Confidence Interval Calculation for Sample Mean: 95% Confidence Level

[JOhnJDC]

Homework Statement

I know the sample size n, the observed sample mean x, and the observed sample standard deviation s. I need to determine a value v such that I'm 95% confident that the average is v or less.

The Attempt at a Solution

If I calculate the 95% confidence interval, then I know that 95% of the resulting intervals will contain the true mean. Does the upper bound of the 95% confidence interval also tell me that this mean will be less than or equal to the upper bound with 95% confidence? Am I thinking about this the wrong way? Thanks

 

[JOhnJDC]

I think the answer is to construct the 90% confidence interval using the data given. Because this interval will be centered on the observed sample mean x, only 5% of averages will be above the upper bound of this interval. Therefore, I can be 95% confident that the upper bound is the value v that I'm looking for.

B% CI = [x-(1.645*s)/sqrt(n), x+(1.645*s)/sqrt(n)]

So, v = x+(1.645*s)/sqrt(n).

Does that logic work?