2008 First Semester STATS课程官方问答贴

~Kurosagi~

所以要怎么explain比较好，
我市弄明白了为什么是68%

但是语言组织部出来。。。

t-multiplier

原帖由 雨轩听风 于 2008-4-17 12:16 发表
(d) Give a formula for the probability that a given match ends in a draw.
现在理解ends in a draw的意思了...但是这个formula应该怎么写...
是不是90/2 P(X=45)来算... 好像不对..

(e) What ...

我感觉e和f的distribution都应该是binomial, where n = 90 and p = c(IP(both teams Goal = 0), IP(Draw)) respectively.

[ 本帖最后由 t-multiplier 于 2008-4-18 16:25 编辑 ]

t-multiplier

原帖由 ~Kurosagi~ 于 2008-4-18 16:21 发表
所以要怎么explain比较好，
我市弄明白了为什么是68%

但是语言组织部出来。。。

你要不下载一个2007 assignment 2看一看正确答案是怎样解答的?

~Kurosagi~

在cecil???

你是说???

t-multiplier

原帖由 ~Kurosagi~ 于 2008-4-18 16:25 发表
在cecil???

你是说???

本贴的第四楼。

~Kurosagi~

谢谢^^
......

kirstens

原帖由 t-multiplier 于 2008-4-18 15:35 发表


Just a thought with part b, in a geometric distribution, X is the number of failures before the first success right?

So does it mean we are actually finding IP(X >= 4) when it is asking us ...

Yea, I'm bit confused. Cos I knew it's Geo, but I don't remember the distribution function, therefore I searched it and it says on Wikipedia:
In probability theory and statistics, the geometric distribution is either of two discrete probability distributions:
A. for k represent the first success trial, pmf = (1-p)^(k-1) and cdf = 1-(1-p)^k
B. for k represent the last failure trial, pmf = (1-p)^k and cdf = 1-(1-p)^(k+1)

in my answer, I picked A, therefore I used k=5; But yea, be caution on this. Thanks to point it out, z-score. BTW, thanks for the answer at #427, too.

~kurosaki~

can someone help me on 101's question 6b... so lostttt T.T

水鱼皇后

原帖由 t-multiplier 于 2008-4-18 15:12 发表




Suddenly, I don't have access to 302, it only showed me the old courses.

水鱼皇后 , could u email me the data sets and the assignment sheet please? I'll upload them here for dk7.

em ...

我email给你啦 收到没有？

z-score

原帖由 水鱼皇后 于 2008-4-19 01:42 发表



我email给你啦 收到没有？

收到了. 上传一下.

z-score

原帖由 ~kurosaki~ 于 2008-4-18 19:05 发表
can someone help me on 101's question 6b... so lostttt T.T

等等阿,我还没看呢.

t-multiplier

原帖由 z-score 于 2008-4-19 12:07 发表


等等阿,我还没看呢.

