Xjyuk

This is purely a hypothetical question. A very common statement is that H0 is never true, it´s just a matter of sample size.

Let´s assume that for real there is absolutely no measurable difference between two means (u0=u1) drawn from normally distributed population (for both u=0 and SD estimated=1). We assume N=16 per group and we use t-test. This would mean that p-value is 1.00000 indicating that there is absolutely no discrepancy from H0. This would indicate that test statistic is 0. Mean difference between groups would be 0. What would be the limits of 95% confidence interval for the mean difference in this case? Would they be [0.0,0.0]?

EDIT: Main point in my question was that when can we really say that H0 is true, ie. u1=U2 in this case?

A confidence interval for a t-test is of the form $barx_1 - barx_2 pm t_textcrit, alphas_barx_1 - barx_2$, where $barx_1$ and $barx_2$ are the sample means, $t_textcrit, alpha$ is the critical $t$ value at the given $alpha$, and $s_barx_1 - barx_2$ is the standard error of the difference in means. If $p=1.0$, then $barx_1 - barx_2 =0$. So the formula is just $pm t_textcrit, alphas_barx_1 - barx_2$, and the limits are just $-t_textcrit, alphas_barx_1 - barx_2$, $t_textcrit, alphas_barx_1 - barx_2$.

I'm not sure why you would think the limits would be $0,0.$ The critical $t$ value is not zero and the standard error of the mean difference is not zero.

Being super-lazy, using R to solve the problem numerically rather than doing the calculations by hand:

Define a function that will give normally distributed values with a mean of (almost!) exactly zero and a SD of exactly 1:

rn2 <- function(n) r <- rnorm(n); c(scale(r)) 

Run a t-test:

t.test(rn2(16),rn2(16))

 Welch Two Sample t-test

data: rn2(16) and rn2(16)
t = 1.7173e-17, df = 30, p-value = 1
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.7220524 0.7220524
sample estimates:
 mean of x mean of y 
6.938894e-18 8.673617e-19 

The means are not exactly zero because of floating-point imprecision.

Your question is somewhat underdefined. There are lots of tests. Chances are you refer to the two-sample t-test. It makes a difference whether the sd is known or estimated (Gauss test vs. t-test). N=16+16 means N=16 in each of two samples? Or did you want to have a really large sample size such as N=16e+16?

In any case, nothing stops you from using standard t- or Gauss-formulae for computing the confidence interval - all informations needed are given in your question (assuming that you know exactly what you mean with your notation even though I don't). p=1 doesn't mean that there's anything wrong with that. Note that p=1 does not mean that you can be particularly sure that the H0 is true. Random variation is still present and if u0=u1 can happen under the H0, it can also happen if the true value of u0 is slightly different from the true u1, so there will be more in the confidence interval than just equality.

It is difficult to have a cogent philosophical discussion about things
that have 0 probability of happening. So I will show you some examples
that relate to your question.

If you have two enormous independent samples from the same distribution,
then both samples will still have some variability, the pooled 2-sample t statistic will
be near, but not exactly 0, the P-value will be distributed as
$mathsfUnif(0,1),$ and the 95% confidence interval will be very short
and centered very near $0.$

An example of one such dataset and t test:

set.seed(902)
x1 = rnorm(10^5, 100, 15) 
x2 = rnorm(10^5, 100, 15)
t.test(x1, x2, var.eq=T)

 Two Sample t-test

data: x1 and x2
t = -0.41372, df = 2e+05, p-value = 0.6791
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.1591659 0.1036827
sample estimates:
mean of x mean of y 
 99.96403 99.99177 

Here are summarized results from 10,000 such situations. First, the distribution of P-values.

set.seed(2019)
pv = replicate(10^4, 
 t.test(rnorm(10^5,100,15),rnorm(10^5,100,15),var.eq=T)$p.val)
mean(pv)
[1] 0.5007066 # aprx 1/2
hist(pv, prob=T, col="skyblue2", main="Simulated P-values")
 curve(dunif(x), add=T, col="red", lwd=2, n=10001)

Next the test statistic:

set.seed(2019) # same seed as above, so same 10^4 datasets
st = replicate(10^4, 
 t.test(rnorm(10^5,100,15),rnorm(10^5,100,15),var.eq=T)$stat)
mean(st)
[1] 0.002810332 # aprx 0
hist(st, prob=T, col="skyblue2", main="Simulated P-values")
 curve(dt(x, df=2e+05), add=T, col="red", lwd=2, n=10001)

And so on for the width of the CI.

set.seed(2019)
w.ci = replicate(10^4, 
 diff(t.test(rnorm(10^5,100,15),
 rnorm(10^5,100,15),var.eq=T)$conf.int)) 
mean(w.ci)
[1] 0.2629603

It is almost impossible to get a
P-value of unity doing an exact test with continuous data, where
assumptions are met. So much so, that a wise statistician will ponder
what might have gone wrong upon seeing a P-value of 1.

For example, you might give the software two identical large samples.
The programming will carry on as if these are two independent samples, and
give strange results. But even then the CI will not be of 0 width.

set.seed(902)
x1 = rnorm(10^5, 100, 15) 
x2 = x1
t.test(x1, x2, var.eq=T)

 Two Sample t-test

data: x1 and x2
t = 0, df = 2e+05, p-value = 1
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval: 
 -0.1316593 0.1316593
sample estimates:
mean of x mean of y 
 99.96403 99.96403