2008 First Semester STATS课程官方问答贴

kirstens

ok. I think:
(b) Give the probability distribution for X1.
-It's not asking distribution (so it's not an answer like binomial or geo, etc), it's asking the probability, so your answer will be something like:
P(X1=1)=p, P(X1=2)=q, P(X1=3)=... or you can just simply draw a table to show that.

(d) Find E[Y ]
-to explain this, let's say, Y=X1+X2 (where X2 is the number shown on the second die). So, what can you get from the experiment? Y cannot be 1, definitely. Y can be 2 iff X1=1 and X2=1. Y can be 3, there are 2 ways - X1=1 and X2=2 OR X1=2 and X2=1, and Y=... and so on. So WHAT IS THE AVERAGE VALUE OF Y? you would've known the probability of Y=2 is 1/24, of Y=3 is 2/24 and so on, so you can find out E(Y)=sum[y*P(Y=y)]=2*1/24+3*2/24+...

kirstens

forgot to metion, level 397 is trying to answer 雨轩听风's questions at #396

I've been thinking how's a 4-side die look like... and then I realize it's a tetrahedron, with each side as a triangle...

[ 本帖最后由 kirstens 于 2008-4-17 09:19 编辑 ]

ミ﹏龍.

阿童木啊~

我在做20X的Assignment最后一道题~

为什么做到Anova table啲时候R就做不出来了，之前每道题都可以做出来。

这是我啲过程：

> library(s20x)
> Survival.df<-read.table(file.choose(),header=T)
> attach(Survival.df)
> interaction.plots(times~treatment+poison)
> interaction.plots(times~poison+treatment)
> levene.test(times~poison+treatment)
                Df  Sum Squares  Mean Square  F-statistic  p-value   
Between Groups  7   0.17707      0.0253       2.82079      0.02709   
Within Groups   24  0.21523      0.00897                             
Total           31  0.3923                                          
> levene.test(log(times)~poison+treatment)
                Df  Sum Squares  Mean Square  F-statistic  p-value   
Between Groups  7   0.21984      0.03141      1.62034      0.17765   
Within Groups   24  0.46517      0.01938                             
Total           31  0.68501                                          
> Survival.fit<-lm(log(times)~poison+treatment+poison*treatment)
> plot(Survival.fit,which=2)
> shapiro.test(residuals(Survival.fit))

        Shapiro-Wilk normality test

data:  residuals(Survival.fit)
W = 0.9775, p-value = 0.7244

> summary.2way(Survival.fit,page="table")
错误于summary2way(fit, page, digit, conf.level, print.out, ...) :
  All explanatory variables should be factors!



-------^  就是到这一步，就提示一个错误，但我看来看去也没看出除了这样写还能怎样写，拜托帮我看看，谢谢咯！

kirstens

yea, you're right.
it's not as difficult as you think. most of them are very basic probabilities.

kirstens

(a)是用poisson P(X=0)?
- yes.

(b) Given that a fixed team scores no goals in a given match, find the probability that they lose
that match.
-important words here from the question: Poisson distribution, independent of the goals scored by the other team!
question asked they lose, which means the other team socre > 0
so you would need to fine P(X NOT= 0)
same in (c), for a team scored 2, they'll win iff the other team scored <2 (which is 0 or 1)

(d)的ends in a draw是什么意思? this means the 2 teams have the same score.

kirstens

think it this way, it's a 15 mark question with 7 semi questions. each question worth only 2 marks on average. So it's not difficult. Don't doubt your answers.

当时的理想

STA 108..de ASSIG 2///

QUESTION 3... do not CACULATE THE PROBABILITIES...    how can i JUSTIFY......    i just don't know how to  described..

and the Q5....   what is x bar stand for?  is it  the population of local pizzeria delivery time ??  what should i say ??

THax,..   

kirstens

(d) Give a formula for the probability that a given match ends in a draw.
your answer is wrong. for a match ends in a draw, you'll need when Team A score x, Team B score x too ->P(A=x)*P(B=x) with x=0 to infinity, so the formula is Sigma(x=0 to infinity)[P(A=x)*P(B=x)] where P(A=x) ~ Poisson(1) as well as P(B=x)

(e)和(f)是问distribution...
not sure about this. don't see the difference between e) and f). but I don't think f) is Geo, why do you consider it to be Geo?

dk7

I'd actually need the data to have a better appreciation for the problem, but I just don't have access to them.

