Question

An experiment had to investigate the effect of two treatments on the formation of tartar in dogs. In addition to the two treatment groups, there was also a control group. The experiment included 26 dogs randomized to one of the three treatment groups: 1. P207 (addition of pyrophosphate to feed) 2. HMP (addition of sodium hexametaphosphate to feed) 3. Control (standard feed) After a period of 4 weeks, the tartar on each dog should be considered as an index, wherein the spread of tartar on the dog's teeth and the thickness of the tartar. A small value of the index represents a small tartar formation. Data is not given here, but Figurel shows the fractal diagrams for the variable Index for the three treatments. The figure does not conflict with a model consisting of three normally distributed observation indexes of the Index, one for each of the three treatment 12 14 14 nd Standard calculations for the three observation lines y,,yin i= 1,2,3 which may used to answer questions alone below is: Gruppe i 1 (P207) 2 (HMP) 3 (Kontrol) 9 0.4375 0.7467 1.0889 0.13655 0.08448 0.17854 1) Show with the help of a test, that it can be assumed that the variance of the Index is the same in the three observationgroups. Help: Barteletts test 2) Show with the help of a test, that it cannot be assumed that the mean value of the Index is that same in the three groups. Help: F-tes 3) Calculate for i 1, 2 estimate and 95% consisteney interval for the difference in mean value of Index in treatment 1 and treatment 3. 4) Based on the results in 3) are there any of the two treatments who show significant effect on tartar?

Answer 1

(a) Null hypothesis, Ho : of o- vs. Alternative hypothesis, H at least one o true variance of group i, i 1,2, 3 (n k) In MSE a is different from other s where, o (n1)n s 2 Test statistic 0.9495 n-1 3(k-1) 1 + where, nin2n3 9+8+926, k = 3, = ni-1)s =0.1353 MSE Mean sum of square due to error = k 2=1 p value P(x> 0.9495 Since p-value the same in three observational groups. 0.6220 0.05 fail to reject Ho and so we can assume that the three variance of index is 0.05 SO we

(b) Null hypothesis, Ho 1 2 3 vs. Alternative hypothesis, Hi at least one u is different from others. where, i of group i, i=1,2, 3 true mean 3 (9 0.7467 8 0.43759 1.0889)/26 0.7700 niyi y _ -1 of square due to between group SSB sum i-1 9* (0.7467-0.7700)2+8* (0.4375-0.7700)2+9* (1.0889-0.7700)2=1.8046 of square between groups _ MSB mean sum sSB/2 - 1.8046/2 0.9023 F - ratio MSB /MSE 0.9023/0.1353 6.6689 p - value P(F> 6.6689 F F2,23) Since p-value < 0.05 in the three groups. 0.0052 < 0.05 so we conclude that the mean value of the inder is not same reject Ho and So we

1 MSE 1 (3) 95% C.I. for pi- iTto.025.231 MSE nj i to.o25,23 is ni Point estimate of , 2,H3 are respectively 0.7467, 0.4375, 1.0889 95% C.I. for 1 2 is (-0.0605, 0.6789) 95% C.I. for 95% C.I. for 2 3 is (-1.0211, 0.2817) where, to025,23 2.0687 (4) Since the C.I. for 2-u3 does not contain zero hence Treatment 2 and Treatment 3 are significantly different. 43 is 0.7009, 0.0165)