrees of freedom. Prof. Tesler χ and F tests Math 83 / Fall / 41

16 Two-sided hypothesis test for F Test at significance level α = 5%: H 0 : σ X = σ Y vs. H 1 : σ X σ Y Data for X: 650, 510, 470, 570, 410, 370 x = , s X = , s X = 103.8, df = 6 1 = 5 Data for Y: 510, 40, 50, 360, 470, 530, 550, 490 ȳ = 481.5, s Y = 401.5, s Y = , df = 8 1 = 7 Test statistic: F = F 7,5 = s Y /s X = = Since our test statistic lies between the cutoffs F.05,7,5 = and F.975,7,5 = , we accept H 0 / reject H 1. Prof. Tesler χ and F tests Math 83 / Fall / 41

17 P-values The CDF is P(F 7, ) = Matlab: fcdf(.376,7,5) R: pf(.376,7,5) To make it two sided, P = min(0.1174, ) = (0.1174) = Since P > α (0.348 > 0.05), we accept the null hypothesis. Prof. Tesler χ and F tests Math 83 / Fall / 41

18 F statistic to compare variances (two-sample data) Theoretical setup for H 0 : σ X = σ Y vs. H 1 : σ X σ Y First sample: x 1,..., x n V = (n 1)s X = σ X Second sample: y 1,..., y m n i=1 (x i x) σ X with n 1 d.f. U = (m 1)s Y = σ Y m j=1 (y i ȳ) σ Y with m 1 d.f. Assuming the null hypothesis σ X = σ Y, the variances cancel: For F = /σ Y U/(m 1) V/(n 1) = s Y s X /σ = s Y X s X H 0 : σ X = Cσ Y vs. H 1 : σ X Cσ Y with m 1 and n 1 d.f. where C > 0 is constant, use F = Cs Y /s X instead. Prof. Tesler χ and F tests Math 83 / Fall / 41

19 k-sample experiments ANOVA (Analysis of Variance) is a procedure to compare the means of k-sample data (analagous to two-sample data, but for k independent sets of data). It involves the F distribution with a formula for F that takes into account all k samples instead of just two samples. Prof. Tesler χ and F tests Math 83 / Fall / 41

20 χ tests for goodness of fit and independence ( ) Prof. Tesler χ and F tests Math 83 / Fall / 41

21 Multinomial test Consider a k-sided die with faces 1,,..., k. We want to simultaneously test that the probabilities p 1, p,..., p k of rolling 1,,..., k are specified values. To test if a 6-sided die is fair, H 0 : (p 1,..., p 6 ) = (1/6,..., 1/6) H 1 : At least one p i 1/6 Decision rule is based counting # 1 s, s, etc. on n independent rolls of the die. For the fair coin problem, the exact distribution was binomial, and we approximated it with a normal distribution. For this problem, the exact distribution is multinomial. We will combine the separate counts of 1,,... into a single test statistic whose distribution is approximately a χ distribution. Prof. Tesler χ and F tests Math 83 / Fall / 41

22 Goodness of fit tests for Mendel s experiments In Mendel s pea plant experiments, yellow seeds (Y) are dominant and green (y) recessive; round seeds (R) are dominant and wrinkled (r) are recessive. Consider the phenotypes of the offspring in a dihybrid cross YyRr YyRr: Expected Observed Type fraction number yellow & round 9/ yellow & wrinkled 3/ green & round 3/ green & wrinkled 1/16 3 Total: n = 556 Hypothesis test: H 0 : (p 1, p, p 3, p 4 ) = ( 9 16, 3 16, 3 16, 1 16 ) H 1 : At least one p i disagrees Prof. Tesler χ and F tests Math 83 / Fall 016 / 41

23 Does the data fit the expected distribution? Expected Observed Type fraction number yellow & round 9/ yellow & wrinkled 3/ green & round 3/ green & wrinkled 1/16 3 Total: n = 556 The observed number of yellow & round plants is O = 315. (Don t confuse the letter O with the number 0.) The expected number is E = (9/16) 556 = The goodness of fit test requires that we convert all the expected proportions into expected numbers. Prof. Tesler χ and F tests Math 83 / Fall / 41