When is the assignment due BTW?

z-score

Sorry sorry 阿童木今天生病了, 卧床休息了一天。突然这么多问题

谢谢kirstens和dk7帮助我回答，希望明天我能好点回答大家的问题。Sorry.

rabby

好好休息养病 可怜的阿木童

dk7

Get well soon!

~Kurosagi~

好好養病把


Stats太亂了...

看的頭暈

不能沒有妳們的幫助啊


謝謝!!!

kirstens

take care.

t-multiplier

原帖由 雨轩听风 于 2008-4-16 15:40 发表
1. An experiment involves rolling two fair dice. The first one is 4-sided and the second one is 6-sided,
so there are 24 possible outcomes. Let X1 denote the number showing on the first die, and ...

b) I think if a 4 sided die is fair, then  getting a 1, 2, 3 or 4 should all have probabilities 0.25. Hence, I think the distribution should be uniform as in X1~Uniform(1,4)

d) Let Y = X1 + X2
X1 ~ Uniform(1,4)
X2 ~ Uniform (1,6)
IE(Y) = IE(X1 + X2) = IE(X1) + IE(X2) = 2.5 + 3.5
as for uniform distributions the expected value is (a + b) / 2, where a is the lowest value, b is the largest value.

[ 本帖最后由 t-multiplier 于 2008-4-18 15:19 编辑 ]

t-multiplier

原帖由 水鱼皇后 于 2008-4-15 23:17 发表
2008 Assignment 2

Download data set 2008Assig2.csv
It contains the same variables as before and some additional ones:
CMR Child Mortality rate (number of deaths per thousand children)
Pop10 ...

原帖由 dk7 于 2008-4-17 19:35 发表
I'd actually need the data to have a better appreciation for the problem, but I just don't have access to them.

When is the assignment due BTW?

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.

email: [email protected]

t-multiplier

原帖由 rabby 于 2008-4-17 23:00 发表
好好休息养病 可怜的阿木童

原帖由 dk7 于 2008-4-17 23:42 发表
Get well soon!

原帖由 kirstens 于 2008-4-18 11:46 发表
take care.

Thank you guys!

t-multiplier

原帖由 kirstens 于 2008-4-17 09:12 发表
ok. I think:
(b) Give the probability distribution for X1.
-It's not asking distribution (so it's not an answer like binomial or geo, etc), it's asking the probability, so your answer will be s ...

Yeah, true true, for part b, I think the question is wanting you to outline the following:

f(x) = {0.25 if x = 1, 0.25 if x = 2, 0.25 if x = 3, 0.25 if x = 4, 0 otherwise}

t-multiplier

原帖由 ミ﹏龍. 于 2008-4-17 11:03 发表
阿童木啊~

我在做20X的Assignment最后一道题~

为什么做到Anova table啲时候R就做不出来了，之前每道题都可以做出来。

这是我啲过程：

> library(s20x)
> Survival.df attach(Survival.df)
> inte ...

This is what I got after I ran the analysis, I put the code you missed in blue.

Unfortunately we have to tell R manually that poison is a factor variable.

>survival.df <- rr()
>attach(survival.df)
> poison<- factor(poison)
>interaction.plots(times~poison+treatment)


>interaction.plots(times~treatment+poison)


>levene.test(times~poison+treatment)

Df
Sum Squares
Mean Square
F-statistic
p-value

BetweenGroups
7
0.17707
0.0253
2.82079
0.02709

WithinGroups
24
0.21523
0.00897

Total
31
0.3923

>levene.test(log(times)~poison+treatment)

Df
Sum Squares
Mean Square
F-statistic
p-value

BetweenGroups
7
0.21984
0.03141
1.62034
0.17765

WithinGroups
24
0.46517
0.01938

Total
31
0.68501


>survival.fit <- lm(log(times)~poison+treatment+poison*treatment)
>plot(survival.fit,which=2)


