Question 1140864
it would be the mean+/-1.96 (the z-value for 95% middle of distribution)*sigma/sqrt(n)
this is 21.6+/-1.96(5.7)/sqrt(86); the half-interval is 1.20
the interval is therefore (20.4, 22.8). Used the formula and checked with calculator "Z-interval" because the SD of sigma implies a population.  If one meant s, the sd of the sample, it needs to be labelled such.
then t is used with 1.99 as the value and the half-interval changes to 1.22, although the final interval to the nearest tenth would still be the same using "T interval." The notation is important.

The interval is a sample of Utah teenagers, not Americans.  It is therefore valid for the population in Utah.  It might be appropriate for Americans in general, but that requires Utahns being a reasonable sample of Americans, which is not likely.