24 Goodness of fit test Observed number Expected number Type O E O E (O E) /E yellow & round 315 (9/16)556 = yellow & wrinkled 101 (3/16)556 = green & round 108 (3/16)556 = green & wrinkled 3 (1/16)556 = Total k = 4 categories give k 1 = 3 degrees of freedom. (The O and E columns both total 556, so the O E column totals 0; thus, any 3 of the (O E) s dictate the fourth.) The test statistic is the total of the last column, χ 3 = The general formula is χ k 1 = k i=1 (O i E i ) E i. Warning: Technically, that formula only has an approximate chi-squared distribution. When E 5 in all categories, the approximation is pretty good. Prof. Tesler χ and F tests Math 83 / Fall / 41

25 Goodness of fit test Smaller values of χ indicate better agreement between the O and E values (so support H 0 better). Larger values support H 1 better. It s a one-sided test pdf Supports H 0 better Supports H 1 better Observed pdf The P-value is the probability, under H 0, of a test statistic that supports H 1 as well as or better than the observed value: P = P(χ ) = Matlab: 1-chicdf( ,3) R: 1-pchisq( ,3) Prof. Tesler χ and F tests Math 83 / Fall / 41

26 Goodness of fit test P = is not too extreme. It means that if H 0 is true and the experiment is repeated a lot, about 7.5% of the time, a χ 3 value supporting H 0 better (lower values of χ 3 ) will be obtained, and about 9.5% of the time, values supporting H 1 better (higher values of χ 3 ) will be obtained. Prof. Tesler χ and F tests Math 83 / Fall / 41

27 Ronald Fisher ( ) He made important contributions to both statistics and genetics. Connection: he invented statistical methods while working on genetics problems. Our way of using the normal, Student t, χ, and F distributions in the same framework, plus ANOVA, is due to him. In genetics, he reconciled continuous variations (heights and weights) with Mendelian genetics (discrete traits), and developed much of population genetics. Prof. Tesler χ and F tests Math 83 / Fall / 41

28 Did Mendel fudge his data? For independent experiments, the values of χ may be pooled by adding the χ values and adding the degrees of freedom. Fisher pooled the data from Mendel s experiments and got χ = with 84 degrees of freedom. Assuming Mendel s laws are true, how often would we get χ 84 supporting H 0 /H 1 better than this? Support H 0 better: P(χ ) = Support H 1 better: P-value P = P(χ ) = = So if Mendel s laws hold and 1 million researchers independently conducted the same experiments as Mendel, about 9 of them would get data with as little or even less variation than Mendel had. Prof. Tesler χ and F tests Math 83 / Fall / 41

29 Did Mendel fudge his data? pdf pdf Supports H 0 better Prob. = = Supports H 1 better Prob. = Prof. Tesler χ and F tests Math 83 / Fall / 41

30 Did Mendel fudge his data? Based on this and similar tests, Fisher believed that something was fishy with Mendel s data: The values are too good in the sense that they are too close to what was expected. At the same time, they are bad in the sense that there is too little random variation. Some people have accused Mendel of faking data. Others speculate that he only reported his best data. Other people defend Mendel by speculating on biological explanations for why his results would be better than expected. All pro and con arguments have later been rebutted by someone else. Prof. Tesler χ and F tests Math 83 / Fall / 41

31 Tests of independence ( contingency tables ) A study in 1899 examined 6800 German men to see if hair color and eye color are related. Observed counts O: Hair color Brown Black Fair Red Total Eye Brown Color Gray/Green Blue Total Hypothesis test (at α = 0.05) H 0 : eye color and hair color are independent, vs. H 1 : eye color and hair color are correlated Meaning of independence For all eye colors x and all hair colors y: P(eye color=x and hair color=y) = P(eye color=x) P(hair color=y) Prof. Tesler χ and F tests Math 83 / Fall / 41

32 Computing E table Hypothesis test (at α = 0.05) H 0 : eye color and hair color are independent, vs. H 1 : eye color and hair color are correlated The fraction of people with red hair is 116/6800. The fraction with blue eyes is 811/6800. Use these as point estimates: P(hair color=red) 116/6800 and P(eye color=blue) 811/6800. Under the null hypothesis, the fraction with red hair and blue eyes would be ( )/6800. The expected number of people with red hair and blue eyes is 6800( )/6800 = ( )/6800 = (Row total times column total divided by grand total.) Compute E this way for all combinations of hair and eye color. As long as E 5 in every cell (here it is) and the data is normally distributed (an assumption), the χ test is valid. Prof. Tesler χ and F tests Math 83 / Fall / 41

