Do we assume a t distribution for the estimate of the difference of normal distributions? I am studying hypothesis testing.
When performing a one sample t test, we assume a t distribution for the sample mean estimates of the true mean.
When conducting a two independent sample test, it seems that we also assume that the estimates of the difference of the means follow a t distribution.
I used the following code to simulated data, which yields the same parameters and results as the R base t.test implementation.
Am I correct? On what grounds do we assume this?
>expected_diff <- 0
>mean_diff <- mean(a) - mean(b) #mean difference
>df_pool <- length(a) + length(b) - 2 # degrees of freedom
>se_pool  <- sqrt(((length(a) - 1) * sd_a^2 + (length(b) - 1) * sd_b^2)/
                   df_pool) # pooled std. error 
>t   <- (mean_diff - expected_diff)/ (se_pool * sqrt(1/length(a) + 1/length(b))) # t-statistic
> result <- c(mean_diff, se_pool, t, p, mean(a), mean(b))
> names(result) <- c("Difference of means", "Std Error", "t", "p-value","Mean A","Mean B")
> result
Difference of means           Std Error                   t
       1.533321e+00        2.808271e-01        4.728513e+01
            p-value              Mean A              Mean B
      1.532661e-140        4.352244e+01        4.198912e+01

> t.test(a,b,var.equal = T)
Two Sample t-test / data:  a and b
t = 47.285, df = 298, p-value < 2.2e-16
Alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:  1.469506 1.597136
Sample estimates:
mean of x mean of y
 43.52244  41.98912

The topic in this forum does not address this question: What is the distribution of the difference of two-t-distributions
 A: 
When conducting a two independent sample test, it seems that we also assume that the estimates of the difference of the means follow a t distribution.

We don't assume this; indeed we know for sure that it's not the case.
The difference in means of two independent samples from normal distributions itself has a normal distribution.
However, when we don't know the variances, we don't have a way to say whether the raw difference in means is larger than we'd expect if the population means is the same. (I get a difference in means of 22.15 -- is that inconsistent with equal population means? We have no way to tell just from that, or even just from that and the sample sizes.)
Consequently, we standardize by an estimate of the standard error of the difference in means. Now all such differences (at a given d.f.) are on the same "scale", and our test statistic is a ratio of two components, an estimate of the mean difference in populations on the numerator and an estimate of the standard error of that difference on the denominator. Now we have a way to say what's surprisingly far from what we should see if the null were true.
Under the assumptions we make, this then yields a pivotal quantity. I'll quote a few sentences from the Wikipedia article:

Pivotal quantities are commonly used for normalization to allow data from different data sets to be compared. It is relatively easy to construct pivots for location and scale parameters: for the former we form differences so that location cancels, for the latter ratios so that scale cancels.
Pivotal quantities are fundamental to the construction of test statistics, as they allow the statistic to not depend on parameters – for example, Student's t-statistic is for a normal distribution with unknown variance (and mean).

If the original distributions we sampled were normal (again with the above independence assumption) then that test statistic would have a t-distribution (when the null is true; when the population means differ, it has a distribution called a non-central t).
There's a description of this for the one-sample t-test here.The case for the two-sample test is analogous.
It's the fact that the denominator varies (each sample would give a different estimate) around the population quantity it estimates that makes the distribution of the t-statistic more spread out and heavier tailed than the normal on the numerator.
For the case where the null is true, this effect is described (intuitively) in this answer Why does the t-distribution become more normal as sample size increases?.
