Analysis of Variance (ANOVA) Homework Solution for STAT 516 - Spring 2004, Assignments of Data Analysis & Statistical Methods

Download Analysis of Variance (ANOVA) Homework Solution for STAT 516 - Spring 2004 and more Assignments Data Analysis & Statistical Methods in PDF only on Docsity! STAT 516 - Spring 2004 - Homework 4 Solutions Pg. 279: #1 False. If for two samples the conclusions from an ANOVA and t test disagree, you should trust the t-test made a mistake! (see bottom of page 230 and top of 231... they must be the same) Pg. 279: #2 True. Since F=MSB/MSW, if MSW is smaller, the F is larger and we would reject more often. Pg. 279: #12 Source df SS Between Factors 2 810 Within (error) 8 720 False: The null hypothesis is that all four three means are equal. False: F=MSB/MSW=(810/2)/(720/8)=405/90=4.5 (1.125 is SSB/SSW) False: The critical value for F for 5% significance is 6.60 4.46. (Table A.4A on page 627) True: It can be rejected at 5% significance because the observed 4.5 is greater than the critical value of 4.46. True: The null hypothesis cannot be rejected at 1% significance because the observed 4.5 is less than the critical value of 8.65. False: There are 10 11 observations in the experiment. (total df = 8+2 = sample size minus 1) 2a) Write down the equation of the one-way ANOVA model that is described by this set-up. Be sure to clearly identify each parameter and the sample sizes. Following page 233, the section called “The Linear Model for Several Populations” we could writte yij = µi + εij where the yij and εij are the observed scores and errors respectively for the jth observer for face i. µi is the average dominance score for that facial expression. The i faces are 1=Angry, 2=Disgusted, 3=Fearful, 4=Happy, 5=Sad, 6=Neutral; and, j goes from 0 to 6 (six observers for each face). Using the section on page 233 called “The Analysis of Variance Model” we would have split the µi into an overall average and treatment effects (τi). DATA faces; INPUT emotion $ rating @@; CARDS; Angry 2.10 Angry 0.64 Angry 0.47 Angry 0.37 Angry 1.62 Angry -0.08 Disg 0.40 Disg 0.73 Disg -0.07 Disg -0.25 Disg 0.89 Disg 1.93 Fear 0.82 Fear -2.93 Fear -0.74 Fear 0.79 Fear -0.77 Fear -1.60 Happy 1.71 Happy -0.04 Happy 1.04 Happy 1.44 Happy 1.37 Happy 0.59 Sad 0.74 Sad -1.26 Sad -2.27 Sad -0.39 Sad -2.65 Sad -0.44 Neut 1.69 Neut -0.60 Neut -0.55 Neut 0.27 Neut -0.57 Neut -2.16
; PROC INSIGHT: OPEN faces; FIT rating=emotion; RUN; PROC GLM DATA=faces ORDER=DATA; CLASS emotion; MODEL rating=emotion; MEANS emotion / HOVTEST=BF; RUN;
b) Check that the assumptions for performing a one-way ANOVA hold, including using Levene's test. -1 0 1 P_rating -2 0 2 R _ r a t i n g -2 0 2 RN_rating -2 0 2 R _ r a t i n g Brown and Forsythe's Test for Homogeneity of rating Variance ANOVA of Absolute Deviations from Group Medians Sum of Mean Source DF Squares Square F Value Pr > F emotion 5 1.6184 0.3237 0.67 0.6477 Error 30 14.4467 0.4816 1) Since the students were randomly divided into the six groups, the errors are independent. 2) The means of the errors are always 0 in a one-way ANOVA (or could look at residual vs. predicted plot) 3) The errors appear to be normally distributed as the q-q plot is very close to a straight line. 4) It is not clear from the residual versus predicted plot if the variance of the errors is constant, but with a p- value of 0.6477 in the modified Levene’s test we accept that the variances of the errors are equal. c) What hypothesis is being tested by the F-statistic in the ANOVA table? State your conclusion at the α=0.05 level. Analysis of Variance Source Model Error C Total DF 5 30 35 Sum of Squares 23.0852 34.9870 58.0722 Mean Square 4.6170 1.1662 F Stat 3.96 Pr > F 0.0071 H0: µAngry=µDisgusted=µFearful=µHappy=µSad=µNeutral HA: At least one is different At α=0.05 we reject H0 because the p-value of 0.0071 is less than α.