看一下course book里面chapter 8的difference between two proportions的example就好了。做题步骤都一样。

true_up_up

326马上就要考试了，请问阿童木们有没有之前的考试题目？非常 谢谢

~Kurosagi~

question 6 b
哪里最后面我做出了 负数。。。。

好奇怪
不知道要怎么做才对

t-multiplier

原帖由 true_up_up 于 2008-4-19 13:56 发表
326马上就要考试了，请问阿童木们有没有之前的考试题目？非常 谢谢

阿童木们好像记得是326的course book最后面有past test and exam. oooOOO

t-multiplier

原帖由 ~Kurosagi~ 于 2008-4-19 14:44 发表
question 6 b
哪里最后面我做出了 负数。。。。

好奇怪
不知道要怎么做才对

我还没有做一遍那道题，但是confidence interval出来的是复数很正常。

~Kurosagi~

哦  i c

我可以把我的ans发给你看看吗？~~

t-multiplier

原帖由 ~Kurosagi~ 于 2008-4-19 15:05 发表
哦  i c

我可以把我的ans发给你看看吗？~~

我还没做呢，你就把你的两个数贴到这里应该没什么问题，看看别人有没有做好的。

~Kurosagi~

pf  - the probability of female
pm - the probability of male
(1) pf - pm = the difference between proportion of female planners whose organisation contributes to costs for professional subscriptions and the proportion of male planners whose organisation contributes to costs for professional subscriptions.
(2) pf - pm = the difference between proportion of female planners whose organisation contributes to costs for professional subscriptions and the proportion of male planners whose organisation contributes to costs for professional subscriptions.
     =170/206-182/235 = 0.8252-0.7745 = 0.0508 (4dp)
(3) The formula estimate ± t ×se (estimate) gives:  pf - pm ± t × se (pf - pm)
(4) This is Situation C: One sample of size n, many yes/no items.
se (pf - pm) =√(minimum(pf+pm,qf+ qm)-(pf – pm)^2)/n
pf + pm = 170/206+182/235 = 0.8252+0.7745 = 1.5997
qf +qm = (1- pf ) + (1-  pm) = (1-0.8252) + (1-0.7745) = 0.4003
se (pf - pm) = √(0.4003-〖0.0508〗^2)/(206+235) = 0.0300
       (5) df = ∞ (working with proportions)
       (6) for a 95% confidence interval with df = ∞, use t = 1.960
       (7) An approximate 95% confidence interval for (pf - pm) is:
             pf - pm ± t × se (pf - pm) = 0.0508 ± 1.96 × 0.0300
                                                     = (-0.008, 0.1096)
       (8) With 95% confidence, we estimate that the proportion of female planners whose organisation contributes to costs for professional subscriptions is somewhere between     than the proportion of male planners whose organisation contributes to costs for professional subscriptions (at the time of the survey).

~Kurosagi~

step 8 还没有做完
因为感觉很奇怪

所以没有继续下去

t-multiplier

原帖由 ~Kurosagi~ 于 2008-4-19 15:13 发表
pf  - the probability of female
pm - the probability of male
(1) pf - pm = the difference between proportion of female planners whose organisation contributes to costs for professional subscrip ...

(1) 最好加上the difference between the underlying proportion........
(2) 最好加上the difference between the sample proportion............

我觉得应该是situation a吧，因为问题是compare male and females, 这两个应该是two independent samples.

true_up_up

哈哈哈哈哈哈哈哈哈，对，想起来了，还是阿童木们有才。。。

~Kurosagi~

謝謝!!!!!


是的

改過來來e

謝謝

dk7

Using Pdf for the data, that's what I call creativity...

dk7

想问你一个问题。以ALB为例，它的儿童总死亡率是１８PER THOUSAND。现在，其中有１０。５％是死于DIABETES。那每一千个小童里面有多少人是死于这种病？

如果你能够解答这个问题，你便应该知道怎样做了。; )

z-score

原帖由 dk7 于 2008-4-21 01:09 发表
Using Pdf for the data, that's what I call creativity...

Could you read the data in ok? I used pdf because I don't have zip on this computer.

dk7

如果我不能把它处理好我便连成为Statistician 的资格也没有了 : )

我已经＂诱導＂她回答自己的问题了...

z-score

原帖由 dk7 于 2008-4-21 17:21 发表
如果我不能把它处理好我便连成为Statistician 的资格也没有了 : )

我已经＂诱導＂她回答自己的问题了...

Good on you mate, you are the man.

阴雨天

给你10个数 算standard division 怎么算来着? 高中学过,忘的一干二净~
谢谢了 阿童木~

水鱼皇后

原帖由 dk7 于 2008-4-21 03:53 发表
想问你一个问题。以ALB为例，它的儿童总死亡率是１８PER THOUSAND。现在，其中有１０。５％是死于DIABETES。那每一千个小童里面有多少人是死于这种病？

如果你能够解答这个问题，你便应该知道怎样做了。; )

18*10.5%...

就是用死亡率per thousand那个column 乘percentage那8个columns...得出来再做PCA？