33 Computing E and O E tables Expected counts E: Hair color Brown Black Fair Red Eye Brown Color Gray/Green Blue In each position, compute O E. For red hair and blue eyes, this is O E = =.95: O E: Hair color Brown Black Fair Red Eye Brown Color Gray/Green Blue Note all the row and column sums in the O E table are 0, so if we hid the last row and column, we could deduce what they are. Thus, this 3 4 table has (3 1)(4 1) = 6 degrees of freedom. Prof. Tesler χ and F tests Math 83 / Fall / 41

34 Computing test statistic χ Compute (O E) /E in each position. For red hair and blue eyes, this is (.95) /47.95 = (You could go directly to this computation after the E computation, without doing O E first.) (O E) /E: Hair color Brown Black Fair Red Eye Brown Color Gray/Green Blue Add all twelve of these to get χ = = There are 6 degrees of freedom, so χ 6 = Prof. Tesler χ and F tests Math 83 / Fall / 41

35 Performing the test of independence χ would be 0 if the traits were truly independent. Smaller values support H 0 better (traits independent). Larger values support H 1 better (traits correlated). It s a one-sided test. At the 0.05 level of significance, we reject H 0 if χ 6 χ 0.95,6 = Indeed, > so we reject H 0 and conclude that hair color and eye color are linked in this data. This doesn t prove that a particular hair color causes one to have a particular eye color, or vice-versa; it just says there s a correlation in this data. Using P-values: P = P(χ ) so P α = 0.05 and we reject H 0. Matlab: can t compute this (gives P = 0). R: pchisq( ,6,lower.tail=false) Prof. Tesler χ and F tests Math 83 / Fall / 41

36 Performing the test of independence pdf 6 pdf pdf Supports H 0 better Prob. = 1 " Supports H 1 better Prob. = " = 1.1e 8 6 = pdf Prof. Tesler χ and F tests Math 83 / Fall / 41

37 Mendel s pea plants revisited: Are loci Y and R linked? We will use the same data as in the goodness-of-fit test but for a different purpose. Consider the phenotypes of the offspring in a dihybrid cross YyRr YyRr: Observed counts O: Seed Shape Round (R) Wrinkled (r) Total Seed Yellow (Y) Color Green (y) Total Hypothesis test (at α = 0.05) H 0 : Seed color and seed shape are independent, vs. H 1 : Seed color and seed shape are correlated Prof. Tesler χ and F tests Math 83 / Fall / 41

38 Mendel s pea plants revisited: Are loci Y and R linked? Seed Color Seed Shape Round (R) Wrinkled (r) Total O: (Observed #) Yellow (Y) Green (y) Total E: (Expected #) Yellow (Y) Green (y) Total O E: (Deviation) Yellow (Y) Green (y) Total (O E) /E: Yellow (Y) (χ contrib.) Green (y) Total Prof. Tesler χ and F tests Math 83 / Fall / 41

39 Mendel s pea plants revisited: Are loci Y and R linked? Using χ as the test statistic: χ 1 cutoff: χ 0.95,1 df = ( 1)( 1) = 1 = = = chiinv(.95,1) = qchisq(.95,1) = < so it s not significant Using P-values: P = P(χ 1 > ) = 1-chicdf(0.1163,1) = pchisq(0.1163,1) P > 0.05 so Accept H 0 (genes not linked) Prof. Tesler χ and F tests Math 83 / Fall / 41

40 Comparison of the two tests At fertilization, if genes R and Y are not linked, then in an RrYy RrYy cross, the expected proportions are RY :Ry:rY :ry = 1:1:1:1. If linked, it would be different. Some genotypes may not survive to the points at which the phenotype counts are made; e.g., hypothetically, 40% of individuals with Rr might not be born, might die before reproducing (affecting multigenerational experiments), etc. This would change the ratio of RR:Rr :rr from 1::1 to 1:1.:1 = 5:6:5, and round:wrinkled from 3:1 to.:1 = 11:5. Prof. Tesler χ and F tests Math 83 / Fall / 41

41 Comparison of the two tests The goodness-of-fit test assumed all genotypes are equally viable. Whether the genes are linked or not should be a separate matter. If you know the yellow:green and round:wrinkled viability ratios, you can use the goodness-of-fit test on 4 phenotypes with 3 degrees of freedom by adjusting the proportions. If you don t know these viability ratios, you can estimate the ratios from data via contingency tables, at the cost of dropping to 1 degree of freedom. Prof. Tesler χ and F tests Math 83 / Fall / 41