Re: Did behavior of "paired" argument in t.test() change?

No, that would look at the mean of value - time. You need to convert to wide format and use both of the resulting value columns (this _is_ wht the current example does). Or do the splitting explicitly, as in

> data_long <- data.frame(
+       value = c(10, 12, 15, 8, 9, 11),
+       time = factor(rep(c("before", "after"), each = 3))
+     )
> groups <- with(data_long, split(value, time))
> t.test(groups$before, groups$after, paired=TRUE)


> On 4 Nov 2025, at 18.39, Petr Pikal <[email protected]> wrote:
> 
> Hallo
> 
> According to recent documentation this works for paired t.test.
> t.test(Pair(value,time)~1, data=data_long)
> 
> It is the last example in help page.
> 
> And your second question seems to be also different from what help page
> says:
> 
> If paired is TRUE then both x and y must be specified and they must be the
> same length. Missing values are silently removed (in pairs if paired is TRUE
> ).
> 
> Cheers.
> Petr
> 
> 
> út 4. 11. 2025 v 18:23 odesílatel Robert Baer via R-help <
> [email protected]> napsal:
> 
>> 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
>> 
>>        [[alternative HTML version deleted]]
>> 
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>> 
> 
> [[alternative HTML version deleted]]
> 
> ______________________________________________
> [email protected] mailing list -- To UNSUBSCRIBE and more, see
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> and provide commented, minimal, self-contained, reproducible code.

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