T-Statistics and s Calculations How can I calculate s in T statistics?
Example: John H. takes one sample of size 20 and finds that the sample mean in 32.8. Calculate a 95% confidence interval for John. (Assume John knows the true standard deviation.)??? 
I know what formula to use but I want to understand how can I calculate s without knowing the sample #s. 
 A: If John knows the "true" standard deviation (i.e. the population standard deviation --- a more truthful term than "true", although the latter word is often used) then he should use a z-statistic rather than a t-statistic.  The t-distribution was introduced for use when the population standard deviation has to be estimated based on a small sample.
Since you're not given any number, you'll just have to call it $\sigma$ and use that in the answer.
A: z Procedures if population SD is known. If $\sigma$ is known, use the statistic $Z = \frac{\bar X - \mu_0}{\sigma/\sqrt{n}}$ to test $H_0: \mu = \mu_0,$ where $\mu_0$ is a
number given or known before you have data. Also, a 95% z confidence interval (CI) is of the form $\bar X \pm 1.96\sigma/\sqrt{n}.$
In your case, you could answer with $``32.8 \pm 1.96 \sigma/\sqrt{20}"$
or "$32.8 \pm 0.438\sigma$", as suggested by @MichaelHardy. If you believe
you know the population standard deviation $\sigma = 9.25$ as mentioned in your comment, you
might use it, referring back that preliminary information.
t procedures if population SD is unknown, estimated by $S$. If $\sigma$ is not known, and must be estimated by the sample
standard deviation $S,$ then use the statistic $T = \frac{\bar X - \mu_0}{S/\sqrt{n}}$ to test $H_0: \mu = \mu_0.\,$ A 95% t confidence interval is of the form $\bar X \pm t^*S/\sqrt{n},$ where $t^*$ cuts 2.5%
from the upper tail of Student's t distribution with $n - 1$
degrees of freedom (from a printed table or software).
If you do not have the individual data values, you have no way
to compute the sample standard deviation $S.$
Notes: Before 1910 and in some elementary textbooks, you may see z methods used
for both tests and CIs if $n$ is sufficiently large that
one can assume $S$ is a very close estimate of $\sigma.$ (Some
people say 'sufficiently large' means $n > 30,$ but that rule
works well only for tests at the 5% level and 95% CIs.)
(a) I say "1910" because that's about when Gossett (writing under
the pseudonym Student) published his paper on what is now
known at Student's t distribution. In the 1920s and 1930s, Fisher played a major role in
popularizing use of the t distribution. [See the Wikipedia article
on 'Gossett-Student'.] 
(b) It's 30 because, if $n > 30,$
then $t^* \approx 2.0 \approx 1.96.$ (For sample size $n = 20$, you
have degrees of freedom $n-1 = 19$ and $t^* = 2.093.$)
(c) $Always$ when using software, regardless of $n$: use z procedures
when $\sigma$ is known (and enter $\sigma$ as prompted); use t procedures when $\sigma$ is
unknown (software will find $S$ from data and use it to estimate $\sigma$).
