I have looked at your other two questions. You never make it entirely clear
(to me anyhow) what your objective is. So, owing to your persistence, I will give you some examples, computations, and explanations that may be helpful. [Most of what is below is introduced in standard elementary textbooks on applied statistics. Such a textbook might be a better-organized guide than online tutorials and videos. Used copies of out-of-date editions of such books are pretty cheap on Amazon and elsewhere.]
Suppose you take a random sample of size $n = 1000$ and find sample mean
$\bar X = 247$ and sample standard deviation $S = 87.$
Your goal might (1) be to find a 95% confidence interval (CI) for the population mean $\mu$, or (2) to test the null hypothesis $H_0: \mu = 250$ against the
alternative $H_a: \mu < 250.$ Then (1) requires a t confidence interval and (2) requires a one-sample t test.
(1) The 95% CI for $\mu$ is of the form $\bar X \pm t^*S/\sqrt{n},$ where $t^* = 1.962$ cuts 2.5% from the upper tail of Student's t distribution with $n - 1 = 999$ degrees of freedom (which is very close to standard normal). My late-night computations, which you should verify, give the interval $(241.60, 252.40
).$
In Minitab statistical software, the output for a t confidence interval procedure is shown below; it agrees with my computation.
One-Sample T
N Mean StDev SE Mean 95% CI
1000 247.00 87.00 2.75 (241.60, 252.40)
(2) The test statistic for the t test is $T = \frac{\bar X - \mu_0} {S/\sqrt{n}} =\frac{247-250}{87/\sqrt{1000}} = -1.090.$ For a left-sided t test specified by $H_0$ and $H_1$ at the 5% level of significance, the 'critical value' is $c = - 1.646.$ That is, you could reject $H_0$ at the 5% level
if $T < -1.646,$ but $T = -1.090$ so you cannot reject.
[The critical value cuts 5% from the lower tail of Student's t distribution
with 999 degrees of freedom.]
The 'P-value' of the test is the probability under
that distribution of a t statistic less than the observed $-1.090.$ You need
some sort of software to find the P-value, which turns out to be $0.1380.$
Using the P-value as a criterion, you could reject $H_0$ if the P-value were smaller than 5% (which it is not).
The Minitab printout for this left-sided t test is shown below (slightly edited for relevance); it shows the same test statistic and P-value as in my computations above.
One-Sample T
Test of μ = 250 vs < 250
N Mean StDev SE Mean T P
1000 247.00 87.00 2.75 -1.09 0.138
Note: In one of your questions you mentioned a sample of size $n = 10,000.$
If you had the same sample mean $\bar X = 247$ and sample SD $S = 87$ a sample of that size, then the 95% CI would be $(245.295, 248.705)$ and the P-value of the one-sided test would have been smaller than $0.0005,$ leading to rejection of the null hypothesis. Sample size matters.
Addendum about a CI for population SD:
For normal data: Because $$Q=\frac{(n−1)S^2}{σ^2} \sim \mathsf{CHISQ}(\text{df} = n−1),$$ one can use a printed table of chi-squared distributions or software to find quantiles .025 and .975, $L$ and $U,$ respectively, of that distribution to get $$P(L<Q<U)= \cdots = P\left(\frac{(n−1)S^2}{U}<σ^2<\frac{(n−1)S^2}{L}\right)=.95.$$
[Notice the 'reversal' of $L$ and $U,$ which results from taking reciprocals in solving
the inequality to 'isolate' $\sigma^2.]$
Hence a 95% CI for $σ^2$ is of the form
$$\left(\frac{(n−1)S^2}{U},\,\frac{(n−1)S^2}{L}\right).$$
Take square roots of endpoints to get 95% CI for $σ.$
For example, if a sample of size $n=50$ from a normal population has sample variance $S^2=34.5,$ then a 95% CI for the population SD $σ$ is $(4.91,\,7.32.).$ [Notice that the point estimate $S=5.87$ is contained within this CI, but not at its midpoint (because chi-squared distribution is skewed.] Computation from R:
v = 34.5; sqrt(49*v/qchisq(c(.975,.025), 49))
[1] 4.906476 7.319376
Output from Minitab:
95% Confidence Intervals
CI for CI for
Method StDev Variance
Chi-Square (4.91, 7.32) (24.1, 53.6)
As you suggest, such intervals tend to get shorter with increasing $n.$ However, intervals can still be disappointingly long, even for moderately large $n.$ If the sample variance $S^2 = 34.5$ had resulted from a sample of size $n=500,$ then the resulting 95% CI for $\sigma$ would be $(5.531,\, 6.262).$ "Variances are very variable."
v = 34.5; sqrt(499*v/qchisq(c(.975,.025), 499))
[1] 5.530786 6.262223
