s*****c
发帖数: 753 | 1 求助万能的学术版
从500个样品中确定了mean = m standard deviation=s
问，随机取9个样品取平均，平均值大于m+s的几率是多少。 |
H******7
发帖数: 34403 | |
s*****c
发帖数: 753 | 3 对呀。可有本书的答案告我说要用students test。然后给了1% 做答案
我凌乱了
【在 H******7 的大作中提到】
: 0.5
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M*P
发帖数: 6456 | 4 书上写的未必都对吧？
什么书？
【在 s*****c 的大作中提到】
: 对呀。可有本书的答案告我说要用students test。然后给了1% 做答案
: 我凌乱了
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l***o
发帖数: 5337 | 5 怎么可能
【在 H******7 的大作中提到】
: 0.5
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y**c
发帖数: 6307 | 6 就是3个sigma之外的几率吧。
【在 s*****c 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
: 问，随机取9个样品取平均，平均值大于m+s的几率是多少。
|
G**Y
发帖数: 33224 | 7 对个P呀
【在 s*****c 的大作中提到】
: 对呀。可有本书的答案告我说要用students test。然后给了1% 做答案
: 我凌乱了
|
g**a
发帖数: 2129 | 8 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
：........... |
s*****c
发帖数: 753 | 9 So what's your answer?
【在 G**Y 的大作中提到】
: 对个P呀
|
s*****c
发帖数: 753 | 10 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
: ：...........
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|
|
g**a
发帖数: 2129 | 11 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.
|
s*****c
发帖数: 753 | 12 The original question says to assume the data from 500 individual are normal
distribution.
【在 g**a 的大作中提到】
: 谁说的？9不够。尤其是你这种要看tail的情况，9太小了。
|
h******e
发帖数: 9616 | 13
赌场扔骰子连续九次都输的也不是少数。
【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*****c
发帖数: 753 | 14 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。
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h******e
发帖数: 9616 | 15
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.
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g**a
发帖数: 2129 | 16 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
发帖数: 753 | 17 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
|
g**a
发帖数: 2129 | 18 哇靠，你搞清楚sample和population的区别没有？前面的问题是population是500，
sample是9，
你现在就变成了sample是500000？那population是多少？你有500000 sample,干嘛还要
做simulation?直接CLT啊。mean=mu，var=delta^2/n。
你不要告诉我，你连population和sample这个基本统计概念都没搞清楚就在这瞎嚷嚷！
【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
发帖数: 753 | 19 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.
【在 g**a 的大作中提到】
: 哇靠，你搞清楚sample和population的区别没有？前面的问题是population是500，
: sample是9，
: 你现在就变成了sample是500000？那population是多少？你有500000 sample,干嘛还要
: 做simulation?直接CLT啊。mean=mu，var=delta^2/n。
: 你不要告诉我，你连population和sample这个基本统计概念都没搞清楚就在这瞎嚷嚷！
|
g**a
发帖数: 2129 | 20 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
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n******r
发帖数: 4455 | 21 是问sample mean > u + s的概率，不是大于u的，所以显然不是0.5
【在 s*****c 的大作中提到】
: 对呀。可有本书的答案告我说要用students test。然后给了1% 做答案
: 我凌乱了
|
s*****c
发帖数: 753 | 22 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
发帖数: 753 | 23 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
发帖数: 4455 | 24 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
发帖数: 753 | 25 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
|
n******r
发帖数: 4455 | 26 你的理解不对
不管你的estimate是从500还是50000个sample里来的，都不能直接当成population
estimate
只有假定population分布已知的情况才能用z-test
你去找本书看看就明白了，好过在这里干争
have
【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
发帖数: 753 | 27 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.
【在 n******r 的大作中提到】
: 你的理解不对
: 不管你的estimate是从500还是50000个sample里来的，都不能直接当成population
: estimate
: 只有假定population分布已知的情况才能用z-test
: 你去找本书看看就明白了，好过在这里干争
:
: have
|
s*****c
发帖数: 753 | 28 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?
【在 n******r 的大作中提到】
: 你的理解不对
: 不管你的estimate是从500还是50000个sample里来的，都不能直接当成population
: estimate
: 只有假定population分布已知的情况才能用z-test
: 你去找本书看看就明白了，好过在这里干争
:
: have
|
s*****c
发帖数: 753 | 29 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
|
T****t
发帖数: 11162 | 30 还是有高手的。
：就是3个sigma之外的几率吧。
：【 在 springc (阿泉) 的大作中提到: 】 |
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k**********i
发帖数: 8706 | |
