The Random Effects Model – Introduction

1. The Random Effects Model – Introduction • Sometimes, treatments included in experiment are randomly chosen from set of all possible treatments. • Conclusions from such experiment can then be generalized to other treatments. • When the treatments are random sample, the treatment effects, τi, are random variables. • This type of model is called a random effects model or a components of variance model. • Note it is important to determine at the planning phase of an experiment whether treatment effects are fixed or random. STA305 week 4

2. Example - Weaving Loom Homogeneity •  Textile Company uses large number of looms to produce fabric. The looms function in uniform manner so fabric produced by company is of uniform strength. • To investigate whether there are any difference between looms, 4 looms will be selected at random from all those owned by company. • Four samples will be obtained from each loom. • Fixed or Random Effects? Since looms are randomly selected, a random effects model must be used. Results obtained from these 4 looms can be generalized to the set of all looms owned by company. STA305 week 4

3. The Random Effect Model • The equation for the statistical model remains the same as for fixed effects model is: Yij = μ + τi + εij . • As in the fixed effects model, the εij are assumed to be i.i.d. N(0, σ2). • However, unlike the fixed effects model, random effects model has treatment effects, τi, which are random variables. • Therefore, instead of assuming that , random effects model requires that E(τi) = 0. • Under this assumption, the overall mean is still μ. To see that this is the case, consider…  • It is also assumed that: 1. the τi are statistically independent of the εij 2. that the τi are i.i.d STA305 week 4

4. Expected Mean Squares for Random Effects Model • To construct hypothesis test concerning treatment effects, we need to find expected mean squares. • Since τi are random, E(MSTreat) is not the same as in fixed effects model. • To find E(MSTreat) start by expressing the treatment sample means in terms of model parameters as follows:…. STA305 week 4

5. Hypothesis Testing • In a random effect model, testing for no treatment effect is in fact testing whether the variance of treatment effects is 0. • The Hypothesis are therefore: • When H0 is true, then it follows that E(MSTreat) = σ2 = E(MSE). • Otherwise, if H0 is false, then E(MSTreat) > σ2. • So the ratio of MSTreat to MSE provides a test statistic which is given by • The calculation of the P-value is then the same as for fixed effect model. That is, P-value = P(F(a-1, n-a) > Fobs). STA305 week 4

6. Analysis of Variance Table • The only aspect of ANOVA table that is different from fixed effects case is E(MSTreat). • The ANOVA table for random effect model is given below STA305 week 4

7. Estimating the Variance Components • In a random effects model, both σ2 and are known as the components of variation. • Point estimates of these quantities can be obtained by using information contained in the ANOVA table. • As we can see the MSE is unbiased estimate of σ2. • Exercise: Show that an unbiased estimate of is: STA305 week 4

8. Proportion of Variance Explained • The total variation in Yij is therefore, the proportion of variance explained by the treatment is: • An estimate of this proportion can be obtained by substituting into the equation above. STA305 week 4

9. Confidence Interval for Variance Component • Confidence interval is useful for assessing amount of treatment variability relative to error variability. • That is, we want a CI for • It can be shown that: • Further, to use Cochrane’s theorem, we divide each of the above by its degrees of freedom and take ratio. We get… • This can be used to derive a confidence interval for by taking the following steps… STA305 week 4

10. Example Weaving Looms Continued • Recall, 4 looms randomly selected from all those owned by the company. • Four cloth samples obtained from each loom & strength is recorded. • The data is given in the following table: STA305 week 4

11. Using SAS to Conduct ANOVA • To conduct ANOVA in SAS we need to add a RANDOM statement. The code is then, proc glm data = looms ; class loom ; model strength = loom ; RANDOM loom ; run ; • The extra line in the code generated information regarding E(MSTreat). • The output is given in the following slide (slide 12). • Exercise, verify that E(MSTreat) is as stated in output… STA305 week 4

12. STA305 week 4

13. Factorial Designs – Introduction • In a factorial design experiments, at least 2 factors are studied for their simultaneous impact on the response. • Suppose there are 2 factors, A and B where A has levels 1, 2, . . . , a and B has levels 1, 2, . . . , b. • Every level of A will be used with every level of B. That is, there are a × b possible treatment combinations. • The experimental units are randomly assigned to the treatments. • In this type of design factors are said to be crossed. STA305 week 4

14. Advantages and Disadvantages of Factorial Designs Advantage s • Factorial designs are more efficient than doing multiple studies to examine one factor at a time. • They provide information about joint impact of factors on response. Disadvantage  • Studies with many factors can be complex and costly to design and execute. • Further, if too many factors are included, it may result in insufficient degrees of freedom for estimating variability due to experimental error. STA305 week 4

15. Example - Toxins and Their Antidotes • Three toxic agents are thought to have varying impacts on survival time. • Four potential poison antidotes are also to be studied to determine if they also had an impact on survival, in the presence of a toxic agent. • Four different animals were randomly allocated to each of the 12 treatments. • The mean survival times for 4 animals exposed to each treatment are given in the Table below: STA305 week 4

16. To see relationship between toxic agent, antidote, and survival time, it is useful to plot mean survival times. The plot is given below: STA305 week 4

17. As we can see, for antidotes B1, B2, and B3, toxic agent A1 has the greatest mean survival time, followed by A2, and then A3. • For antidote B4 there is a different pattern. • Also notice that the plots for antidotes B1 and B3 are almost parallel but those for B2 and B4 are very different. • It appears that there is an interaction between toxic agents and antidotes. STA305 week 4

18. Understanding Interaction • If there is no interaction between factors A and B then plot should contain approximately parallel curves. • Variability in data can be explained as a contribution due to factor A and a contribution due to factor B. • However, in presence of interaction, contributions by factors A and B don’t explain all of the differences seen in the data. • There is also a contribution associated with each particular combination of A and B. • The following figures illustrate what might be expected when there is or isn’t an effect of Factor A, Factor B, or an interaction effect. STA305 week 4

19. STA305 week 4

20. STA305 week 4

21. STA305 week 4

22. STA305 week 4

23. STA305 week 4

24. STA305 week 4