Edexcel S3 - Wednesday 25th May AM 2016

Papers seem quite regurgitative:

Random Variables
Confidence intervals/unbaised estimators
Contingecy Table
Sampling - probably stratified
spearman's rank
poisson distribution
CLT/2 independent ND

Seems the general trend of every paper tbh

Reply 681

Zacken

22

Original post by L'Evil Wolf

Papers seem quite regurgitative.

Every A-Level module ever.

Reply 682

L'Evil Wolf

6

Original post by Zacken

Every A-Level module ever.

Yeah haha.

btw if you and others havent checked this out already;

http://drfrostmaths.com/page.php?page=21

good site, especially for S3 theory.

Reply 683

username1162972

18

Original post by L'Evil Wolf

Papers seem quite regurgitative:

Random Variables
Confidence intervals/unbaised estimators
Contingecy Table
Sampling - probably stratified
spearman's rank
poisson distribution
CLT/2 independent ND

Seems the general trend of every paper tbh

a question to s3 is that it can be taught within days and one can do papers and get A's. ridiculously easy material to get a hang of.

Reply 684

L'Evil Wolf

6

Original post by physicsmaths

a question to s3 is that it can be taught within days and one can do papers and get A's. ridiculously easy material to get a hang of.

I agree, the concept of the 2 tailed test used prevelantly in chapter 5 should be used if possible in Chapter 4 with chi squared.

Shame, we don't do wilcoxon, f,t distributions, not even in yates correction is used aha

Reply 685

anndz3007

2

is there any particularly hard s3 paper except from the june 15 one ??

Reply 686

Euclidean

11

Could someone please explain how and why to perform the variance calculation on part (d) of Q5 2015?

Reply 687

Ayman!

15

Original post by Euclidean

Could someone please explain how and why to perform the variance calculation on part (d) of Q5 2015?

Express U-bar in terms of what it actually means. i.e. U-bar = 0.2(U_1 + ... + U_5).

By the way, did you manage to find your hard papers? :tongue:

Reply 688

math42

20

Original post by Euclidean

Could someone please explain how and why to perform the variance calculation on part (d) of Q5 2015?

Well the how has been answered...but as to why, have you answered c? Essentially, it's all about dependence

Reply 689

Euclidean

11

Original post by Ayman!

Express U-bar in terms of what it actually means. i.e. U-bar = 0.2(U_1 + ... + U_5).

By the way, did you manage to find your hard papers? :tongue:

Nah I think S3 is pretty standard except for this one. I fluffed up this part (d) though so lost 5 marks there... :colonhash:

I understand writing the mean in terms of 0.2(sum) but I don't understand where this variance calculation has come from...

Reply 690

Euclidean

11

Original post by 1 8 13 20 42

Well the how has been answered...but as to why, have you answered c? Essentially, it's all about dependence

I answered (c):

The variance combinations formula only applies if the samples are independent, which is not necessarily true for any sample hence the assumption of the same variance as in part (b) cannot be made

I gave myself that mark

Reply 691

math42

20

Original post by Euclidean

Nah I think S3 is pretty standard except for this one. I fluffed up this part (d) though so lost 5 marks there... :colonhash:

I understand writing the mean in terms of 0.2(sum) but I don't understand where this variance calculation has come from...

It's just the normal combination of variables. Each of the Ui has variance sigma^2, and when combining you use the rule (here extending to more than two variables of course) Var(aX + bY) = a^2Var(X) + b^2Var(Y)

Reply 692

Ayman!

15

Original post by Euclidean

Nah I think S3 is pretty standard except for this one. I fluffed up this part (d) though so lost 5 marks there... :colonhash:

I understand writing the mean in terms of 0.2(sum) but I don't understand where this variance calculation has come from...

Well, what does

\mathrm{Var\left ( F \right )}= \mathrm{Var\left ( U_{1}-\bar{U} \right )}= \mathrm{Var\left ( \frac{5U_{1}-U_{1}-\cdots -U_{5}}{5} \right )}

compute to? Remember that we can't directly use Var(aX + bY) = Var(aX) + Var(bY) on Var(U - U-bar) because U and U-bar are not independent.

