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I'm doing the t.test from a paired sample, only considering as alternative hypothesis that the mean is greater than 0. Why the confidence interval has a minimum value but not a maximum?

The only reason I can think of it is that the maximum value of the mean is not as important as the minimum value if your restriction is Ho>0, but seems strange.

This is the code:

healthy<- c(-0.9914, 1.471, 1.2459, 0.4024, 0.0325, -0.6396, 0.7246, 0.0604)
lame<-c(4.3541, 4.7865, 6.1945, 10.7383, 3.3007, 4.8678, 7.8965, 3.9338)

t.test(lame, healthy, paired = TRUE, alternative='greater')

And this is the outcome:

Paired t-test

data:  lame and healthy
t = 6.5639, df = 7, p-value = 0.0001574
alternative hypothesis: true mean difference is greater than 0
95 percent confidence interval:
 3.891736      Inf
sample estimates:
mean difference 
         5.4708**
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A 95% confidence interval doesn't have to be symmetric about the mean. It just needs to define an interval based on a calculation that would contain the true value 95% of the time in repeated experiments.

Your idea that "the maximum value of the mean is not as important as the minimum value if your restriction is" $H_\text{alt}>0$* is pretty much what's going on here. Your asking for a one-sided test has effectively ruled out the idea that the alternative could include $H_\text{alt}<0$. Any finding of a value below 0 would be attributed to random variation rather than a true negative value. So there's no need to consider the upper confidence limit for the mean; you just want to make sure that the lower 5% doesn't include the value of 0.

The result in practice is that the lower confidence limit in such a 1-sided t-test is farther away from 0 than it would be with a 2-sided t-test and its symmetric CI about the mean. For your data and a 2-sided test, the lower limit is at 3.50 instead of 3.89.

@BruceET has a nice illustration of the difference between 2-sided and 1-sided tests here.


*In standard terminology, $H_0$ stands for the null hypothesis, so I use $H_\text{alt}$ to represent the alternate hypothesis.

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