GENSTAT Linear Statistical Modelling Math Problem

GENSTAT Linear Statistical Modelling - mathematics Problem ExampleThe histogram for the folate levels appears also to assemble the assumption of normality. However, the variances for the three groups do not satisfy the assumption of homogeneity. The variance of Group I is very large comp bed to the variances of Group II and III.(c) heedless of what you reason about the assumptions for analysis of variance, use the GENSTAT analysis of variance commands to test the hypothesis that ventilation sermon has no picture on mean red cell folate level. Include appropriate GENSTAT printout to survive your conclusions.***** depth psychology of variance ***** Variate folate Source of variation d.f.s.s. m.s. v.r. F pr.ventil 2 15516. 7758. 3.71 0.044Residual 19 39716. 2090. occur 21 55232.***** Tables of means ***** Variate folate Grand mean 283.2 ventil I II III 316.6 256.4 278.0 rep. 8 9 5*** Standard errors of differences of means *** Table ventilrep. unequald.f. 19s.e.d. 28.92X min.rep 25.50 max-min 21.55X max.rep(No comparisons in categories where s.e.d. marked with an X) The results of analysis of variance test show that there is a significant difference between the three groups. public exposure has an effect on mean red cell folate levels. Furthermore, the probability of F was 0.044, which is less than the alpha level, 0.05.(d) (e) levy appropriate residual plots to check further the appropriateness of the analysis of variance model. Comment, in the light of these plots, on the adequacy of the model.ANSWERThe histogram...The measures were independent of the researchers judgment. The histogram for the folate levels appears also to satisfy the assumption of normality. However, the variances for the three groups do not satisfy the assumption of homogeneity. The variance of Group I is very large compared to the variances of Group II and III.(c) Regardless of what you concluded about the assumptions for analysis of variance, use the GENSTAT analysis of variance commands to test the hypothesis that ventilation treatment has no effect on mean red cell folate level. Include appropriate GENSTAT printout to support your conclusions.The results of ANOVA test show that there is a significant difference between the three groups. Ventilation has an effect on mean red cell folate levels. Furthermore, the probability of F was 0.044, which is less than the alpha level, 0.05.The histogram shows that the residuals are not normally distributed. Also, the normal plot shows that the residuals do not fit a now line. In the light of these observations, it can be said the model is not adequate. The assumptions for the use ANOVA are violated.The model included only 4 of the original 9 variables. It discarded the other 5 instructive variables. With these 4 variables, the equation can account for the observed data. This is shown by the fact that the mean of the residuals for the model is 0.