I i=1 1 I(J 1) j=1 (Y ij Ȳi ) 2. j=1 (Y j Ȳ )2 ] = 2n( is the two-sample t-test statistic.

Size: px

Start display at page:

Download "I i=1 1 I(J 1) j=1 (Y ij Ȳi ) 2. j=1 (Y j Ȳ )2 ] = 2n( is the two-sample t-test statistic."

Jayson Ball
5 years ago
Views:

1 Serik Sagitov, Chalmers and GU, February, 08 Solutions chapter Matlab commands: x = data matrix boxplot(x) anova(x) anova(x) Problem.3 Consider one-way ANOVA test statistic For I = and = n, put F = MS A MS E = I I i= (Ȳi Ȳ ) I( ) I i= j= (Y ij Ȳi ) Ȳ = X, Ȳ = Ȳ, Ȳ = X + Ȳ. In this two-sample setting, the F-test statistic F = n[( X X+Ȳ ) + (Ȳ X+Ȳ ) ] [ n (n ) j= (X j X) + n j= (Y j Ȳ ) ] = n( X Ȳ ) s p = ( X Ȳ s ), p n equals T, where T = X Ȳ s p n is the two-sample t-test statistic. Problem.5 Derive the likelihood ratio test for the one-way layout and show that it is equivalent to the F-test. The null hypothesis says that the data Y ij comes from a single normal distribution H 0 : µ =... = µ I = µ described by two parameters µ and σ, so that dim Ω 0 =, while dim Ω = I +. The likelihood ratio where putting n = I, L(µ,..., µ I, σ ) = I Λ = i= j= L 0 (ˆµ, ˆσ 0) L(ˆµ,..., ˆµ I, ˆσ ), πσ e (Y ij µ i ) σ σ n exp{ ΣΣ (Y ij µ i ) }, σ L 0 (µ, σ ) = L(µ,..., µ, σ ) σ n exp{ ΣΣ (Y ij µ) σ }.

2 We find the maximum likelihood estimates to be ˆµ = Ȳ, ˆσ 0 = SS T n, ˆµ i = Ȳi, ˆσ = SS E n, which yields Λ = ˆσ n 0 exp{ ΣΣ (Y ij ˆµ) } ˆσ 0 ˆσ n exp{ ΣΣ (Y ij ˆµ i ) } = ( ˆσ 0 ) n/. ˆσ ˆσ The likelihood ratio test rejects the null hypothesis for small values of Λ or equivalently for large values of ˆσ 0 = SS ˆσ T SS E = + SS A SS E = + (I )MS A I( )MS E = + (I ) F I( ) that is for large values of F-test statistics. This leads to an asymptotic approximation of the F (I ),I( ) in terms of the chi-square distribution with df = I. Problem.0 One-way layout with I = 0, = 7, X ij N(µ i, σ ). Pooled sample variance uses df = I( ) = 60. s p = MS E = I( ) (X ij X i. ) i j (a) A 95% CI for a single difference µ u µ v X u X v ± t 60 (0.05)s p has the half-width of.8 sp. (b) Bonferroni simultaneous 95% CI for ( ) 0 = 45 differences µu µ v X u X v ± t 60 ( )s p has the half-width of giving the ratio 4.79 sp, 4.79 =.7..8 (c) Tukey simultaneous 95% CI for differences µ u µ v X u X v ± q 0,60 (0.05) sp has the half-width of giving the ratio Bonferroni Tukey 4.65 sp, = =.03.

3 Problem. For I = 4 control groups of = 5 mice each, test H 0 : no systematic differences between groups. Significant differences among the control groups, although not expected, might be attributable to changes in the experimental conditions Values Column Number One way ANOVA table Source SS df MS F P Columns Error Total Do not reject H 0 at 0% significance level. Boxplots show non-normality. The largest difference is between the third and the fourth boxplots. Control question: why the third boxplot has no upper whisker? Kruskal-Wallis test. Pooled sample ranks Kruskal-Wallis test statistic group I R. = 8.4 group II R. = 0.6 group III R3. = 7. group IV R4. = 5.8 K = 5 ( ( ) + ( ) + (7. 0.5) + ( ) ) = Since χ 3(0.) = 6.5, we do not reject H 0 at 0% significance level. Problem.6 I = 3 treatments on = 0 subjects with K = observations per cell. H 0 : no treatment effects. Results of anova(x): 3

4 Source SS df MS F Columns (blocks) Rows (treatments) Error Total Two P-values: columns = 0.877, rows = Reject H 0 at 5% significance level. Friedman s test. Ranking within blocks: The observed value of the Friedman test statistic Q = R. = R. = R3. = ( (.8 ) + (.9 ) + (.3 ) ) =.4. Since χ (0.) = 4.6, we can not reject H 0 even at 0% significance level. Problem Values Group Number I = 3 types of stopwatches, different sample sizes. H 0 : no systematic differences between groups. One way ANOVA table gives the P-value of We do not reject H 0. Source SS df MS F Columns Error Total

5 Kruskal-Wallis test. Pooled sample ranks group I:,, 3, 4, 7, 0.5, 4, 5, 0, R. = 8.5 group II: 6, 8,, 6.5, 6.5, 9, R. = 3.0 group III: 5, 9, 0.5, 3, 8, R3. =. The observed value of the test statistic K = ( 9 ( ) + 6 ( ) + 5 (. 0.5) ) =.5. 0 Since χ (0.) = 4.6, we do not reject H 0 even at 0% significance level. Problem.34 Forty eight survival times: I = 3 poisons and = 4 treatments with K = 4 observations per cell. Cell means for the survival times A B C D I II III Draw three profiles: I and II cross each other, and profile III is more flat. Three null hypotheses of interest H A : no poison effect, H B : no treatment effect, H AB : no interaction. (a) Survival in hours x data matrix. Results of anova(x,4) Source SS df MS F Columns (treatments) Rows (poisons) Intercation Error Total Three P-values: columns = , rows = , interaction = Reject H A and H B at % significance level, we can not reject H AB even at 0% significance level: 3 poisons act differently, 4 treatments act differently, some indication of interaction. Analysis of the residuals Y ijk Ȳij normal probability plot reveals non-normality, skewness = 0.59, kurtosis = 4.. 5

6 Normal Probability Plot Normal Probability Plot Probability 0.50 Probability Data Data Figure : Left panel: survival times. Right panel: death rates. (b) Transformed data: death rate = /survival time. Cell means for the death rates A B C D I II III Draw three profiles: they look more parallel. New data matrix y=x.ˆ(-). Results of anova(y,4): Source SS df MS F Columns (treatments) Rows (poisons) Intercation Error Total Three P-values: columns = , rows = , interaction = Reject H A and H B at % significance level, accept H AB at 0% significance level. Conclusions 3 poisons act differently, 4 treatments act differently, no interaction, the normal probability plot of residuals reveals a closer fit to normality assumption. 6

Chapter 12. Analysis of variance

Chapter 12. Analysis of variance Serik Sagitov, Chalmers and GU, January 9, 016 Chapter 1. Analysis of variance Chapter 11: I = samples independent samples paired samples Chapter 1: I 3 samples of equal size J one-way layout two-way layout