Let's say we are asked to determine the 99% confidence interval of a sample. For the 99% confidence interval, we know that there is a 99% chance the sample mean falls within 2.58 standard deviations of our population mean.
A typical textbook reaches the conclusion that if the P of the sample mean falling within 2.58 standard deviations of the TRUE population mean is 99%, then the reverse is also true - there's a 99% chance that the TRUE population mean falls within 2.58 st. devs of our sample mean. But I don't see how this can be equal (I'll elaborate but I maxed out space).

Question

Let's say we are asked to determine the 99% confidence interval of a sample.  For the 99% confidence interval, we know that there is a 99% chance the sample mean falls within  2.58 standard deviations of our population mean.
A typical textbook reaches the conclusion that if the P of the sample mean falling within 2.58 standard deviations of the TRUE population mean is 99%, then the reverse is also true - there's a 99% chance that the TRUE population mean falls within 2.58 st. devs of our sample mean.  But I don't see how this can be equal (I'll elaborate but I maxed out space).

kropot72 · Answer

$$0.99=P(\bar{X}-2.58\frac{\sigma}{\sqrt{n}}<\mu<\bar{X}+2.58\frac{\sigma}{\sqrt{n}})$$
The above equation can be interpreted as "the probability that the population mean lies between $$\bar{X}\pm 2.58\frac{\sigma}{\sqrt{n}}$$
is 99%". However we must be careful, the reason being that at this point the random variable in the expression is X-bar. When we replace the random variable X-bar by the sample mean, x-bar, we can no longer use a probability statement. Rather we must make a statement such as "a 99% confidence interval for the population mean is (A, B), and 99% of such intervals will contain the population mean".