95% confidence interval

[obscene]

Hey all,

I finished studying a whole chapter on confidence interval and was feeling happy to be finally able to answer most of the questions at the end of the chapter, but there are still few that I have no clue about. It would be great if you can help me with this one especially:

"For the case of a 95% confidence interval, the following statement is true from the properties of the sampling distribution of the sample mean x bar. Statement 1: Pr(m-z(se) < x bar < m+z(se)) = 0.95. This implies that once a sample had been selected and the corresponding x bar found, the following is true: Statement 2: Pr(x bar-z(se) < m < x bar + z(se)) = 0.95. Prove that Statement 2 follows from Statement 1."

Does any of you have an idea about how to prove this? I feel so dumb. -_-

[Mean Joe]

Just work with the stuff inside: m-z(se) < xbar < m+z(se)
Turn it into xbar-z(se) < m < xbar+z(se), using 8th grade algebra.
Don't think so hard, it's just adding m's and xbar's.

[squareandrare]

Remember that when you multiply both sides of an inequality by a negative number, the inequality sign flips.