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Comparing more than two population means We can use two sample test to test the equality of more than two population means, but this procedure – Require large number of two sample tests – Performing many two sample tests at α tends to inflate the overall α risk. For example, To test the equality of 10-population means, we have to perform 45 test If the tests are independent and each test use α =0.05, then overall α=45(0.05)=2.25 we require a procedure for carrying out test of hypothesis about the equality of several population means simultaneously –we can use F-distribution in ANOVA that yields a single test statistic for comparing all means. Analysis of Variance (ANOVA) Analysis of Variance is a procedure that partitions the total variability present in the data set into meaningful and distinct components. Each component represents the variation due to a recognized source of variation, in addition, one component represents the variation due to uncontrolled factors and random errors associated with the response measurements NORMALITY:-The K-populations from which sample are drawn should be normal INDEPENDENCE:-The k-samples should be independent Randomness: The k-Samples should be random Four groups of students were subjected to different teaching techniques and tested at the end of a specified period of time. Due to drop outs in the experimental groups (sickness, transfers etc) the number of students varied from group to group Method 1 Method 2 Method 3 Method 4 65 75 59 94 87 69 78 89 73 83 67 80 79 81 62 88 81 72 83 69 79 76 90 454 549 425 351 1779 Do the data provide sufficient evidence to indicate a difference in the mean achievements for the 4 teaching techniques? Construction of hypotheses Ho : 1=2=3=4(i.e Mean achievements from 4 methods are same) H1: At least two ’s are different Level of significance = 5% Test Statistic 2 F S b 2 S w A N O V A TABLE Source Of Variation Degree of Freedom Sum of Squares Mean Sum of Squares Fcal (S.O.V) DF SS MSS=SS/df Between Methods 4-1 =3 712.6 237.5 S2 b 3.77* Within Methods 22-3=19 1196.6 63.0 S2w(MSE) (Error) TOTAL 23-1=22 1909.2 CALCULATION FOR ANOVA TABLE ? Method 1 Method 2 Method 3 Method 4 65 75 59 94 Correction Factor=(G.T)2/Obs= (1779)2/23 = 137601.8 87 69 78 89 73 83 67 80 Total SS=(65)2+(87)2 …(88)2 – CF= 139511 – 137601.8 79 81 62 88 81 72 83 = 1909.2 69 79 76 90 454 549 425 351 1779 (454 ) 2 (549 ) 2 (425 ) 2 (351 ) 2 BetweenMet hodsSS CF 138314 .4 137601 .8 712 .6 6 7 6 4 Within Methods SS =Total SS – Between Methods SS=1909.2 – 712.6 = 1196.6 SOV DF SS MSS=SS/d f F cal Between Methods 4-1=3 712.6 237.5 237.5/63=3.77 Within Methods 19 1196.6 63 TOTAL 23-1=22 1909.2 Decision Rule:- Reject Ho if F cal F(3,19) Result:-As F cal =3.77 > F.05(3,19) =3.10 So reject Ho and conclude that there is difference in the mean achievements for the four teaching methods. • Let 6 different wheat varieties with three plots each was observed and measured the larval population (%).The data is given below. • Test the Hypothesis that all Varieties have the same Larval Population? H 0 ; U1 U 2 U3 U 4 U5 U 6 H1; At least one mean is not equall Now, the layout is as follows: 1-V3 2-V6 3-V4 4-V2 5-V2 6-V2 74.00 48.57 49.24 59.75 45.75 60.22 7-V5 8-V3 9-V5 10-V4 11-V5 12-V6 50.30 75.28 36.70 60.84 36.06 45.33 13-V1 14-V1 15-V6 16-V3 17-V1 18-V4 42.93 31.20 39.15 61.53 33.28 60.58 V1 V2 V3 V4 V5 V6 42.93 59.75 74.00 49.24 50.30 48.57 31.20 45.75 75.28 60.84 36.70 45.33 33.28 60.22 61.53 60.58 36.06 39.15 TOTAL 107.41 165.72 210.81 170.6 123.06 133.05 910.71 1. Group the data by treatment and calculate the treatment totals (T) and grand total (G). 2. Construct an outline of the analysis of variance as follows: SOURCE OF DEGREE OF SUM OF MEAN COMPUTED TABULAR F VARIATION FREEDOM SQUARES SQUARE F 5% 1% TREATMENT t-1 EXPERIMENTAL ERROR t(r-1) TOTAL (r)(t)-1 3. Calculation of Degree of Freedom Total d.f = (r) (t) – 1 = (3) (6) -1 = 18 – 1 = 17 Treatments d.f. = t – 1 =6–1=5 Error d.f. = t(r-1) = 6(3 – 1) = 6(2) = 12 The error d.f. can also can also be obtained through subtraction as: Error d.f. = total d.f. – treatment d.f. = 17 – 5 = 12 4. Using X i to represent the measurement of the ith plot, Ti as the total of the ith treatment, and n as the total number of experimental plots [i.e., n =(r)(t)], Calculate the correction factor and the various sums of squares (SS) as: STEP-1 2 Correction factor (C.F.) = G n Where G indicates X (42.93 31.20 33.28 59.75 45.75 ........ 39.15)2 C.F. 18 2 (910.71) 18 = 46077.37245 STEP-2 n Total SS (sum of squares) = X i2 C.F i 1 (42.93)2 (31.20)2 .......... .. (45.33)2 (39.51)2 C.F = 49062.475-46077.37 = 2985.105 STEP-3 t 2 Ti SS treatment = r C.F. i 1 i (107.41)2 (165.72)2 .......... . (133.05)2 46077.37 3 =48470.59-46077.37 =2393.22 Note: Treatment = Variety STEP- 4 SS error = Total SS – Treatment SS SS error = 2985.10 – 2393.22 SS error = 591.88 We can calculate the men square (MS) for each source of variation by dividing each SS by its corresponding d.f: TreatmentSS Treatment MS = t 1 2393.22 2393.22 6 1 5 = 478.64 ErrorSS Error MS = t(r 1) 591.88 591.88 591.88 6(3 1) 6(2) 12 = 49.32 ANOVA Table S.O.V. D.F. SS MSS F-value Treat 5 2393.22 478.64 Error 12 591.88 49.32 9.70 total 17 2985.10 Calculate the F value for testing significance of the treatment difference as: TreatmentMS F= ErrorMS 478.64 49.32 = 9.70 The F value should be computed only when the error d.f. is large enough for a reliable estimate of the error variance. We can calculate the tabular F value from F-distribution table where f1 = treatment d.f. = (t-1) = 6-1 = 5 f 2 = error d.f. = t(r-1) = 6(3-1) = 6(2) =12 For our example, tabular F values are 5.06 for the 1% level of significance. Since Fcal(9.70) is greater than Ftab = F(5,12;0.99) = 5.06, so result is significant and we Conclude their is evidence that varieties have different larval Populations. V1 V2 V3 V4 V5 V6 35.81 55.24 70.27 56.88 41.02 44.35

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Comparison.ppt

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