ECMT quiz 2 (1 Viewer)

[Lainee]

Lainee to the rescue... you're my hero! hehehe Do you know where to find the proof for 3k?

Ooh no wait, I change my mind about 2(d) I thought you said 3(d). For 2(d) I think we were meant to use standard error of the mean to calculate that. ie. Sigma/sqrt.n
But I'm not too sure about that one.

3(k) is in one of the additional word docs from our lecture notes. Basically:

E(X-bar)=E(sigma Xi/n)
=1/n * sigma E(Xi)
=1/n * n * u
=u

X-bar is said to be an unbiased estimator of mean because the average of all sample means of size n (ie. E(X-bar)) will be equal to the population mean (ie. u).

 

[ToO LaZy ^*]

I did in this way:
P(X>2000)=0.9
P(X>6000)=.03
In the other words:
P(X<2000)=0.1
P(X<6000)=0.97
Then
1:2000-u/SD=-2.33(Z value from normal table)
2:6000-u/SD=1.88(As above)

Slove 1 and 2, I got u=4213.78; SD=950.12
Any commente?

hmm..i don't get the -2.33 bit.
coz when Z = -2.33, the area equals 0.0099...and i cant see any connection with that ..
care to explain anyone?

 

[Lainee]

Should be:

2000-u/SD=-1.28
6000-u/SD=1.88

 

[ToO LaZy ^*]

ohh@.
i thought so.!

thx dear.

 

[04er]

hmm..i don't get the -2.33 bit.
coz when Z = -2.33, the area equals 0.0099...and i cant see any connection with that ..
care to explain anyone?

same... isn't it meant to be -1.28?????

 

[Lainee]

ohh@.
i thought so.!

thx dear.

Heheheh have you been constantly refreshing this page for the last half hour hoping for a responce? Any more things I can help with, cause I'm going to crash soon... last call.

 

[ToO LaZy ^*]

yeah..im done
could u just post ur answers for i) and j) tho..no need for working

 

[04er]

Heheheh have you been constantly refreshing this page for the last half hour hoping for a responce? Any more things I can help with, cause I'm going to crash soon... last call.

thank you all your help tonight Lainee, you've been really kind

 

[Lainee]

(i) 126.507,129.493
(j) 2.9993,3.0207

 

[ToO LaZy ^*]

sweeeeeeeeeeeeet~.
thx again love/

 

[ToO LaZy ^*]

oh..g) ii) is 2248 yeah?

 

[Lainee]

thank you all your help tonight Lainee, you've been really kind

sweeeeeeeeeeeeet~.
thx again love/

Good luck guys.




EDIT: (g)(ii) 2488 isn't it? 9(16^2)+4(4^2)-2(3)(-2)(-10)

EDIT EDIT: Using Var(aX+bX)=a^2Var(X)+b^2Var(Y)-2abCov(X,Y)

 

[ToO LaZy ^*]

2304 + 64 + (-120) = 2248?

 

[Lainee]

Alright, so I suck at calculator work. 2248 it is then.

 

[absolution*]

Good luck guys.




EDIT: (g)(ii) 2488 isn't it? 9(16^2)+4(4^2)-2(3)(-2)(-10)

EDIT EDIT: Using Var(aX+bX)=a^2Var(X)+b^2Var(Y)-2abCov(X,Y)

Why is the covariance a minus?

I have a few additional questions..

What answers did people get for 3. a) c) i) j)
How do you do 2 d) and 3 l)

cheers

 

[spin spin sugar]

the covariance is a minus cos i said so

 

[absolution*]

crusin' for a bruisin'

 

[spin spin sugar]

wellllll it answered your question. also i have the answers to the questions you wanted. come on msn and i'll tell youu

 

[absolution*]

confidence interval 100%

P(me going onto msn)=0

 

[Lainee]

Why is the covariance a minus?

I have a few additional questions..

What answers did people get for 3. a) c) i) j)
How do you do 2 d) and 3 l)

cheers

The covariance is given to you in the question: cov (x, y) = -10