Is a t distribution for a certain degree of freedom equivalent to the sample mean distribution for the corresponding sample size?

Question

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 $15$?

Answer 1

It seems you may be confusing $Z = \frac{\bar X - \mu_0}{\sigma/\sqrt{n}}$ and $T = \frac{\bar X - \mu_0}{S/\sqrt{n}}.$ The crucial difference is that only the numerator of $Z$ is random, while both numerator and denominator of $T$ are random. So you cannot talk about the number of SDs $\bar X$ is from $\mu_0$ without being clear whether the "SD" is $\sigma$ or its estimate $S.$

Suppose we are testing $H_0: \mu = 100$ against $H_a: \mu \ne 100$ at the 5% level and that $n = 5$ and $\sigma = 15.$ If we know that $\sigma = 15,$ then we reject $H_0$ when $|Z| > 1.96$. If we do not know the value of $\sigma,$ then we reject when $|T| > 2.776.$

The figure below shows $S$ plotted against $\bar X$ for 200,000 tests in which the data are sampled from $Norm(100, 15)$. For the t tests, the 5% of the points in red are the $(\bar X, S)$ points for which $H_0$ is incorrectly rejected. It is not just the distance between $\bar X$ and $\mu_0 = 100$ that matters; the random variable $S$ must be taken into account.

By contrast, for the z tests, the 5% of the points outside the vertical green lines are the ones for which $H_0$ is incorrectly rejected. (The green lines are vertical because the value of $S$ is not relevant.)

Notes: (1) As a bonus, the figure illustrates that for normal data the random variables $\bar X$ and $S$ are independent. (2) The R code to make the figure is provided below.

m = 200000;  n = 5;  x = rnorm(m*n, 100, 15)
DTA = matrix(x, nrow=m);  a = rowMeans(DTA);  s = apply(DTA, 1, sd)
plot(a, s, pch=".", xlab="Sample Mean", ylab="Sample SD")
t = (a-100)*sqrt(n)/s;  t.crit = qt(.975, n-1);  cond = (abs(t) > t.crit)
points(a[cond],s[cond], pch=".", col="red")
pm = c(-1,1); abline(v=100+pm*1.96*15/sqrt(n), col="darkgreen", lwd=2)

Answer 2

Note: the t-statistic that you calculate is supposed to be compared to the t-value from the t-table with your selected $\alpha$ level. Also, the Central Limit Theorem can only be applied when you have a large enough sample size.

1) 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?

Let's say our sample size $n = 100$. Then yes, this is true. If there's a value for t in the two-tail table with $\alpha = 32\%$, it would be around 1. However, for small sample size, say $n = 4$, this statement no longer holds.

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?

I'm not sure what you mean here, but the t-statistic you compute from this sample will need to be compared with the t-value from the table with df = 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?

Since $n = 15$, it would be close, but once again, as a rule of thumb, we need $n>30$ to apply the CLT.