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# Edexcel S2 Exam Paper January 2011 14/01/11 Watch

1. (Original post by Spongy)
So was the student guessing or not guessing !!!

Are we all agreed that we had to accept H0 which is p=0.2

The student was guessing, and we accept h0
(I did the same thing as you btw)
2. darn that wordy question No.2 .....flushes A* down the drain...;(
3. (Original post by LUCYPOOCEY)
Agreed with most things you say, although I'm confused with the teacher one.. Surely if 0.12>0.05 you don't reject hypothesis because it's insufficient evidence? I thought it was only if it was within the critical region that you reject Everyone seems to be saying what you're saying though so i don't doubt you're right, just would like some explanation
Everyone at my college did not reject H? because the result wasn't significant. Therefore the teacher couldn't prove anything.
4. (Original post by snowflakesblues)
The student was guessing, and we accept h0
(I did the same thing as you btw)
I put the answer is the student was NOT guessing simply because we accepted H0. However I can also it see it the other way because we accepted H0 the student was guessing because they did not get a high score - which I now believe is correct !!

So seems I have lost one mark for putting the wrong statement.

I think this will definately be discussed in the examiner's meeting because the question was not worded properly.
5. (Original post by jit987)
I'm very sure it said atleast 2. So there are 2 ways of doing this:
1 - P(x+0)+P(x=1)
Or P(x+2)+P(x=3)

Also how did everyone explain how the f(x) is 4-8x from the diagram given.
I said because the f(x) is a straight line and a straight line is in the form y=ax+b, where b is the y-intercept and 'a' is the gradient(change in y/x). So y intercept is 4(from the sketch) and the 'a' is 4/0.5 = 8. Also because the slope is going from left to right it is representing negative gradient so -8. Therefore f(x) is -8x+4 (4-8x).
I think that should be alright, i did
when x=0, f(x)=4, put this into the form f(x)=mx + c
4=0 + c
c=4
therefore, f(x)=mx + 4
when x=0.5, f(x) = 0, put this into f(x)=mx + 4
0=0.5m + 4
m=-8
therefore, y=4-8x
I didn't know if i had to do anything to show it was only within the region 0<x<0.5 and what i did didn't seem like 4 marks worth.
6. (Original post by Helsy)
Everyone at my college did not reject H? because the result wasn't significant. Therefore the teacher couldn't prove anything.
Yes - that is what i put down too - that the result was not significant so there was insufficient evidence!

The question now is does that mean the student was NOT GUESSING or GUESSING !!!!!!!
7. (Original post by Spongy)
I put the answer is the student was NOT guessing simply because we accepted H0. However I can also it see it the other way because we accepted H0 the student was guessing because they did not get a high score - which I now believe is correct !!

So seems I have lost one mark for putting the wrong statement.

I think this will definately be discussed in the examiner's meeting because the question was not worded properly.
Fingers crossed they do, everyone who came out of the exam said that it was badly worded...but I suppose when you think about it, it makes sense. I just hope that the grade boundaries aren't ridiculously high....because as far as I know, i've already lost 3 marks 1 from that teacher thing and that 2 marks...from that E(X^2) thing (when all I needed to dad was add the variance and mean ^2)
8. WTF?! but you can argue the student was not guessing as the expected value using that binomial distribution is 2/10 and the studen got higher and then the Teachers hypothesis (H1) is that the student got higher than the expected value because she was guessing but then you prove her wrong by accepting H0.
9. (Original post by Andy Ftw)
Yeah, I was thinking the same when working it out! I thought it was quite a "clever" question though, hadn't seen that type before
There is a type of question that separates the men from the boys type of speak that I read in my S2 book:

The trunk of a small tree varies in diameter from 10cm at the bottom to 2cm at the top. A small child is asked to measure the diameter of the trunk. The random variable R is the radius of the cross section of the tree as measured by the child. R-U[1,5].

Find the expected value of the area, A, of the cross-section of the tree.
Most people, I guess would simply find E(r) which would be 3 and put it into pi(r^2) which would give 9pi as the estimated area, however this is how it's done:

A=pi(r^2)
E(A)=E(pi r^2)
E(A)=pi E(r^2)

E(r^2)=Var(R)+[E(R)]^2

Var(R)= 1.33333333...
E(R)=3
E(R^2)=9+1.3333333333 = 10.3333333

E(A)=(31pi)/3

Don't know why I just posted that .
10. (Original post by Spongy)
Overall I found the paper very good.

