Official Edexcel S3 thread Wednesday 20 May 2015 Morning [6691]

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Reply 320

username1200599

Original post by nayilgervinho

It just doesn't feel like 5 marks, I think it should be more.

I felt like 5 marks was too much. You can use a known result format for that sort of thing.

Reply 321

DCMed96

Original post by Mutleybm1996

Neither.....

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then how it is done then?
I realise all of this is just rote learning.... never understood any of them lol

Reply 322

STATER

Original post by V0ldemort17

Omg someone please explain this

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Part a) bias = estimator - true value

Hence the bias is 1/2a.

Part b) K= 2/3 since multiplying X bar by this constant = a (unbiased estimator).

Part c) max value means you need to find the mean of the sample and substitute it to find a (using result from part b) and then substitute it into the upper limit of the interval (2a+3).

Hope the explanation is clear enough.
Edit: alpha not a

(edited 8 years ago)

Reply 323

username1200599

Original post by V0ldemort17

ImageUploadedByStudent Room1432062410.714921.jpg

Omg someone please explain this

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Use Y as an unbiased estimator for alpha and substitute into the upper limit of the distribution.

Reply 324

Mutleybm1996

Original post by Iridann

I felt like 5 marks was too much. You can use a known result format for that sort of thing.

What's that?

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Reply 325

Mutleybm1996

Original post by DCMed96

then how it is done then?
I realise all of this is just rote learning.... never understood any of them lol

We don't need to know

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Reply 326

V0ldemort17

ImageUploadedByStudent Room1432063910.172133.jpg

Part d???

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Reply 327

nayilgervinho

Any last minute tips.

Reply 328

Mutleybm1996

Original post by nayilgervinho

Any last minute tips.

Take your time
Do the 11-13 markers first

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Reply 329

Mutleybm1996

Original post by Iridann

I felt like 5 marks was too much. You can use a known result format for that sort of thing.

What's the known result

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Reply 330

Damien_Dalgaard

How do you know whether to use pearsons ranks or spearmans rank the edexcel book is misleading.

do we always have to do spearmans?

Reply 331

nayilgervinho

Original post by V0ldemort17

Part d???

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p(|x-mean|<1) = 0.98
(x-mean)/(standard deviation/square root on n)
therefore: p(z<1/(3/square root on n)) = 0.98
1/(3/square root of n) = 2.3263
square root on n = 3(2.3263)
n=48.7 therefore n =49

Reply 332

nayilgervinho

Original post by Damien_Dalgaard

How do you know whether to use pearsons ranks or spearmans rank the edexcel book is misleading.

do we always have to do spearmans?

Unless it asks otherwise such as with tied ranks where it'll say to use PMCC.

Reply 333

Damien_Dalgaard

Original post by nayilgervinho

Unless it asks otherwise such as with tied ranks where it'll say to use PMCC.

Thank you, pearsons being the long crap from s1 with the sxx etc.

yao gervinho? roma>?

Reply 334

V0ldemort17

How did you get 2.3263 from 0.98

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Reply 335

V0ldemort17

Original post by nayilgervinho

p(|x-mean|<1) = 0.98
(x-mean)/(standard deviation/square root on n)
therefore: p(z<1/(3/square root on n)) = 0.98
1/(3/square root of n) = 2.3263
square root on n = 3(2.3263)
n=48.7 therefore n =49

How did you get 2.3263 from 0.98

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Reply 336

nayilgervinho

Original post by V0ldemort17

How did you get 2.3263 from 0.98

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It's the z value for 1% each way.

Reply 337

Mutleybm1996

Original post by nayilgervinho

p(|x-mean|<1) = 0.98
(x-mean)/(standard deviation/square root on n)
therefore: p(z<1/(3/square root on n)) = 0.98
1/(3/square root of n) = 2.3263
square root on n = 3(2.3263)
n=48.7 therefore n =49