OCR S2 (non-mei)

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Yeah, +0.5...

38-50/sq rt(50) = -1.697, and I thought you needed to compare it to -1.645?

Isnt the test supposed to be (X+0.5)-50/rt(50) ? So you'd 38.5-50/rt(50) ?

I'm not 100% sure, you may very well be right :smile:

Reply 81

lllllllllll

Original post by Molly_xox

Yeah, +0.5...

38-50/sq rt(50) = -1.697, and I thought you needed to compare it to -1.645?

I think the continuity correction had to be -0.5 since we were finding P(X<x)

Reply 82

iwishiwasclever

can someone please write out the 2 simultaneous equations they got for question 1, i think i made a very stupid error
thanks!

Reply 83

Younjaxx

Original post by tokopoko

I felt pretty good about all my answers, ill post as many as i can remember.

a=6 then a=1.5 for cont dist qn

Mean = 146, S.D=2 for normal qn

Very last qn was something like:
1/3(0.6 + 0.03) <-- i really cant remember
So P(type 2 error) = 0.23 ish

So P(X at least 1 out of 4) = 1 - P(X=0) = 1- (1-0.23)^4

Yep I got exactly the same as you for all the questions.
7ii I got 37 (critical value was like 38.4 and then continuity correction so its 37.9 and then you round down to 37)

Reply 84

rich1334

Original post by lllllllllll

I think the continuity correction had to be -0.5 since we were finding P(X<x)

I thought we were finding P(X less than or equal to x) ?

Reply 85

tokopoko

Original post by lllllllllll

I think the continuity correction had to be -0.5 since we were finding P(X<x)

ye i thought so too. i.e. P(X<38) would change to P(X<37.5)?

(edited 8 years ago)

Reply 86

Archer61

Can someone give the equation for question 6) I think I looked at the equation wrong because I got a = 1/6 and 6

Reply 87

TOxp68

Original post by Archer61

Can someone give the equation for question 6) I think I looked at the equation wrong because I got a = 1/6 and 6

Think it was something like 1.5a^-3 * x^2

Reply 88

tokopoko

Original post by Younjaxx

Yep I got exactly the same as you for all the questions.
7ii I got 37 (critical value was like 38.4 and then continuity correction so its 37.9 and then you round down to 37)

please explain why you got 37.9, i thought C.Correction was done before not after the test

Reply 89

ms8061

just realised for the first one I did it as sigma^2, when the formula means it'd be sigma, so I got sigma^2 = 2 and thus sigma = sqrt(2)
would I lose a mark for this? Suppose I would do, looking at a past paper with a similar question doesn't seem to allow sigma^2

Reply 90

Younjaxx

Original post by AMDD

What did everyone get as the value for a for each part of the questions? I got a=6 for both....

part two was 1.5 and I know the mistake you made because I made the same mistake, when calculating E(X) it was 1/(2a^3) * ( (1/4)x^4) between a and -a so if you put that in you get 1/(2a^3) 8 ((1/4)a^4 - 1/4a^4) the mistake you made was to put ((1/4)a^4 + ((1/4)a^4) but remember that -a^4 is a^4 so if you integrate it you get E(X) as 0. hope it makes sense

Reply 91

lllllllllll

Original post by rich1334

I thought we were finding P(X less than or equal to x) ?

Original post by tokopoko

ye i thought so too. i.e. P(X<38) would change to P(X<29.5)?

Possible...I don't remember Q7 off the top of my head. All I remember is a continuity correction with -0.5

Reply 92

Younjaxx

Original post by tokopoko

please explain why you got 37.9, i thought C.Correction was done before not after the test

Because you're working backwards. So what I did was say ok so what value of x gives us z of 0.05 and that gave me x as 38.4 or something. And then I was like ok if we're saying that x=38.4 is the critical region, we must've had 37.9 which we then corrected to 38.4 so the actual value of x is 37.9 and the answer is 37. So imagine you wanted to work out the P(x=<38) you would actually work out P(x=<38.5) so if you have P(x=<38.4) the original thing you wanted to find out was P(x=<37.9) makes sense?

Reply 93

iwishiwasclever

were the 2 simultaneous equations for question 1:
150-u=2(sigma)
and
u-146=2(sigma)
thanks

(edited 8 years ago)

Reply 94

Molly_xox

Original post by rich1334

Isnt the test supposed to be (X+0.5)-50/rt(50) ? So you'd 38.5-50/rt(50) ?

I'm not 100% sure, you may very well be right :smile:

I hadn't even thought about that...but then you can't have half a signal failure?

Original post by lllllllllll

I think the continuity correction had to be -0.5 since we were finding P(X<x)

Yeah that might be where I went wrong, I thought it was strange that I was rounding up...

Reply 95

AMDD

Original post by Younjaxx

Thatnkyou so much for replying and i think i followed that but why would it not be (1/2a^3)*((1/4)a^4- - (1/4)a^4) inside the brackets which is where i seem to have gone wrong

Reply 96

tokopoko

Original post by Younjaxx

hmm, ive never seen that before. you may be right but i would never think to do that. hopefully someone else can confirm

Reply 97

alexjackson1711

Original post by thickmaster

I stored all my answers in my calculator and here they are (I think 8ii is wrong):
1. Mean = 146, sigma = 2
2. i) The rabbits occur independent of each other
ii) 0.222 (Using poisson)
3. i) a = 6
ii) a = 1.5
4. 0.8311
5. 0.2536
6. 0.9807
7. i) 0.1247
ii) 38
8. i) 0.0316
ii) 0.961 (I did 1 - (1/3((1-0.9685)+(1-0.6047)+(1-0.0933)))^4)

I think 8ii is wrong as the probability of getting a type II error when p=0.3 was 0 as you would've accepted H0 correctly rather than erroneously so there was no error.

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Reply 98

Archer61

Original post by AMDD

Thatnkyou so much for replying and i think i followed that but why would it not be (1/2a^3)*((1/4)a^4- - (1/4)a^4) inside the brackets which is where i seem to have gone wrong