(edited 7 years ago)

Reply 693

Euclidean

11

Original post by 1 8 13 20 42

Var(aX + bY) = a^2Var(X) + b^2Var(Y)

Doesn't this only work for independent variables? :cry2:

Original post by Ayman!

Well, what does

\mathrm{Var\left ( F \right )}= \mathrm{Var\left ( U_{1}-\bar{U} \right )}= \mathrm{Var\left ( \frac{5U_{1}-U_{1}-\cdots -U_{5}}{5} \right )}

compute to? Remember that we can't use Var(aX + bY) = Var(aX) + Var(bY) on Var(U - U-bar) because U and U-bar are not independent.

I honestly do not know

Unparseable latex formula:

\displaystyle \mathrm{Var\left(\dfrac{4U_1 - U_2 - U_3 - U_4 - U_5}{5}}\right)}

How should I evaluate this for any sample? :frown:

I think I just don't understand what conditions the formulae require properly...

(edited 7 years ago)

Reply 694

anndz3007

2

I dont really get part c of this question, the MS said 2 x "their alpha " + 3

Reply 695

Euclidean

11

Original post by Ayman!

Remember that we can't directly use Var(aX + bY) = Var(aX) + Var(bY) on Var(U - U-bar) because U and U-bar are not independent.

I think I understand your statement now.

We can't go straight into

Var(aX + bY) = Var(aX) + Var(bY)

but we can apply it to the individual

X_i

's right?

Reply 696

coolguy123456

8

Original post by anndz3007

I dont really get part c of this question, the MS said 2 x "their alpha " + 3

alpha is the mean you calculate and the +3 is cause if you look at the question at the top, the max would be 2a + 3 hence that's what they've done for the max value.

Reply 697

Ayman!

15

Original post by Euclidean

I think I understand your statement now.

We can't go straight into

Var(aX + bY) = Var(aX) + Var(bY)

but we can apply it to the individual

X_i

's right?

Apologies for the slow replies, my 'tex is not that good. :tongue:

But yes - that is precisely what I meant. You can't jump straight into that because the mean is dependent on the individual X_is, but each X_iisn't dependent on each other - they're all independent units. That's why we can go

\mathrm{Var\left ( \frac{4U_{1}-U_{2}-\cdots -U_{5}}{5} \right )} = \frac{1}{25} \left[ \mathrm{ Var\left (4U_{1} \right )+Var\left ( U_{2} \right )+\cdots +Var\left ( U_{5} \right ) } \right]

. That's my interpretation of what's going on here, at least. I'm also curious as to what @13 1 20 8 42 says.

Reply 698

math42

20

Original post by Euclidean

Doesn't this only work for independent variables? :cry2:

Original post by Ayman!

Apologies for the slow replies, my 'tex is not that good. :tongue:

But yes - that is precisely what I meant. You can't jump straight into that because the mean is dependent on the individual X_is, but each X_iisn't dependent on each other - they're all independent units. That's why we can go

\mathrm{Var\left ( \frac{4U_{1}-U_{2}-\cdots -U_{5}}{5} \right )} = \frac{1}{25} \left[ \mathrm{ Var\left (4U_{1} \right )+Var\left ( U_{2} \right )+\cdots +Var\left ( U_{5} \right ) } \right]

. That's my interpretation of what's going on here, at least. I'm also curious as to what @13 1 20 8 42 says.

Yeah it's basically that the individual Ui are independent so the point of expressing the mean in terms of them is that you can then use the standard expectation algebra

Reply 699

anndz3007

2

Original post by coolguy123456

alpha is the mean you calculate and the +3 is cause if you look at the question at the top, the max would be 2a + 3 hence that's what they've done for the max value.

lol thank you, i thought it's a new formula, didnt notice the max value in the distribution.

Edexcel S3 - Wednesday 25th May AM 2016

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you