Timeline for Back-transforming contrast lstrends results in r
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 15, 2016 at 21:08 | vote | accept | Peter Albertson | ||
| Aug 15, 2016 at 21:08 | comment | added | Peter Albertson | Ok, thanks a lot!! I'm also struggeling getting the confidence levels out of the back-transformed lstrends, but the new lstrends would sure save me a lot of trouble! | |
| Aug 15, 2016 at 19:15 | comment | added | Russ Lenth | It is not right to use the pairs results. Basically I think I'd better try to get lstrends to do the right thing and send it to you. This'll take a day or two at least. | |
| Aug 15, 2016 at 16:36 | comment | added | Peter Albertson | I edited my entry post and calculated the estimated values for the timepoint time=4. Is the calculation the right interpretation of your solution mentioned above? And is it right to use the lsmeans from the pairs() argument as i did in my calculations? | |
| Aug 15, 2016 at 13:01 | comment | added | Russ Lenth | I was talking about specific time values. If you want to work on the response scale, the trends are not linear: the slope is different at each time, and if you want numerical estimates, you have to say which time(s) to use. | |
| Aug 15, 2016 at 12:56 | comment | added | Peter Albertson | Using the pairs comment, i already to a sub-group analysis (young race 1, old race1, young race2, old race2, having 1 slope and SE for each combination of race and age). So the pairs-comment, compares the slope of dv over time for the how you call them specific key values of age and race. Instead of having a slope for all dv over time-values, i'm having now with your method different slopes for each timepoint, right? So the slope would not be just "2.4 +/- 0.8", but something like "0.3*t +/- 0.5*(sqrt(t))"? So i need to add specifc t-value, or can i get a slope in function of the time? | |
| Aug 15, 2016 at 1:21 | history | answered | Russ Lenth | CC BY-SA 3.0 |