Thanks, I will test if the heterogeneous variances model fits better than the homogeneous variances model.   What is the difference between birdtype "CFL laye" and "LED laye"? the difference between CFL layer and LED layer is the lighting experience of the birds, i.e., before using these birds in the preference test, we raised them in either LED or CFL light since day-old, we want to see if early lighting experience will influcence the preference of light in the later period.   In this case, I understand that the p-value is testing whether the intercept=0, or proportion=0.5. As all three birds prefered CFL light, how can I know the p-value for each type of birds? for example, if I want to know exactly p-value for pullets (proportion of time for pullet as compared to 0.5). I mean, without include birdtype as a factor, by running three separated sas code, I will have p-value of intercept for each type of bird.   ... View more

proc glimmix data=have method=laplace; class pair birdgroup day birdtype; model proportionLED = pair | birdtype | day / dist=beta intercept s; random birdgroup*pair*birdtype /group=birdtype; estimate "Intercept" intercept 1 / ilink cl; lsmeans birdtype /pdiff adjust=tukey ilink cl; run; Here is the code I used according to your answers,  /group=birdtype I used this option as I want each type of birds had respective standard error, then I got results as followings. As we had three types of birds included in the model, the DF became 30 for the intercept.  But for the leaset squares means for each type of birds, DF=30, is that right? Besides, with birdtype being included, the intercept became the overall preferece of three type of birds, right? Here P<0.0001 means that birds (overall of three types) don't prefer LED light, what about each individual type of birdsy?      I really appriciate your consistent assistance. Thanks very much!!! ... View more

Thanks very much for your time, Yes, for each trial and each day, the testing time is 10 hour with lights-on proc glimmix data=have method=laplace; class pair group day; model proportionLED = pair day pair*day / dist=beta intercept s; random group*pair; estimate "Intercept" intercept 1 / ilink cl; run; I tried this new code for the dataset, as you/we expected, all pair, day, and pair*day effects are not significant. The DF=10 for the intercept. I think this code will be suitable to analyze my data now.   Indeed, time spent is just one of the response we measured. there is another response value for the preference, i.e., LEDnum (num of birds under LED at night, as birds won't move at night, so we count number of birds in each cage, we have total 3 birds for each group). for time spent or proportion of time, the value is continous data, how about the number of birds (0,1,2, or 3) or the corresponding proportion? Can we use the same model to analyze it?     What if we have three types of birds tested, i.e., pullet, turkey and broiler. For each type of birds, we run the exactly the same procedure (12 groups of birds, 2 groups of birds tested each time in the two pairs of testing comparments (A, B); each group of birds tested for 4 days), can we test the effects of bird type? For example, if I run the code reparatedly for each type of birds, we find pullets and boriler like LED, but turkey does not. How can we include Birdtype as a fixed factor in the model you have told me?                 ... View more

Thanks very much,   Yes, I will use a same dataset structure as you described "One format for a dataset that I would expect has 48 observations, one for each day for each group of birds. Variables would be (1) group identification (with 12 unique values); (2) pair (A or B); (3) day (1, 2, 3, or 4); (4) amount of time spent under LED; and (5) amount of time spent under CFL. Subsequently in a datastep, you could compute the proportion of the total time (amount under LED + amount under CFL) that a group spent under LED; this proportion is the response variable that you would use in the model I proposed."   In terms of your Random statement Random pen day; I would thiink it should be Random group day*group; repesenting random group effect among 12 groups of birds, and random day effect among 4 days within each group of birds. Thus we will have DF=11 for the estimates of intercept, am I right?     In the study, the objective is to determine which light is prefered by the birds, but when we present results, we want to present time spent or proportions of time for both lights, then tell which one is prefered, so we would also like to determine mean values, stderr and 95% ci for both LED and CFL lights.            ... View more

My dataset looks like the following tables, we have two identical testing chamber (pair), named A and B, each testing camber has two cages. we have 12 groups of birds. with this dataset, could you help to classify the Random statement? Still, I am not quite understand about your statment: The proportion of time under LED is not just (negatively) correlated with the proportion of time under CFL; these two metrics are totally redundant--there is no information in one metric that is not already in the other. If you told me that in one trial, the proportion of time under LED was 0.62 (62%), I could tell you that the proportion of time under CFL in that trial was 0.38, because these two values must sum to 1 (100%). althtough the sum is fixed, but why can't we compared this two values directly?     proc glimmix data=have method=laplace; where light="LED"; class pair group day day cage; model lightpre = / dist=beta intercept s; random pen day; estimate "Intercept" intercept 1 / ilink cl; run;   flock pair group day cage light lightdur lightpre 1 A 1 1 1 LED 4.07 40.74 1 A 1 1 2 CFL 5.93 59.26 1 A 1 2 1 CFL 5.85 58.46 1 A 1 2 2 LED 4.15 41.54 1 A 1 3 1 LED 4.37 43.71 1 A 1 3 2 CFL 5.63 56.29 1 A 1 4 1 CFL 5.05 50.46 1 A 1 4 2 LED 4.95 49.54 1 B 2 1 1 LED 2.9 29 1 B 2 1 2 CFL 7.1 71 1 B 2 2 1 CFL 5.57 55.71 1 B 2 2 2 LED 4.43 44.29 1 B 2 3 1 LED 3.58 35.75 1 B 2 3 2 CFL 6.42 64.25 1 B 2 4 1 CFL 5.08 50.79 1 B 2 4 2 LED 4.92 49.21   ... View more

Thanks very much for your reply and explanation   I applied your code in my data analysis proc glimmix data=preference method=laplace plots=all; where light="LED"; class birdgroup day; model timeproLED=/ dist=beta link=logit intercept s; random day*birdgroup ; estimate "Intercept" intercept 1 / ilink cl; run; With this SAS code, I am able to test if the timeproLED (proportion of time spent under LED light) equals to 0.5. The code gives a estimated Mean value, and 95% ci values. However, with this method, the "Mean" does not match the mean calucalated by Proc Means procedure.    As you told, for each day, time spent in LED and time spent in CFL were not independent responses (correlated), what do you think about the paired t-test for this study? To do test, I tried to average all values over the four testing days for each birdgroup (12 groups in total), then run code as followings?  proc ttest data=preference; paired LEDdur*CFLdur; run; Thanks very much ... View more