Confidence interval question when population SD known Im trying to understand why we can have situations where when trying to find a confidence interval we can know the population standard deviation but not the mean ?
If I know the standard deviation of the population wouldn't I have had to have known the population mean to calculate it ? Or I am I getting my terminology jumbled ?
 A: Your understanding is correct. 
This scenario is purely fabricated in textbooks to demonstrate how to calculate Confidence Interval's using Z-tables. In practice, I've never encountered a situation in which somehow I knew the population standard deviation and not the population mean. If such a case exists, I'd be curious to hear about it.
More realistically, you'll have data from which you can calculate a sample mean and a sample standard deviation, and use a t-test to produce Confidence Intervals.
A: It's a good question, and in general such a situation (knowing the population sd but not the mean) doesn't arise.
It might occur if you have past experience of a process in which you assume that an intervention may have caused only the mean to shift but the previously known (or effectively known) sd would be unchanged.
Note, however, that it is in fact possible to compute a variance without computing a mean - the population variance is half the average squared pairwise distance between points, and a similar calculation can be done for the sample variance, but it is not efficient to calculate it that way.
I have had at least one or two occasions to use a test where the variance was treated as known -- where sample means were the only thing known about the samples, but an upper bound on the population standard deviation was available.
(For example, with proportions, it's not unusual to take the worst case for the variance to compute margin of error; the same could be done with a test of proportions.)
