4 21 98 252z9932 4 a A bank wishes to compare deposit size at 4 branches The result of the analysis of variance appears below Sample means Branch 1 4 87 Branch 2 2 20 Branch 3 4 31 Branch 4 1 46 Source SS DF MS F Between 56 33 3 18 78 67 56 Within 6 67 24 0 278 Total 63 00 27 i Do mean deposits differ between banks Why 2 ii Assuming that this table is based on four random samples of equal size how many numbers are in each column 1 iii Do a confidence interval for the mean deposit in branch 1 2 iv Do a Scheffe confidence interval for the difference between deposits in branches 2 and 4 and explain why this type of interval might be preferred to one using t 3 b In a study of inflation 3 company sizes Factor A 5 degrees of industrial concentration Factor B and two types of product Factor C are distinguished An ANOVA table was generated using the average price rise over 5 years for a random sample of 4 firms in each category Generate an ANOVA table showing all possible interactions using the following data SSA 22 SSB 227 SSC 32 SSAB 54 SSABC 136 SST 1219 Using a 1 significance level which of the differences and interactions are significant 8 Correction SSAC 10 SSBC 98 Solution a 3 24 3 01 i This is a simple ANOVA Use 05 H 0 1 2 3 4 Since F 05 is less than 67 56 reject H 0 ii Since n 1 27 n 28 If we divide that by 4 we get 7 n m iii As explained in class 1 x 1 t 2 MSW where MSW and the degrees of freedom n1 come 24 from the within line of the ANOVA If we use 05 t 025 2 064 and 1 4 87 2 064 0 278 7 4 87 0 38 iv From the outline 2 4 x 2 x 4 m 1 F m 1 n m MSW 1 1 where the n2 n4 degrees of freedom are the same as those used in the F test in the ANOVA m is the number of columns so that 3 24 3 01 we use F 05 So 1 1 0 74 0 85 7 7 This has a 95 confidence level together with any other intervals you might do The intervals using t have a 95 confidence level alone 2 4 2 20 1 46 3 3 01 0 278 7 4 21 98 252z9932 b There are 3 5 2 30 groups with 4 observations in each group so n 30 4 120 s means significant difference H 0 rejected ns means no significant difference H 0 accepted Source SS DF MS F F 01 Factor A 22 2 11 00 1 547 F 2 90 4 85 ns Factor B 227 4 56 75 7 980 Factor C 32 1 32 00 4 500 Interaction AB 54 8 6 75 0 949 Interaction AC 10 2 5 00 0 703 Interaction BC 98 4 24 50 3 445 136 8 17 00 2 391 640 1219 90 119 F 4 90 3 54 s F 1 90 6 93 ns F 8 90 2 72 ns F 2 90 4 85 ns F 4 90 3 54 ns Interaction ABC F 8 90 2 72 ns Error Within Total 7 1111 8 4 21 98 252z9932 5 An airline wishes to explain the number of passengers it carries over 10 months as a consequence of advertising in the previous month It collects data as follows Observation Advertising Passengers The xy column is added here xy 1000 1000s You were given at least 8 1 10 16 160 examples of this calculation 2 12 18 216 3 8 14 112 4 17 24 408 5 10 17 170 6 15 22 330 7 10 15 150 8 11 21 231 9 19 25 475 10 10 18 180 2432 xy For your convenience the following values are given x 122 x 2 1604 y 190 y 2 3740 n 10 a Compute the regression equation Y b0 b1 x to predict the number of passengers 6 b Compute R 2 4 c Compute se 3 Solution Spare Parts Computation x 122 12 2 x n y n 115 6 Sxy 10 y x SSxx x nx y 2432 10 12 2 19 0 114 0 y Sxy SSxx nx 2 1604 10 12 2 2 xy 190 19 0 10 SSyy a b1 2 2 ny 2 3740 10 19 0 2 130 0 TSS xy nx y 2 nx 2 114 0 0 9862 115 6 b0 y b1 x 19 0 0 9862 12 2 6 9689 Y b0 b1 x becomes Y 6 9689 0 9862 x b RSS b1 Sxy b1 R2 R 2 xy nx y 0 9862 114 0 112 4221 RSS 112 4221 0 8648 or TSS 130 0 Sxy 2 SSxx SSyy xy nxy x nx y 2 2 2 2 ny 2 114 0 2 115 6 130 0 8648 0 R 2 1 always c ESS TSS RSS 130 0 112 4221 17 5779 ESS 17 5779 s e2 2 1973 or n 2 8 s e2 SSyy b1 Sxy n 2 y 2 ny 2 b1 n 2 xy nxy 130 0 0 9862 114 0 2 1972 8 9 2 1 R 2 y 2 ny 2 or s e2 1 R TSS or s 2 e n 2 n 2 s e 2 1973 1 4823 s e2 y 2 ny 2 b12 x 2 nx 2 n 2 is always positive 10 4 21 98 252z9932 6 Continuing the previous problem 02 a Compute s b0 and do a significance test on b0 4 b Do a confidence interval for b1 3 c Do a prediction interval for passengers when the expenditure on advertising is 10 thousand 5 d Using your SST etc put together the ANOVA table 05 6 a 1 1 x 2 s b20 s e2 s e2 n SSxx n x2 x nx 2 3 0488 b0 6 9689 s b0 3 0480 1 7461 H 0 0 0 2 2 2 1972 1 12 2 10 115 6 t H 1 0 0 b0 00 b0 0 6 9689 3 991 Since this is not s b0 s b0 1 7461 between 8 t n 2 t 01 2 896 reject H 0 and conclude that 0 is significant 2 2 b s b1 s e2 SSxx x s e2 2 nx 2 2 1972 0 01900 115 6 s b1 0 01900 0 1379 so 1 b1 s b1 0 9862 2 896 0 1379 0 986 0 389 c If Y 6 9689 0 9862 x and x 0 10 then Y 0 6 9689 0 9862 10 16 8309 1 s 2y s e2 0 n x0 x 2 x 2 nx 2 1 10 12 2 2 1 2 1972 1 2 5089 115 …
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