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Summary

This study is a controlled experiment. The three groups were subjected to three different treatments. Then their red cell folate levels were measured. The fact that they were exposed to different treatments makes the study a controlled experiment rather than a simple observational study.…

- Subject: Miscellaneous
- Type: Math Problem
- Level: College
- Pages: 6 (1500 words)
- Downloads: 0
- Author: eleonoreheaney

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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.

ANSWER:

The histogram...

The measures were independent of the researcher's 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 straight 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 explanatory 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. ...Download file to see next pagesRead More

(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.

ANSWER:

The histogram...

The measures were independent of the researcher's 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 straight 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 explanatory 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. ...Download file to see next pagesRead More

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