Second question the answer is A but maybe the book is wrong
florinpop Wrote: ------------------------------------------------------- > The appropriate teste statistic for constructing > confidence intervals for the population mean of a > non normal distribution when the population > variance is unknown and the sample size is large > >30 is: > a. z statistic or t statistic > b. z statistic at alpha with n degrees of > freedom > c. t statistic at alpha with 29 degrees of > freedom > d. t statistic at alpha/2 with n degrees of > freedom I think it is A. Since n >= 30, so there is not much difference between t and z statistics. I think it should be n-1 degree of freedom for t statistic. so D is not right.
florinpop Wrote: ------------------------------------------------------- > Second question the answer is A but maybe the book > is wrong From Schweser: The z-statistic is theoretically acceptible for a non-normal distribution with unknown variance and a large sample, but the t-statistic is more conservative. So C was also correct, but A is technically more correct. That was in a footnote though. Pretty bogus.
They are just saying that the t statistic will provide a more conservative range but they both are acceptable
I think it should be: t statistic at alpha/2 with n-1 degrees of freedom.
- Which of the following in not a property of the student’s t distribution? A it’s symmetrical B as the degrees of freedom get larger the variance approaches zero C it is defined by a single parameter, the degrees of freedom which is equal to n-1 D. it has more probability in the tail and less as the peak than standard normal distribution
florinpop Wrote: ------------------------------------------------------- > 4. Which of the following in not a property of the > student’s t distribution? > A it’s symmetrical > B as the degrees of freedom get larger the > variance approaches zero > C it is defined by a single parameter, the degrees > of freedom which is equal to n-1 > D. it has more probability in the tail and less as > the peak than standard normal distribution D
B. It should be closer to be normal distribution when degrees of freedom becomes very large.
disptrayou are right. B is the answer if the variance became 0 that would mean that the t distribution would flatten right? if you increase the df then it approaches a normal distribution and the variance should increase. Am I correct?
yeah. stupid me.
u mean 10 -18?
florinpop Wrote: ------------------------------------------------------- > disptrayou are right. > B is the answer > if the variance became 0 that would mean that the > t distribution would flatten right? > if you increase the df then it approaches a normal > distribution and the variance should increase. Am > I correct? I am not sure. But my intuition is that variance should decrease when df increases. But it will not decrease to zero. It will just converge to be a normal distribution. For the variance to be zero, it should converge to just a single vertical line. I think the normal distribution has mean 0 and variance 1.
hm I see I thought that if you increase the df then t distribution will have more kurtosis, and variance will increase, although i have the same feeling that the more df the less variance
i definitely agree that an increase in df will result in more kurtosis. i would also think that as df increases, std and varience would decrease. think about it, the denominator in std. deviation is n - 1. an increase in n would result in a lower quotient. also, std. dev. is a measure of risk… think about a portfolio with 2 securities vs. a portfolio with 2000 securities. the diversification (increase in n) would lead to lower risk and lower std. dev. does anything i wrote make sense? F quant.
For some reason i was comparing to the horizontal line but the mean for the distribution is vertical. so more kurtosis means that distribution si closer to the mean, which means that variance and std is lower. makes perfect sense and opposite less df… the tails are fatter which means are farther from the mean threfore more variance. Thanks apcarlso
For #4, re: Florinpop, no _real_ probability distribution can have zero variance because a zero-variance distribution would be a Dirac Delta Function, which is an infinitely narrow peak (with infinite height). Meaning in such a function that the probability of finding a value anywhere other than that spike is zero, hence all values are at the spike with zero variance. Well beyond the scope of the CFA level I. Since the student-T, since student-T approaches normal distribution at large N, this is definitely not true. You are sort of close, as a perfectly-flat distribution extending to infinite values would have infinite variance, as any value from negative infinity to positive infinity is equally likely. [I hope this doesn’t make anyone’s brain hurt reading this] For #2, it should definitely be A, since for large N the student-T approaches the normal distribution. If you have a sample of 100 datapoints, you shouldn’t use N=29 (although the difference wouldn’t be so great), it’s better to use the normal distribution.
A stock has a mean return of 10% per month, with a variance of 25%. Which of the following is the closest to the probability that the stock will earn a return of less than 20% per month for two months in a row? Assume that returns are normally distributed and independent from month to month. A. 43.0% B. 65.5% C. 95.0% D. 97.5% What do you think??
I’d guess A, but I think you need a Z-table. This is the probability of having an observation below +(10/25) stdev. If +/-1 std. is 68%, then that’s 34 on either side of the line, in this case we are taking the .5 prob to the left of the mean, +0.4 stdev (estimate .4*34=13.6%) on the right side of the line. (.5+.136)^2=40.4% close to A, all the other answers are too high Edit: Whoops it’s variance. So stdev =5%, so the chance of <20% are .5 + .95/2=.975 and .975^2=.951 hope I got that right
disptra Wrote: ------------------------------------------------------- > florinpop Wrote: > -------------------------------------------------- > ----- > > The appropriate teste statistic for > constructing > > confidence intervals for the population mean of > a > > non normal distribution when the population > > variance is unknown and the sample size is large > > > >30 is: > > a. z statistic or t statistic > > b. z statistic at alpha with n degrees of > > freedom > > c. t statistic at alpha with 29 degrees of > > freedom > > d. t statistic at alpha/2 with n degrees of > > freedom > > I think it is A. Since n >= 30, so there is not > much difference between t and z statistics. > I think it should be n-1 degree of freedom for t > statistic. so D is not right. If the distribution is non-normal, then your C.I. for the population mean is approximate anyway. You are relying on the central limit theorem here to be able to do anything (unless you actually know what the distribution is). While a t may be more conservative, the t distribution is relying on the independence of s^2 and X-bar (true only for normal distn’s of observations) and the ratio of s^2/sigma^2 to be a chi-square dist’n (true only for a normal distribution of observations). If you know the distribution is non-normal use the z but recognize it is an approximation.
P(X<20%) = P(Z<20-10/5) = P(Z<2) = 0.9772 P(S) = .0228 P(F) = .9772 Binomial… Probability P(X=2) = .9772^2 = .9549 Choice C