So, tell me if this makes sense to any of you...
If the individual observations have the N(mu, theta) distribution, then the sample mean (x bar) of n independant observations has the N(mu, theta/ Square root of n) distribution.
Yeah. Cuz I have no idea and I'm about to go face my doom in.... 10 minutes.
I'm not even kidding.... I have been going to class, but this shit is so boring, no joke, I don't even know what theta is, numerically. I think it might be the standard of deviation of a set of data... but even if that was it... I DONT KNOW WHAT THE STANDARD OF DEVIATION IS OR HOW TO COMPUTE IT. This will be pathetic and possibly my demise.
Guess I'll just go get it over with.
Yeah. Cuz I have no idea and I'm about to go face my doom in.... 10 minutes.
I'm not even kidding.... I have been going to class, but this shit is so boring, no joke, I don't even know what theta is, numerically. I think it might be the standard of deviation of a set of data... but even if that was it... I DONT KNOW WHAT THE STANDARD OF DEVIATION IS OR HOW TO COMPUTE IT. This will be pathetic and possibly my demise.
Guess I'll just go get it over with.