Hi all,
Did the behavior of base t.test() subtly change with respect to the
paired argument?
I think this code used to work with long form data, but now it seems not
to work:
# Example long format data
data_long <- data.frame(
value = c(10, 12, 15, 8, 9, 11),
time = factor(rep(c("before", "after"), each = 3))
)
t.test(value ~ time, data = data_long, paired = TRUE)
# Errorin t.test.formula(value ~ time, data = data_long, paired = TRUE)
: # cannot use 'paired' in formula method
A quick Google seems to suggest that the above code should still work.
What am I missing?
-------------------------------
Less sure about this one ...
Also, I thought it had always been possible to use x = "difference
scores" in t.test(..., paired = TRUE) as long as y = NULL. My memory
about this is less secure than the behavior above.
Control = c(10, 12, 15) Treat = c(8, 9, 11) data_wide <- data.frame(
Control, Treat, Diff = Treat - Control) t.test(data_wide$Diff, NULL,
"two.sided", 0, paired = FALSE) # works but y = NULL, mu = 0 mostly
implies difference scores
t.test(data_wide$Diff, NULL, "two.sided", 0, paired = TRUE) # fails but
if y = NULL the value of paired should not matter should it?
For y, help says "y an optional (non-empty) numeric vector of data
values". I take this to imply that x = compared to a population mean
hypothesis whenever y = NULL. Why should the value of paired influence
the outcome?
Just bad memory on my part?
R.version _ platform x86_64-w64-mingw32 arch x86_64 os mingw32 crt ucrt
system x86_64, mingw32 status major 4 minor 5.1 year 2025 month 06 day
13 svn rev 88306 language R version.string R version 4.5.1 (2025-06-13
ucrt) nickname Great Square Root
--
---
Robert W. Baer, Ph.D.
Professor of Physiology
Kirksville College of Osteopathic Medicine
A.T. Still Univerisity of Health Sciences
800 W. Jefferson St.
Kirksville, MO 63501
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