ECMT1010 Computer Ass. 2 (2 Viewers)

[stazi]

In 2(a), in your first table, you seem to have a count of only 11. I'm pretty sure there should be 12 items in there. Just check that again?

Did you mean 3a? Yes, it does seem like i pooped that one. Thanks

 

[04er]

your sample variance in the first table is 2.83038E+11 ...

 

[ToO LaZy ^*]

Stas to the rescue of ECMT1010. Together we will win.

YES! NO MORE GAY 4MARK COMP ASSES.

was there meant to be something on the last few pages?...there seems to be fragments of words..

 

[stazi]

was there meant to be something on the last few pages?...there seems to be fragments of words..

Nope, dont know how those got there

 

[Lainee]

Did you mean 3a? Yes, it does seem like i pooped that one. Thanks

Yeah, I meant 3(a).

Had a closer look at your answers after dinner again and noticed a few more little errors at some points. Apart from the fact that you spelled 'alleged' wrong ( ) I think you need to clarify your reason in 3(c) why you don't believe that the results don't support the allegations of theft.

I don't think that a lower upper limit of the CI during the alleged theft period would support theft allegations. I would think that the whole CI for the period of alleged theft should be reasonably lower than the CI for the period where theft was not alleged. For example, if the CI for period where theft was not alleged was a,b , the CI for the period where theft was alleged would have to be c, d - where d is less than b and and c is less than a.

On a number line:
_________c_________d______a________b________

or

________c______a______d______b________


...if that makes any sense?

 

[011]

Did you use excel to get those CI printouts 1Time?

If so what in excel can do that?

 

[sarevok]

zzzzzzzzzzzzzzz

 

[absolution*]

Lainee > all

nuff said.

 

[ToO LaZy ^*]

what are we suppose to look for in Q3 c) and d) to determine whether there was theft or not?..

 

[absolution*]

Changes in con/city ratio id assume( over the 3 periods/).

 

[ToO LaZy ^*]

still don't know what to write

 

[Lainee]

what are we suppose to look for in Q3 c) and d) to determine whether there was theft or not?..

In (c) - so you worked out the C.I. for one period where no theft was alleged and another period where theft is alleged. If theft did occur then the average money collected in a certain period would be significantly less compared to the other period. So, the question you're asking yourself is: Is the C.I. for the period of alleged theft significantly lower than that for the period where no theft was alleged?

In (d) - you're looking for the same thing. After ommitting the outlier, and thus changing the C.I. for one period, is the C.I. for one period significantly lower than that of the other period?

 

[ToO LaZy ^*]

when you are saying if the CI for one period is lower than the other...do you mean like..the distance apart?>

 

[Lainee]

I drew this number line a few posts up ^

On a number line:
_________c_________d______a________b________

or

________c______a______d______b________

where a,b is your C.I. for the period where there is no allegation of theft; and
c,d is the C.I. for the period where theft is alleged.

You need to be able to show one of the two situations to support an allegation of theft.

As an example, in case my 'number line' isn't too clear

You find that C.I. for period where theft is not alleged = 100,200
You find that C.I. for period where theft is alleged = 60,120
This supports allegation of theft (1st number line) because C.I. for period where theft is alleged is significantly lower than the period where theft is not alleged.

OR, similarly

You find that C.I. for period where theft is not alleged = 100,200
You find that C.I. for period where theft is alleged = 20,70
This supports allegation of theft (2nd number line) because C.I. for period where theft is alleged is significantly lower than the period where theft is not alleged.

 

[ToO LaZy ^*]

did you end up saying that theft had occured?..
i found that P1 mean > P2 mean...and that the whole CI shifted to the left..?.

 

[Lainee]

did you end up saying that theft had occured?..
i found that P1 mean > P2 mean...and that the whole CI shifted to the left..?.

Well, I could post up my answers, but I pretty much got what 1Time posted up a few pgs back.

For (c) my conclusion was: As the above results show, the confidence interval for the period of alleged theft is not reasonably lower than the confidence interval for the period where no theft was alleged. Thus, these results do not completely support the allegation of theft.


For (d) when omitting outlier, conclusion is different. The results can now be used to support the allegation of theft. This is because during the period of alleged theft the upper and lower limits of the confidence interval are reasonably lower than those during the period where theft was not alleged. This would indicate that compared to the collections made by the city workers near city hall, the collections made by Brink, on average, is substantially less than during the other given periods.

 

[ToO LaZy ^*]

I <3 Lainee

 

[Lainee]

I <3 Lainee

Awww thanks dear.

 

[Lainee]

1Time, I know you've submitted your assignment already, but I looked at the solutions to Q1 you posted before and I think there are a few problems. I wish I saw it earlier!

1.(a)(i) What you're trying to find should be P(750<X<800) and NOT P(X>750). This is because the range of scores is between 200 and 800. It doesn't go up to infinity, so you must limit the area under the curve to between 750 and 800. ie:

P(750<X<800) = P(X<800)-P(X<750)
“=NORMDIST(800,520,96.3,TRUE)-NORMDIST(750,520,96.3,TRUE)”
= 0.006640395


Same thing with (ii)

P(200<X<560)=P(X<560)-P(X<200)
“=NORMDIST(560,520,96.3,TRUE)-NORMDIST(200,520,96.3,TRUE)”
= 0.660618519

All the rest are unaffected by this.

 

[myg0t]

1Time, I know you've submitted your assignment already, but I looked at the solutions to Q1 you posted before and I think there are a few problems. I wish I saw it earlier!

1.(a)(i) What you're trying to find should be P(750<X<800) and NOT P(X>750). This is because the range of scores is between 200 and 800. It doesn't go up to infinity, so you must limit the area under the curve to between 750 and 800. ie:

P(750<X<800) = P(X<800)-P(X<750)
“=NORMDIST(800,520,96.3,TRUE)-NORMDIST(750,520,96.3,TRUE)”
= 0.006640395


Same thing with (ii)

P(200<X<560)=P(X<560)-P(X<200)
“=NORMDIST(560,520,96.3,TRUE)-NORMDIST(200,520,96.3,TRUE)”
= 0.660618519

All the rest are unaffected by this.

Im not so sure that this is right. The probability is affected by the mean score (520) and the s.d (96.3), therefore I think the original way was right.