llOne-Way AnovA, cont Format for data: Data appear in separate columns or rows, organized as treatment groups. Sample size of each group may differ Calculations SST= SSTR SSE definitions follow) Sum of squares total sst)= sum of squared differences between each individual data value (regardless of group membership) minus the grand mean, x, across all data. total variation in the data not variance SST=∑∑(x1,-x)2 o 2002 The Wadsworth Group
One-Way ANOVA, cont. • Format for data: Data appear in separate columns or rows, organized as treatment groups. Sample size of each group may differ. • Calculations: – SST = SSTR + SSE (definitions follow) – Sum of squares total (SST) = sum of squared differences between each individual data value (regardless of group membership) minus the grand mean, , across all data... total variation in the data (not variance). SST = ( – x)2 ij x x © 2002 The Wadsworth Group
ll One-Way AnovA,cont Calculations, cont Sum of squares treatment (SSTr)=sum of squared differences between each group mean and the grand mean, balanced by sample size.between groups variation (not variance) SSTR=∑n.(x Sum of squares error sse)=sum of squared differences between the individual data values and the mean for the group to which each belongs.within group variation(not variance) SSE=∑∑(x1;-x o 2002 The Wadsworth Group
One-Way ANOVA, cont. • Calculations, cont.: – Sum of squares treatment (SSTR) = sum of squared differences between each group mean and the grand mean, balanced by sample size... betweengroups variation (not variance). – Sum of squares error (SSE) = sum of squared differences between the individual data values and the mean for the group to which each belongs... withingroup variation (not variance). SSTR ( – x)2 j x j = n SSE = (x ij – x j )2 © 2002 The Wadsworth Group
lLOne-Way ANOVA, cont Calculations, cont Mean square treatment(MSTR)=SstR/(t-1) where t is the number of treatment groups. between groups varlance Mean square error(Mse)=SSE/(N-t)where n is the number of elements sampled and t is the number of treatment groups. within-groups variance F-Ratio= MStR/MSE, where numerator degrees of freedom are t-1 and denominator degrees of freedom are N-t o 2002 The Wadsworth Group
One-Way ANOVA, cont. • Calculations, cont.: – Mean square treatment (MSTR) = SSTR/(t – 1) where t is the number of treatment groups... betweengroups variance. – Mean square error (MSE) = SSE/(N – t) where N is the number of elements sampled and t is the number of treatment groups... within-groups variance. – F-Ratio = MSTR/MSE, where numerator degrees of freedom are t – 1 and denominator degrees of freedom are N – t. © 2002 The Wadsworth Group
l One-Way ANoVA- An example if occupancy of a vehicle might be related to the speed r da Problem 12.30: Safety researchers, interested in determinin which the vehicle is driven, have checked the following speed(MPH)measurements for two random samples of vehicles Driver alone. 64507155676180565974 1+ rider(s:445254486967545758516267 a. What are the null and alternative hypotheses? 11=12 where group 1= driver alone H1:A1≠A2 Group 2= with rider(s) o 2002 The Wadsworth Group
One-Way ANOVA - An Example Problem 12.30: Safety researchers, interested in determining if occupancy of a vehicle might be related to the speed at which the vehicle is driven, have checked the following speed (MPH) measurements for two random samples of vehicles: Driver alone: 64 50 71 55 67 61 80 56 59 74 1+ rider(s): 44 52 54 48 69 67 54 57 58 51 62 67 a. What are the null and alternative hypotheses? H0 : µ1 = µ2 where Group 1 = driver alone H1 : µ1 µ2 Group 2 = with rider(s) © 2002 The Wadsworth Group
ul One-Way aNOva- An example b Use ANOVA and the 0.025 level of significance in testing the appropriate null hypothesis 1=637,S1=9.357 56.916.S~=7.806.n、=12 x=60.0 SSTR=10(637-602+12(56917-602=250983 SSE=(64-637)2+(50-63.7)2+…+(74-63.7)2 +(44-56917)2+(52-56917)2+…+(67-56917)2 1487.017 SSTotal=(64-60)2+(50-60)2+…+(74-60)2 +(44-60)2+(52-60)2+…+(67-60)2 1738 o 2002 The Wadsworth Group
One-Way ANOVA - An Example b. Use ANOVA and the 0.025 level of significance in testing the appropriate null hypothesis. SSTR = 10(63.7 – 60)2 + 12(56.917 – 60)2 = 250.983 SSE = (64 – 63.7 )2 + (50 – 63.7 )2 + ... + (74 – 63.7 )2 + (44 – 56.917) 2 + (52 – 56.917) 2 + ... + (67 – 56.917) 2 = 1487.017 SSTotal = (64 – 60 )2 + (50 – 60 )2 + ... + (74 – 60 )2 + (44 – 60) 2 + (52 – 60) 2 + ... + (67 – 60) 2 = 1738 x 1 = 63.7, s 1 = 9.3577, n 1 = 10 x 2 = 56.916 , s 2 = 7.806, n 2 = 12 x = 60.0 © 2002 The Wadsworth Group