I agree with your answers above with one exception - the teacher's question. I worked out the same probability as you however I put down accept H0 therefore the teacher's claim is rejected because the result was not significant. ?!?!?

I have definately lost one mark because I got mixed up with the negative skew and positive skew.
Yeh i accepted it too even though I got the same But everyone on here has said that 0.12>0.05 means you reject it, i thought if it was bigger it's not within the critical region so you don't reject... Very confused now.
11. (Original post by Helsy)
Everyone at my college did not reject H? because the result wasn't significant. Therefore the teacher couldn't prove anything.
Well the teacher stated that the student just guessed.

Gives H0: P = 0.2.

Expected value of x = 2.

but the student got 4. so X = 4

So H1: P > 0.2

So Reject if P(X_>4{X ¬ B(10, 0.2)}) <_ 0.05

work out and you get 0.12 something which is > 0.05

So accept H0,

There is no reason to reject the teachers statement. Basically what he said isn't rejected.

Ballsy wording leads to a tricky question.
12. for the teacher question, i think if you said Ho: p = 0.2 then you accept Ho - the teacher was right and she was guessing
but if you used p = 0.4 then you reject Ho
both ways ends up with saying that she was guessing? but both methods should be in the mark scheme? :s
13. unofficial markscheme anyone?
14. (Original post by snowflakesblues)
Fingers crossed they do, everyone who came out of the exam said that it was badly worded...but I suppose when you think about it, it makes sense. I just hope that the grade boundaries aren't ridiculously high....because as far as I know, i've already lost 3 marks 1 from that teacher thing and that 2 marks...from that E(X^2) thing (when all I needed to dad was add the variance and mean ^2)
(Original post by Ultimate1)
WTF?! but you can argue the student was not guessing as the expected value using that binomial distribution is 2/10 and the studen got higher and then the Teachers hypothesis (H1) is that the student got higher than the expected value because she was guessing but then you prove her wrong by accepting H0.
15. (Original post by Ultimate1)
WTF?! but you can argue the student was not guessing as the expected value using that binomial distribution is 2/10 and the studen got higher and then the Teachers hypothesis (H1) is that the student got higher than the expected value because she was guessing but then you prove her wrong by accepting H0.
I don't know if I'm right but I put the null hypothesis is accepted because the result wasn't significant.

So p=0.2 and then I concluded there is evidence to support the teachers claim
that the student is guessing...
16. (Original post by Ultimate1)
Within the range of what's asked of the question,I don't think they'll consider that because they didn't ask us to find E(X)...maybe if they did, then we might get compensated...otherwise....I guess h0 it is, and the student was guessing :|
17. (Original post by snowflakesblues)
Within the range of what's asked of the question,I don't think they'll consider that because they didn't ask us to find E(X)...maybe if they did, then we might get compensated...otherwise....I guess h0 it is, and the student was guessing :|
Oh well that just means one mark lost. But I still think the wording of that question was atrocious.
18. Question 2 seems to be causing a lot of hassle especially for the conclusion part.

I put:

H0: p=0.2
H1: p>0.2

and you reject H0 if P(X>(or equal to)4) <(or equal to) 0.05

which gives you 1 - P(X<(or equal to)3) <(or equal to) 0.05

end result was something like 0.12... <(or equal to) 0.05 which is not true, so there is insufficient evidence to reject H0, meaning we cannot conclude whether the teacher's claim was right or wrong.

Did anyone else put that?
19. (Original post by Ultimate1)
Oh well that just means one mark lost. But I still think the wording of that question was atrocious.
I know So was it just 1 mark u lost? Who knows it might still be 100UMS
20. (Original post by 20100)
Question 2 seems to be causing a lot of hassle especially for the conclusion part.

I put:

H0: p=0.2
H1: p>0.2

and you reject H0 if P(X>(or equal to)4) <(or equal to) 0.05

which gives you 1 - P(X<(or equal to)3) <(or equal to) 0.05

end result was something like 0.12... <(or equal to) 0.05 which is not true, so there is insufficient evidence to reject H0, meaning we cannot conclude whether the teacher's claim was right or wrong.

Did anyone else put that?
Your testing to see if the teachers claim is right or wrong, if you accept H0 then what he claims is right, if you reject H0 and accept H1 you are rejecting what he said.

However, "saying their is no significant evidence to reject H0" would be considered right

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