>shapiro.test(residuals(survival.fit))


Shapiro-Wilk normality test

data:
residuals(survival.fit)
W = 0.9775,p-value = 0.7244
>summary.2way(survival.fit,page="table")
ANOVATable:

Df
Sum Squares
Mean Square
F-statistic
p-value

poison
1
0.27874
0.27874
3.66476
0.06757

treatment
3
3.09688
1.03229
13.57192
2e-05

poison:treatment3
0.20897
0.06966
0.91579
0.44811

Residuals
24
1.82546
0.07606

Total
31
5.41005

>summary.2way(survival.fit,page="nointeraction")


poisoncomparisons:


Estimate Tukey.L Tukey.U Tukey.p
1
-
20.186663 -0.0146
0.3879
0.0676


treatmentcomparisons:


Estimate Tukey.L Tukey.U Tukey.p
A
-
B-0.8250359 -1.2054 -0.4446
0.0000
A
-
C-0.2373069 -0.6177
0.1431
0.3353
A
-
D-0.5413803 -0.9218 -0.1610
0.0033
B
-
C
0.5877290
0.2073
0.9681
0.0014
B
-
D
0.2836556 -0.0967
0.6641
0.1959
C
-
D-0.3040734 -0.6845
0.0763
0.1505

t-multiplier

原帖由 雨轩听风 于 2008-4-17 11:04 发表
谢谢...kirstens ....你答得很详细......

(b) Give the probability distribution for X1.
你说P(X1=1)=p, P(X1=2)=q  
fair dice不是说明每一个的prob.都等于1/4吗?

(c) Find E[X1] and V ar(X1).
如 ...

Hehe, I just realised you've already answered it.

Thank you kirstens, I would hate to give out the wrong advice.

t-multiplier

原帖由 当时的理想 于 2008-4-17 16:50 发表
STA 108..de ASSIG 2///

QUESTION 3... do not CACULATE THE PROBABILITIES...    how can i JUSTIFY......    i just don't know how to  described..

and the Q5....   what is x bar stand for?  is ...

Questions 3

Do you think the explanations in 309 and 310 are helpful to you?

309:

ACT ~ Normal(21, 4.7)

SAT ~ Normal(500, 100)

Statement 1 - FALSE

21 - 2 * 4.7 = 11.6
21 + 2 * 4.7 = 30.4

因为95%的学生已经在11.6和30.4的中间了, 那么在11和30中间的学生会比95%更多, 而不是更少.

Statement 2 - TRUE

IP(ACT > 28) = IP(Z > (28-21)/4.7) = 1.489362

IP(SAT < 300) = IP(SAT < (300 - 500)/100) = -2

luvwei你说的是不是z-scores > 1? ATM大于1 is OK.

310:

Statement 4 - TRUE

IP(ACT = 25) = IP(Z = (25 - 21)/4.7) = 0.8511 (4 d.p.)

IP(SAT = 610) = IP(Z = (610 - 500)/100) = 1.1 (4 d.p.)

是这道吧?

Question 5

lower case x bar represents the sample mean time taken for a pizza delivered to you from the local pizzaria.

[ 本帖最后由 t-multiplier 于 2008-4-18 15:31 编辑 ]

dk7

Cheers, I was about to ask her actually : )

t-multiplier

Cool, cheers dk7, I thought it was something to do with my login.

t-multiplier

原帖由 qian.qian 于 2008-4-15 18:25 发表
3. Suppose that births in a family are female with probability .48 and male with probability .52,
independently of all previous births. A couple who are currently childless decide to continue
h ...

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 the probability that the family will have at least 5 children?

~Kurosagi~

Stats 108 asisgment 2
question 3 d

statment 3 不明白为什么是68-95-99.7 rule的那68%obbse

t-multiplier

原帖由 ~Kurosagi~ 于 2008-4-18 15:58 发表
Stats 108 asisgment 2
question 3 d

statment 3 不明白为什么是68-95-99.7 rule的那68%obbse

嗯，这是一个很强的trick, 68%是中间的区域那就证明两边的区域是32%, 所以一般的区域就是32%/2 = 16%。