We use cookies to create the best experience for you. Keep on browsing if you are OK with that, or find out how to manage cookies.
Nobody downloaded yet

GENSTAT Linear Statistical Modelling - Math Problem Example

Comments (0)
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.
Download full paper
GRAB THE BEST PAPER

Extract of sample
GENSTAT Linear Statistical Modelling

Download file to see previous pages... 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
Comments (0)
Click to create a comment
CHECK THESE SAMPLES - THEY ALSO FIT YOUR TOPIC
Patterns Withing Systems of Linear Equations
Algebra is one of the fascinating fields of mathematics, because algebra allows the finding of unknown numbers from information given. In algebra, letters are used in place of numbers that are not known. These letters are then manipulated in accordance with certain rules until an answer appears.
16 Pages(4000 words)Math Problem
Counseling
This then limits the values for a function where there is only value of x for every y whereas, the linear function gives unlimited values for both x and y as one tries to solve for such values by substituting assumptions for x to get the value of y. As mentioned earlier, functions and linear equations form lines when they are plotted on graphs so that from this point of view, it is concluded that a function is a linear equation and vice versa.
1 Pages(250 words)Math Problem
Statistical Reasoning in public Health
Acute otitis media is an infection that causes inflammation of the middle ear canal. In the study, children were randomized to receive either the influenza vaccine or a placebo. (randomization was done in a 2 to 1 ratio, meaning that two times as many children were randomized to the vaccine treatment as were randomized to the placebo group).
3 Pages(750 words)Math Problem
HW
2. Report a 95% confidence interval for this difference. The 95% CI for this difference ranges from -2.98 to -1.22 3. Write a sentence interpreting both the unadjusted mean difference and the corresponding confidence interval. When not considering other variables, women in 1985 earned approximately 2.1 dollars less than the men did; and 95% of the times, the actual values for this estimate would fall between 2.98 dollars to 1.22 dollars less then what the men earned on an average.
4 Pages(1000 words)Math Problem
Mathematical Modelling
The questions apply the accurate descriptions of the motion under various motions of situations. The work also introduces a lot of results and notions that are equally applicable in things that oscillate and vibrate in similar manner. For instance, the current analysis in fundamental circuit is analogous to analysis involving the spring’s mass.
4 Pages(1000 words)Math Problem
Energy efficiency rating
In the event that it goes beyond that, the unit may be considered too expensive. Furthermore, if manufacturers discover that customers are not willing to pay for air conditioning units beyond 449, then they may be compelled to produce units that do not go beyond an 11.0 EER to limit costs.
2 Pages(500 words)Math Problem
Statistical Methods
The confidence interval is the result from the test statistic N (0, 1). Test statistic: Under H0, tn - 2 distribution where a is the sample slope parameter, A is the population slope parameter, s is the sample estimate for the standard deviation. The results from Q4 are a = 0.0867, s = 1.4495 and Cxx = 267.6 and A = 0.
8 Pages(2000 words)Math Problem
Statistical computation of maximum likelihood estimates using R
SMA do not account for seasonal changes. The duration of the moving average can best be determined according to the type of application data to forecast. Long time periods gives smoother response by removing random variations but react slower to changes in the data as it lags the trend.
10 Pages(2500 words)Math Problem
Linear Programming in relationship to the Profit Maximization of the Business
The co-ordinates of A and B are (5250000, 5250000) and (6, 6) respectively. The value of the objective function at these points is 0.45 X 5250000 = 2362500 and 2.7 respectively. The value of the objective function at the points of ray AD beyond point A would be 0.2x + 0.25(10500000 - x) i.e.
3 Pages(750 words)Math Problem
A Linear Programming
4 Pages(1000 words)Math Problem
Let us find you another Math Problem on topic GENSTAT Linear Statistical Modelling for FREE!
Contact us:
+16312120006
Contact Us Now
FREE Mobile Apps:
  • About StudentShare
  • Testimonials
  • FAQ
  • Blog
  • Free Essays
  • New Essays
  • Essays
  • The Newest Essay Topics
  • Index samples by all dates
Join us:
Contact Us