I’m having trouble with the example given in Shweser’s LOS review: SS3 R10 LOS J. I know it must be something really stupid I’m overlooking, but I don’t know how they get 2.575 as the z(alpha/2) value in the answer! If the confidence interval is 99%, then alpha is .005 but there is no value for this in the Z-Table. Why is this? Example: Suppose we administer a practice exam to 100 CFA Level I candidates, and we discover the mean score on this practice exam for all 36 of the candidates in the sample who studied at least 10 hours a week in preparation for the exam is 80. Assume the population standard deviation is 15. Construct a 99% confidence interval for the mean score on the practice exam of candidates who study at least 10 hours a week. Answer: 80 +/- 2.575 * 15/sqrt(36)
the number stated in most textbook is 2.58 (which may look familiar) what they did is took the values in a Ztable and did the average. (table value .9949 [2.57] and value .9951 [2.58]) hope this helps
Z from the table at 0.9951, but why 15/sqrt(36)? 15% is the given population std dev., why divide by 6?
Because it’s population std dev,thus divide by 6, otherwise would divide by (36-1)^0.5. I think
no that’s not it…you should divide by Sqrt(n) only if you have a sample std dev…I guess I have to go back and review quant.
hopetobeat Wrote: ------------------------------------------------------- > Because it’s population std dev,thus divide by 6, > otherwise would divide by (36-1)^0.5. I think Yes
Dreary Wrote: ------------------------------------------------------- > no that’s not it…you should divide by Sqrt(n) > only if you have a sample std dev…I guess I have > to go back and review quant. No
even more confused now!
Sorry - I was talking on the phone… You’re right of course.
if you know the population mean AND stdev then you’re confidence interval is X-bar+/-z(sigma) you don’t know the population mean here so you have to use sigma/(sqrt(n)) since the sample is large
…waiting for joey to confirm…
but that’s how the answer was from Shweser’s LOS review: SS3 R10 LOS > Answer: 80 +/- 2.575 * 15/sqrt(36)
So for C.I.'s for mu sigma known, observations normal: X-bar ± z*sigma/sqrt(n) sigma unknown, observations normal X-bar ± t*s/Sqrt(n) where t is (n-1) df sigma unknown, large sample, approximate C.I. X-bar ± z*s/sqrt(n) (or use t but it’s approximate anyway)
right sample mean +/- z * sigma/sqrt(n) the population stdev is known but the mean is a sample, both statistics need to be the population statistic to drop the sqrt(n) adjustment
…so, you have to use the standard error all the time, except when you have no samples to deal with, and the standard error will be either sigma or sample std dev depending on which one you have…no population std dev use sample’s.
Dreary Wrote: ------------------------------------------------------- > …so, you have to use the standard error all the > time, except when you have no samples to deal > with correct , and the standard error will be either sigma > or sample std dev depending on which one you > have…no population std dev use sample’s. not sure what you’re asking here…
I’m talking about how you calculate the standard error…take std dev of population if available, or of sample otherwise and divde by Sqrt(n). The distribution of the sample means which is approximately normally distributed will have sampling errors that we need to adjust for. So, that’s why we don’t take sigma as is…this is what you alluded to in your explanation above, which is correct.
jo_l Wrote: ------------------------------------------------------- > (table value .9949 [2.57] and value .9951 [2.58]) > still not seeing how .9949 == 2.57
figured it out, writing it down to burn this stupid thing into my mind: if you have Z=1.65 for example, you find that the Z table says .9505. This is the probability of a value under +1.95 SDs from the mean, but you have to take into account the probability of -1.95 SDs as well. So: P(+1.95 SDs)=.9505 P(-1.95 SDs)=1 - .9505 => P(-1.95 < x < +1.95) = .9505 - (1-.9505) => .901