This may seem like a weird question, but hear me out. I'm essentially struggling to see the connection between a t-value from a t-table and a t-value that is calculated.
The following formula is used to calculate the value of a t-score: $$t= \frac{\bar{X}-\mu }{\frac{S}{\sqrt{n}}}$$
It requires a sample mean, a hypothesized population mean, and the standard deviation of the distribution of sample means (standard error).
According to the Central Limit Theorem, the distribution of sample means of a population is approximately normal and the sample distribution mean is equivalent to the population mean.
So the t-score formula is essentially calculating the magnitude of difference between the sample mean in question and the hypothesized population mean, relative to the variation in the sample data. Or in other words, how many standard errors the difference between sample mean and population mean comprise of. For example: If t was calculated to be 2, then the sample mean in question would be $2$ standard errors away from the mean of the sample distribution.
1.) Phew, ok. So question 1: Let's just say a t-score of 1 was calculated for a sample mean and since a distribution of sample means is normal according to the CLT, does that mean that the sample mean in question is part of the $68\%$ (because of the $68-95$ rule)of all sample means that are within $1$ standard error of the sample mean distribution?
2.) Let's say we have a distribution of sample means of sample size $15$. Is this distribution equivalent for a t-distribution of degrees of freedom $14$? Or more importantly: Is the t-value from a t-table for $14$ degrees of freedom and $95%$ confidence EQUIVALENT to a calculated t-value using a sample mean that is $2$ standard errors away from the mean of a distribution of sample means with sample size $15$?