Edexcel S3 - Wednesday 25th May AM 2016

This is a necessary condition for PMCC

June 2009, Question 3(c)

Why does the mark scheme answer mean SRCC is better than PMCC? Why does the finishing position need to be normally distributed?

Mark Scheme:

Reply 661

L'Evil Wolf

6

In contngency tables I believe gagaface asked something similar, but is H0 There is no association beetween the two variables OR is it The two variables are independent

Reply 662

Krollo

17

Original post by L'Evil Wolf

In contngency tables I believe gagaface asked something similar, but is H0 There is no association beetween the two variables OR is it The two variables are independent

Either should be perfectly fine but I generally go with whatever it says in the question. Failing that I normally go dor the association stuff.

Posted from TSR Mobile

Reply 663

economicss

3

Original post by Euclidean

Thanks! I didn't know about the normally distributed thing, had a google and found this for anyone who was wondering the same thing:

No problem, thanks for that :smile:

Reply 664

Rkai01

6

https://771a1ec81340d97ae9ed29694f73dd633b1c7c70.googledrive.com/host/0B1ZiqBksUHNYV1BTbDkxWXZCVmc/CH5.pdf

For q 14 c exD shouldn't it be two tailed as it is asking for any agreement meaning either positive or negative agreement?

Reply 665

economicss

3

Original post by Rkai01

https://771a1ec81340d97ae9ed29694f73dd633b1c7c70.googledrive.com/host/0B1ZiqBksUHNYV1BTbDkxWXZCVmc/CH5.pdf

For q 14 c exD shouldn't it be two tailed as it is asking for any agreement meaning either positive or negative agreement?

I see what you mean, but I think it wants you to infer from the previous part of the question that because there was quite strong positive correlation between the results then you should test for positive correlation as we can see that there won't be negative correlation so a two-tailed test would be pointless :smile:

Reply 666

paradoxequation

2

Why is there only 1 constraint here? Calculating the mean is one, but what about the usual one where the expected frequency is equal to the observed frequency? I have a feeling this was asked before but I can't find it lol.

Original post by paradoxequation

Why is there only 1 constraint here? Calculating the mean is one, but what about the usual one where the expected frequency is equal to the observed frequency? I have a feeling this was asked before but I can't find it lol.

When testing for any kind of uniform distribution- so continuous uniform or discrete uniform, the degrees of freedom are always the number of cells after combining-1, if we had been testing for a Poisson distribution then you would have been right and we would have been subtracting 2, hope that helps :smile:

Reply 668

paradoxequation

2

Original post by economicss

When testing for any kind of uniform distribution- so continuous uniform or discrete uniform, the degrees of freedom are always the number of cells after combining-1, if we had been testing for a Poisson distribution then you would have been right and we would have been subtracting 2, hope that helps :smile:

So because we are not testing for a poisson, but rather if the rate of accidents is constant, then there is no constraint for the expected f = observed f? I guess it makes sense considering we are not testing a model. Thanks! :smile:

Reply 669

economicss

3

Original post by paradoxequation

So because we are not testing for a poisson, but rather if the rate of accidents is constant, then there is no constraint for the expected f = observed f? I guess it makes sense considering we are not testing a model. Thanks! :smile:

Yep that's it :smile:

No problem!

Reply 670

username1162972

18

Original post by economicss

When testing for any kind of uniform distribution- so continuous uniform or discrete uniform, the degrees of freedom are always the number of cells after combining-1, if we had been testing for a Poisson distribution then you would have been right and we would have been subtracting 2, hope that helps :smile:

is this not poisson?

Reply 671

Kevin De Bruyne

OP

21

The day is approaching ...

Out of questions to do? Try madasmaths statistics booklets.

Reply 672

economicss

3

Original post by physicsmaths

is this not poisson?

I don't think so, because we're not testing whether the Poisson distribution is a suitable model for the data, we're testing whether the probability is constant :smile:

Reply 673

Euclidean

11

Does anyone have an idea why 2015 isn't on physicsandmathstutor anymore?

Reply 674

math42

20

Original post by Euclidean

Does anyone have an idea why 2015 isn't on physicsandmathstutor anymore?

Nah but I'm pretty sure it'll be up on Edexcel anyway

Reply 675

Zacken

22

Original post by Euclidean

Does anyone have an idea why 2015 isn't on physicsandmathstutor anymore?

Anymore? I think he was just too lazy/cba to put it up? It's on the Edexcel website anyhow.

Reply 676

username1162972

18

Original post by Zacken

Anymore? I think he was just too lazy/cba to put it up? It's on the Edexcel website anyhow.

its weird week before m3 ial june 15 was up then before the exam it was gone....
his website is so big i think he has to be very careful on what he can put up and stuff.

Reply 677

L'Evil Wolf

6

Original post by Krollo

Either should be perfectly fine but I generally go with whatever it says in the question. Failing that I normally go dor the association stuff.

Posted from TSR Mobile

Thanks, checked with the edexcel questions that seems a solid approach.

Can I ask you, on the issue of the Poisson model testing stuff, on Chapter 4, page 79 Ex7 Edexcel obtain the expected value by multiplying the probability for some r, with the total number of observations,

yet on Ex4b q7 they incline to multiply the probablity for some r, with the mean, lander.

So which is it. Essentially when calculating the observed value in a poisson test, what do we multiply the probabiilty with to obtain the expected value.