You can assume it is normal distribution. Actually, based on central limit theorem, 9 samples average will be very close to normal distribution regardless.
【在 g**a 的大作中提到】 : 你得先说明population是什么distribution。n(0,1)之类的。 : [在 springc (阿泉) 的大作中提到:] : :求助万能的学术版 : :从500个样品中确定了mean = m standard deviation=s : :...........
l*g
20 楼
嘻嘻 谁骂大街了? 把你吵醒了?
【在 I*D 的大作中提到】 : 但那种骂大街似的好像还真从没发生过。。。。
g*a
21 楼
谁说的?9不够。尤其是你这种要看tail的情况,9太小了。
【在 s*****c 的大作中提到】 : You can assume it is normal distribution. : Actually, based on central limit theorem, 9 samples average will be very : close to normal distribution regardless.
【在 s*****c 的大作中提到】 : You can assume it is normal distribution. : Actually, based on central limit theorem, 9 samples average will be very : close to normal distribution regardless.
S*8
26 楼
围观吵架。。。
s*c
27 楼
Well, that has nothing to do with what I said. I am also not sure about your definition of 少数. I can only assert that 连 续九次都输 will occur less frequently than otherwise.
【在 h******e 的大作中提到】 : : 赌场扔骰子连续九次都输的也不是少数。
h*e
28 楼
找500个赌徒做个问卷,连续输9+把的目测一半以上。
【在 s*****c 的大作中提到】 : Well, that has nothing to do with what I said. : I am also not sure about your definition of 少数. I can only assert that 连 : 续九次都输 will occur less frequently than otherwise.
g*a
29 楼
你这不回到我原来说的了吗?对population distribution需要一个假设。你原题里面 没有写population distribution是什么。如果你限制了population只有500,那完全可 以把mean distribution simulate出来。choose(500,9)也就是5e18个点,均匀取1/ 10e10就能看出来了。
normal
【在 s*****c 的大作中提到】 : The original question says to assume the data from 500 individual are normal : distribution.
s*c
30 楼
Good idea. If I have 500000 samples, I might apply some funding to get a good computer to run this simulation.
1/
【在 g**a 的大作中提到】 : 你这不回到我原来说的了吗?对population distribution需要一个假设。你原题里面 : 没有写population distribution是什么。如果你限制了population只有500,那完全可 : 以把mean distribution simulate出来。choose(500,9)也就是5e18个点,均匀取1/ : 10e10就能看出来了。 : : normal
【在 s*****c 的大作中提到】 : Good idea. If I have 500000 samples, I might apply some funding to get a : good computer to run this simulation. : : 1/
s*c
32 楼
No, my population is really the population on earth. The assumption given by the question is that the quantity I measure (say height) follows normal distribution. Now I measured 500 individuals (500 samples). From that we computed the mean (m) and standard deviation (s) from the 500 samples. The question is the probability of the average of the 9 randomly sampled individual to be greater than m+s. The solution you provided is basically monte carlo method. To get accurate answer, I have to run many iterations to reduce the variance. Actually what you described will contain bias from the 500 sample itself. What you really need to do is to get another 500 samples and simulate. I am just pointing out that it is absurd to resort to this brute-force approach if we can analytically solve the problem.
9 samples out of 500 or 9 samples out of population? If it is the second one, assuming your 500 samples are iid and random, you should be able to use m and s to estimate sample mean and variance. The dist for 9 samples mean should also follow normal dist with mean and variance calculated from estimated population mean and variance.
The accurate what
【在 s*****c 的大作中提到】 : No, my population is really the population on earth. The assumption given : by the question is that the quantity I measure (say height) follows normal : distribution. Now I measured 500 individuals (500 samples). From that we : computed the mean (m) and standard deviation (s) from the 500 samples. The : question is the probability of the average of the 9 randomly sampled : individual to be greater than m+s. : The solution you provided is basically monte carlo method. To get accurate : answer, I have to run many iterations to reduce the variance. Actually what : you described will contain bias from the 500 sample itself. What you : really need to do is to get another 500 samples and simulate. I am just
0.5% basically 9 sample mean will have standard deviation of s/3. So 3 sigma gives you 99%. they only ask you the part that is greater. So 0 .5%. I don't think we need to use t-distribution because the m and s from 500 samples (of a normal distribution) should be good enough estimation of the u and sigma of the underlying population. I will need to use t-distribution if I get m and s from the 9 samples.
【在 n******r 的大作中提到】 : 是问sample mean > u + s的概率,不是大于u的,所以显然不是0.5
s*c
36 楼
That part I am fuzzing about. I don't have that book with me. What's the difference? If it is randomly drawn from the 500 samples (I guess we can assume it is our finite population).
dist
【在 g**a 的大作中提到】 : 9 samples out of 500 or 9 samples out of population? : If it is the second one, assuming your 500 samples are iid and random, you : should be able to use m and s to estimate sample mean and variance. The dist : for 9 samples mean should also follow normal dist with mean and variance : calculated from estimated population mean and variance. : : The : accurate : what
n*r
37 楼
这题用t-test是对的 在实际中假定知道population distribution的情况很少,除了题目明确给定的理想情况 如果population u 和 s 未知,sample size较小(30以下),应该用t-test而不是z- test 你找本统计推断的书或者google一下t-test和z-test的选择
0 u distribution
【在 s*****c 的大作中提到】 : 0.5% : basically 9 sample mean will have standard deviation of s/3. : So 3 sigma gives you 99%. they only ask you the part that is greater. So 0 : .5%. : I don't think we need to use t-distribution because the m and s from 500 : samples (of a normal distribution) should be good enough estimation of the u : and sigma of the underlying population. I will need to use t-distribution : if I get m and s from the 9 samples.
s*c
38 楼
I know. But the m and s are estimated from 500 samples! My understanding is that if we obtain m and s from <30 samples, then we have to use t-test.
情况
【在 n******r 的大作中提到】 : 这题用t-test是对的 : 在实际中假定知道population distribution的情况很少,除了题目明确给定的理想情况 : 如果population u 和 s 未知,sample size较小(30以下),应该用t-test而不是z- : test : 你找本统计推断的书或者google一下t-test和z-test的选择 : : 0 : u : distribution
【在 s*****c 的大作中提到】 : I know. : But the m and s are estimated from 500 samples! : My understanding is that if we obtain m and s from <30 samples, then we have : to use t-test. : : 情况
s*c
40 楼
but my sample size is 500. Or do you think my population is the 500 cases, and my sample size is 9? If so, my population's u and s is known.
I read those books. It just I interpret it differently from you. I really like to know where my mistakes are. Seems you believe 9 is the sample size and you have to look up the t-test table with degree of freedom of 8. And it doesn't matter whether my estimate is from how many samples. I really can't agree. If I got my estimate from 5 billion samples, wouldn't the mean and standard deviation really close to population? And you still tell me I can't use z- test but to use t-test and will get the same answer?
so 500 measurements is my population. The randomly selected 9 cases are from the population. Given that we know the sigma of the population, we should use z test and the answer is the 3 sigma. I made a mistake earlier regarding the probability of 3 sigma. Should be 99 .97%. So the answer should be 0.015%. Is that right?
dist
【在 g**a 的大作中提到】 : 9 samples out of 500 or 9 samples out of population? : If it is the second one, assuming your 500 samples are iid and random, you : should be able to use m and s to estimate sample mean and variance. The dist : for 9 samples mean should also follow normal dist with mean and variance : calculated from estimated population mean and variance. : : The : accurate : what