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sorry.... not something I'm strong at...
Would you be able if I gave you the answer?
and you, @glgan1 ?
I learned some about confidence interval before but it was a long time ago. now i'm viewing back my notes, hopefully i can help u.
what's the answer?
i got 12.5% which is 1-87.5%. anyway i will tell you how i got it. |dw:1338115879203:dw| A and B are the lower boundary and the upper boundary for the interval respectively. so within the interval, the values which are smaller than the mean would be from A to mean. as alpha% represents the probability from A to B and the mean is the midpoint of it, then the probability from A to mean would be alpha/2. so equate alpha/2 = 1/16 and you will get 0.125.
So, why would you minus it from 1?
the answer i got is 12.5 which isn't the same as yours. I am not sure why we minus it from 1. sorry. :(
Hm Ok.... no problem :D
Assuming that a normal distribution applies, if the probability that an α % confidence interval includes only values that are lower than the population mean is 1/16 then the probability that the α % confidence interval includes only values that are higher than the population mean is also 1/16. Therefore the probablity that a confidence interval will include only values that do not include the mean is 1/16 + 1/16 = 1/8 or 12.5%. The require value for the confidence interval is 100 - 12.5 = 87.5%