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1. Ive just done the OCR june 2001 paper, but realised i dont have the answers! can anyone help!
2. yeh what questions?

3. (Original post by confused?)
yeh what questions?

i dont have answers to any of the questions at all.

could you just post up the final kind of answers for each question if you dont mind. I'll check them through and post back if i cant get any.
Thank you.
4. 1. 5
2.(i) yes
(ii) List all people
assign number
use random number
3. 0.887
4.(ii) R>=5
(iii)0.630
5.(i) doesnt matter if rope too strong
(ii)Z=-1.77 reject Ho.
(iii)CLT
6.(ii) 1/8
(iii) 2.75
(iv) 2.79
7.(ii)0.554
(iii)0.184
(v) 1.05

hope that helps
5. yh they did, still got some wrong though.
my teachers going to mail me the markscheme through next week, so ill check the working out twith them to see where i went wrong.
thank you!
6. Hey.... can anyone tell me how to do Question 4 on that paper (Jan 2001)?

The hypothesis question... I don't know what the question is asking for... Binomial hypothesis? Normal? Poission?

Thanks.
7. Well these are the answers anyway....

Statistics 2 – June 2001 – Answers

1 5.

2 (i) Distribution of R is B(60, 0.05); since n > 50 and np < 5, a Poisson
approximation (R ~ Po(3)) would be appropriate.
(ii) Assign a number to each pupil in the school and use random number tables to
choose the sample of size 60.

3 0.887.

4 (i) ;
(ii) r  5; if r = 4, accept the null hypothesis – there is no evidence that
employee is late on more than 1 day in 10;
(iii) 0.630.

5 (i) Only a breaking strength less than that claimed is a cause for concern;
(ii) Reject null hypothesis (acceptance region is > 12953) – the company’s
claim that the breaking strength is 13000 N is not justified.
(iii) Central Limit Theorem needed because breaking strengths are not given to
be normally distributed.

6 (i) 3;
(ii) 0.125;
(iii) 2.75;
(iv) 2.79.

7 (i) The typing errors occur independently of each other;
(ii) 0.554;
(iii) 0.184;
(iv) 0.0923;
(v) 1.05 minutes.
8. (Original post by Saiyan)
Hey.... can anyone tell me how to do Question 4 on that paper (Jan 2001)?

The hypothesis question... I don't know what the question is asking for... Binomial hypothesis? Normal? Poission?

Thanks.
i didnt get the answers correct for question 4 calculations.
but i reckon it will be poisson distribution for the hyposthesis, because he says he's late no more than 1 in 10 days, λ=1, therefore in 20 days, λ=2!

so for the hypothesis
H0 : λ=2
H1 : λ>2
9. (Original post by Tazzie)
i didnt get the answers correct for question 4 calculations.
but i reckon it will be poisson distribution for the hyposthesis, because he says he's late no more than 1 in 10 days, λ=1, therefore in 20 days, λ=2!

so for the hypothesis
H0 : λ=2
H1 : λ>2

haha... I've done it now..

it was
Ho: Mu = 1/10
H1: Mu > 1/10

los 5%

onetail test
Under Ho
X~B(20,1/10)

P(X<=3) = 0.8670
<---- 0.9
P(X<=4) = 0.9568

therefore,

P(X>3) = 0.133
<----0.1
P(X>4) = 0.0432

Decision,
Reject Ho if X>4 (X=>5)

r days = 4
not in rejection zone so... no evidence
etc etc (i think it was that binomial proportion thingy topic)

hmmm... that poisson way looks kind of dodgy? lol. Don't think it would work like that. I think you've thought of lamba value wrong.

Bye....
10. saiyan is correct
11. (Original post by Saiyan)
haha... I've done it now..

it was
Ho: Mu = 1/10
H1: Mu > 1/10

los 5%

onetail test
Under Ho
X~B(20,1/10)

....
see i thought it was that too! i wrtoe it up, and then deleted the post!
imconfused!
can u guys please clarify why its one and the not the other!
12. (Original post by Tazzie)
see i thought it was that too! i wrtoe it up, and then deleted the post!
imconfused!
can u guys please clarify why its one and the not the other!
hehe...I have the same problem... but in this case they didn't give us one average CONSTANT value for being away (poisson)... and ho and h1 were in fractions... so I guessed it would be binomial.... proportion testing..
13. (Original post by Saiyan)
hehe...I have the same problem... but in this case they didn't give us one average CONSTANT value for being away (poisson)... and ho and h1 were in fractions... so I guessed it would be binomial.... proportion testing..
and this binomial proportioney thing. i can work it in s3, but dont remember doing it in s2. would you just look up the tables to do it?
14. (Original post by Tazzie)
and this binomial proportioney thing. i can work it in s3, but dont remember doing it in s2. would you just look up the tables to do it?
oh wait sorry... I just realized that was just a normal binomial hypothesis question!!..

Hypothesis test of proportion is different.. here is an example question of it.

ho: p=1/4
h1: p>1/4

one tail test
los 5%

Carried out 100 times leading to 32 successes
under ho

X~B(100,0.25)

using normal approx
X~N(25,18.75)

to find C so that P(X>C-½) remember? you have to minus ½... can't remember why but i just know you have to

standardising

P(Z> (c-½-25)/√18.75) = 0.05

P(Z>1.645) = 0.05
Hence

(C-½-25)/(√18.75) = 1.645
therefore c = 32.6

decision

X>=33 Reject Ho
X<33 Accept Ho

Observed value x = 32
32<33 therefore accept Ho
No evidence that p is less than 1/4

15. oh right!
yh, thats just approximating the normal to the binomial distribution right?

and you add ½, due to the continutiy correction (discrete to continuous).

Thats a bit of sly question. what did they ask you to find?

Thanx.
16. that example, is it from an exam paper? if so, do you know which one.
17. (Original post by Tazzie)
that example, is it from an exam paper? if so, do you know which one.
its not from a past paper lol... it was an example from my teacher...

If I come across one in the pastpapers I will let you know
18. (Original post by Saiyan)
its not from a past paper lol... it was an example from my teacher...

If I come across one in the pastpapers I will let you know
Thanx!
19. hey any of you had practise on june 2004 yet? I just did it. It was ok. Anyone got answers?
20. nope, havent done the june 04 yet.

but i have a question anyway.

I have X, distributed by poisson, with mean 20.

PART 1: use the exact distribution to find P(X=22)
That's fine, just from the tables.

PART 2: use a suitable appoorximation to find P(X=22)

I need help with the seond part. Iaproximate it the normal, so thats
X-N(20,20)....But then?

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