ECMT1010 - Revision Questions (1 Viewer)

s.m.i.t.h

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yvonne_710 said:
i think it will be:
H0:B=0
H1:B<0
and then use test stat. which is t = (b1-B1)/ S(b1) and then u can get -0.87, isnt it?
s (b1) is the standard error which is 0.7637.. anyways this is the way Tig told us, i went to see him 2day as well.................
errrrrrrrrrrr......but still dont know how to do condtional distribution, can someone tell? the qs like 45 (b)

ohh ok..but what is B1 and s???
 

kow_dude

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s.m.i.t.h said:
ohh ok..but what is B1 and s???
From memory, B1 is the slope and s is the standard error of the slope. You'll find the information on chapter 13 regarding regression etc... There's also notes on the last few slides of MDS notes. :uhhuh:
 

yvonne_710

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s.m.i.t.h said:
ohh ok..but what is B1 and s???
u can see there are 2 numbers under the equation, they are standard errors, 5.8166 is standard error for B1, 0.7637 is standard error for b1, btw B1 is 0............
 

ressul

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yvonne_710 said:
i think it will be:
H0:B=0
H1:B<0
and then use test stat. which is t = (b1-B1)/ S(b1) and then u can get -0.87, isnt it?
s (b1) is the standard error which is 0.7637.. anyways this is the way Tig told us, i went to see him 2day as well.................
errrrrrrrrrrr......but still dont know how to do condtional distribution, can someone tell? the qs like 45 (b)
Well, I got the same answer as yours. But do u reject H0 or not. Because I found the critical value was 3.3554, I didn't reject it, so it means the B is 0, rather than negative. I think this conclusion is wrong(It should be negative as shown). What do u think?
 

yvonne_710

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ressul said:
Well, I got the same answer as yours. But do u reject H0 or not. Because I found the critical value was 3.3554, I didn't reject it, so it means the B is 0, rather than negative. I think this conclusion is wrong(It should be negative as shown). What do u think?
i think it should be negative...
 

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yvonne_710 said:
the question:
suppose the average charge for out-of-court work by solicitors in NSW is $200 per hour with a variance of 860. a random sample of 49 solicitors in NSW was selected.
if the population variance were not known, but you knew that 1% of solicitors charge less than $193 what would be the value of the population variance?
that one is confusing.
I guess a proportion confidence interval or hypothesis test is necessary for something first.
But i don't understand how we can get the population variance.
 

yvonne_710

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1Time4thePpl said:
that one is confusing.
I guess a proportion confidence interval or hypothesis test is necessary for something first.
But i don't understand how we can get the population variance.
i dont know how to do it ...........before, it just asked the probablity of a sample mean, i really dont know how to solve it!
 

yvonne_710

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ressul said:
Does your answer support it? What's your critical value?
i didnt find critical value, the way i did is from Tig, so i only did test stat. and he said one thing : B1=0, there is no relationship between experience and comission, (its only a special case), i am not sure about that, i havent done all of them, i only know these information,...srry
 

yvonne_710

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i found the way to work out critical value, but i dont know the way is right or not, coz it told u α = 0.01, then i α/2 is 0.005, and 1-0.005 = 0.995. and i found the value from Z table which is 2.58, and -0.87 is obivously smaller than 2.58,so dont reject H0, this is what i do, i really dont know its right or not.......but it makes sense to me
 

ressul

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Anyone

Anyone knows how to do Q11, I); Q21,h); Q48 and the full process for Q57c)?
 

ressul

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yvonne_710 said:
i found the way to work out critical value, but i dont know the way is right or not, coz it told u α = 0.01, then i α/2 is 0.005, and 1-0.005 = 0.995. and i found the value from Z table which is 2.58, and -0.87 is obivously smaller than 2.58,so dont reject H0, this is what i do, i really dont know its right or not.......but it makes sense to me
You set H0: meul=0 and H1: meul>0, right? if you do not reject H0, how ca u prove the slop is positive? That's where i'm confused of
 

yvonne_710

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ressul said:
Anyone knows how to do Q11, I); Q21,h); Q48 and the full process for Q57c)?
for Q11(I), to use E(X*Y)- E(X)*E(Y).
E(X)=1*.11+2*.31+3*.46+4*.12
E(X*Y)=1*1*.4+1*2*.14+1*3*.23+1*4*.07
and so on


Q57(C):
H0:B1=0
H1:B1>0
α IS 0.01, and distribution of test stat, Tn-2=10
and test stat. is (b1-B1)/Standard error of b1,
so (.575-0)/.2514=2.287 (B1 is always =0)
the critical value (i am not sure) : α=0.01, α/2=0.005,
1-α/2=0.995, and then T critical value is 3.1693
T obsevered < T critical value, then dont reject H0...
errrrrrr.......i just reliase T and Z thing!
 

yvonne_710

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ressul said:
You set H0: meul=0 and H1: meul>0, right? if you do not reject H0, how ca u prove the slop is positive? That's where i'm confused of
i am confused now..........errrrrrrrrrrr...but using Z table or T table?
 

avster

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with the rejection or non rejection

if its ho = 0
h1 > 0

tobs < tcritical
we do not reject

if its ho = 0
h1 < 0

tcritical < tobs
we do not reject

is that correct?
cheers
 

leila

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i finished it but can not do
Q2 d)
Q7
i do not agree with the answer for
Q51,Q54,Q55
(all about reject or not reject)
can someone help me.
 

leila

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avster:
for any circumstance (less than or greater than)
if t observe>t-critical(if it is positive number) or t observe<t-critical(if it is negative number)
we reject t observe

because in these two condition, the t observe area from the t-diagram is less than the t critical area, we have to reject them.
 

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tig said question 2)d, question 7, and q 25 not in syllabus...
 

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