Answered You can hire a professional tutor to get the answer.
I need some assistance with these assignment. genstat linear statistical modelling Thank you in advance for the help!
I need some assistance with these assignment. genstat linear statistical modelling Thank you in advance for the help! 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.
***** Analysis of variance *****
Variate: folate
Source of variation d.f.s.s. m.s. v.r. F pr.
ventil 2 15516. 7758. 3.71 0.044
Residual 19 39716. 2090.
Total 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 ventil
rep. unequal
d.f. 19
s.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 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.
(d)
(e) Produce 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.
