I have a daily diary dataset, i.e. not consisting of RT or accuracy data. I want to calculate the Spearman Brown splithalf reliability, odd-even-method, for a single item called day_sitnumberneg. I've cleaned the dataset and computed an variable assigning each row either "even" or "odd", named "odd_even_between". However, there always occurs an error:
"error in splithalf(data = mds, outcome = "RT", score = "average", halftype = "random", : one or more of the conditions do not exist in the condition variable"
I've tried following guidelines (https://cran.r-project.org/web/packages/splithalf/readme/README.html; https://sdparsons.github.io/splithalf_documentation/examples.html) and a youtube-videos (https://www.youtube.com/watch?v=FhqyKxH4kyk) but I didn't manage to figure out why it doesn't work.
This is the code I've tried to run:
splithalf(data = mds,
outcome = "RT",
score = "average",
halftype = "random",
permutations = 5000,
var.compare = "odd_even_between",
compare1 = "odd",
compare2 = "even",
var.condition = "odd_even_between",
var.participant = "vp_initials",
var.RT = "day_sitnumberneg")
This is the tibble mds:
| vp_initials | odd_even_between | day_sitnumberneg |
|---|---|---|
| 222002 | odd | 7 |
| 212003 | even | 2 |
| 222002 | odd | 4 |
| 212004 | even | 6 |
| 222002 | odd | 3 |
| 212003 | even | 4 |
| 212004 | odd | 1 |
| 212004 | even | 1 |
Thank you very much in advance!