OCR S2 (non-mei)

lllllllllll

13

How did everyone end up doing S2?

Reply 1

ChoccyPhilly

13

I've barely started s2 yet. Because it's only for the AS, I don't need it si I've neglected it. Only hoping for a C tbh

Reply 2

lllllllllll

OP

13

Original post by ChoccyPhilly

I've barely started s2 yet. Because it's only for the AS, I don't need it si I've neglected it. Only hoping for a C tbh

You can learn S2 in quite a short time, it's only about 7 chapters and the ones on hypothesis tests and sampling are pretty much the same.

Reply 3

ChoccyPhilly

13

Original post by lllllllllll

You can learn S2 in quite a short time, it's only about 7 chapters and the ones on hypothesis tests and sampling are pretty much the same.

Yeah I'm gonna spend a day or two during half term JUST doing s2 work from scratch. Should be... Interesting

Reply 4

Sam596

2

Best revision places online? I find S2 so freaking hard; I can't do Hypothesis testing, which is basically half of the paper.

Reply 5

lllllllllll

OP

13

Original post by Sam596

Best revision places online? I find S2 so freaking hard; I can't do Hypothesis testing, which is basically half of the paper.

I was looking at exam solutions, I can't believe Edexel don't have this!! I just learnt it by bashing my way through the book. Wasn't the most time-effective way to do it but it eventually worked :tongue:

Edit: Oh wait.. They do have it

http://www.examsolutions.net/maths-revision/syllabuses/Edexcel/period-1/S2/module.php
but it seems they only do discrete random variables. Still worth looking at though :smile:

(edited 8 years ago)

Reply 6

Sam596

2

Original post by lllllllllll

I was looking at exam solutions, I can't believe Edexel don't have this!! I just learnt it by bashing my way through the book. Wasn't the most time-effective way to do it but it eventually worked :tongue:

Edit: Oh wait.. They do have it

http://www.examsolutions.net/maths-revision/syllabuses/Edexcel/period-1/S2/module.php
but it seems they only do discrete random variables. Still worth looking at though :smile:

http://www.furthermaths.org.uk/ocr-revision is what i've found and using now.

Reply 7

lllllllllll

OP

13

Original post by Sam596

http://www.furthermaths.org.uk/ocr-revision is what i've found and using now.

That guy is also very good :smile:

Haven't watched any of his videos in a while though

Reply 8

James_Tait

Hey guys, can anyone help me with deciding when to divide the varience of something by the sample size? Simple I know but I can't get my head around it

Reply 9

TOxp68

6

Original post by James_Tait

Hey guys, can anyone help me with deciding when to divide the varience of something by the sample size? Simple I know but I can't get my head around it

You divide by the sample size when you're asked about probabilities regarding a sample. No matter how the population is distributed, providing n is large, the central limit theorem says the sample mean can be approximated by X-N(mean, variance/sample size).

Reply 10

lllllllllll

OP

13

For continuity corrections, how do we know when we should use

\dfrac{1}{2n}

as the continuity correction.…and which scenario do we use

Var(n(X)) = \dfrac{s^2}{n}

instead of

Var(n(X)) = \dfrac{\sigma^2}{n}

?

(edited 8 years ago)

Reply 11

TOxp68

6

You use 1/2n when the initial distribution is discrete. So if the initial distribution was X-B(40,0.2) you can approx that to X-N(8, 6.4) then to get the distribution for the sample of size n you get X(bar)-N(8,6.4/n) and say you were looking for P(X(bar) <=10), because the distribution at the start was binomial you use a 1/2n continuity correction instead of 0.5.

(edited 8 years ago)

Reply 12

lllllllllll

OP

13

Original post by Tomh97

You use 1/2n when the initial distribution is discrete. So if the initial distribution was X-B(40,0.2) you can approx that to X-N(8, 6.4) then to get the distribution for the sample of size n you get X(bar)-N(8,6.4/n) and say you were looking for P(X(bar) <=10), because the distribution at the start was binomial you use a 1/2n continuity correction instead of 0.5.

Oh I see, so it's CLT for discrete variables?

Posted from TSR Mobile

Reply 13

TOxp68

6

Original post by lllllllllll

Oh I see, so it's CLT for discrete variables?

Posted from TSR Mobile

The CLT says that if a sample is sufficiently large the distribution of the sample mean is approximated by X-N(mean, variance/n), no matter what the underlying distribution is.

Reply 14

lllllllllll

OP

13

Original post by Tomh97

The CLT says that if a sample is sufficiently large the distribution of the sample mean is approximated by X-N(mean, variance/n), no matter what the underlying distribution is.

Think I understand it slightly better now. June 2013 Q6 is a very good example to practice. Thanks :smile:

Reply 15

Molly_xox

16

Hiya

I'll be sitting this tomorrow, feeling okay about it so fingers crossed!

Would someone possibly be able to help me with this question though? Question 4 on Jan 2012. I have the right method but I don't understand why or how they've standardised it?

Thank you :smile:

Original post by Molly_xox

Hiya

I'll be sitting this tomorrow, feeling okay about it so fingers crossed!

Would someone possibly be able to help me with this question though? Question 4 on Jan 2012. I have the right method but I don't understand why or how they've standardised it?

Thank you :smile:

Really hope this goes well after the train wreck that was M2. :frown:

It's standardised because it's the sample mean of H so you divide the variance by 50. Hope that helps! :smile:

Reply 17

TOxp68

6

Original post by Molly_xox

Hiya

I'll be sitting this tomorrow, feeling okay about it so fingers crossed!

Would someone possibly be able to help me with this question though? Question 4 on Jan 2012. I have the right method but I don't understand why or how they've standardised it?

Thank you :smile:

Because the initial R.V was said to be discrete you use a continuity correction of 1/2n instead of the usual 0.5. Because they want <2.6 you do 2.6-1/(2*50), because there were 50 observations.

Reply 18

Molly_xox

16

Original post by tbhlouise

Really hope this goes well after the train wreck that was M2. :frown:

It's standardised because it's the sample mean of H so you divide the variance by 50. Hope that helps! :smile:

Here's hoping! Fingers crossed for your M2 result too.

Okay...this concept is completely new to me! I know that the variance has to be divided by 50, so it becomes

H(sample) ~ N(2.5, 1.25/50)

But I'm not sure why that applies to the 2.6? And why does it get standardised to 2.59 as opposed to, say, 2.55?

I'm sorry if I'm being really thick here :redface:

thanks for your help so far!

Edit:

Original post by Tomh97

Because the initial R.V was said to be discrete you use a continuity correction of 1/2n instead of the usual 0.5. Because they want <2.6 you do 2.6-1/(2*50), because there were 50 observations.

I didn't see this before I posted, thank you! So every time the variable is discrete you just use that? Thank you!

(edited 8 years ago)

Reply 19

tbhlouise

1

Original post by Tomh97

Because the initial R.V was said to be discrete you use a continuity correction of 1/2n instead of the usual 0.5. Because they want <2.6 you do 2.6-1/(2*50), because there were 50 observations.