1. ok one question has asked me to find the product moment correlation coefficient, I have done this, but the next part of the question asks to find Spearmans Rank. However, the question is worth 1 mark, therefore is their a quick method to finding it given ive found the product moment correlation?? As doing all the calculations surely is worth more than 1 mark?
2. Using { = sum of
{(x-200) = 17000
{(x-200)^2 = 86000000
n = 8
(a) Find the mean
(b) Find the Standard Deviation
(c) Find {(x-300)
Thanks!
-
Sang
- Follow
- 0 followers
- 0 badges
- Send a private message to Sang
- Thread Starter
Offline0ReputationRep:- Follow
- 1
- 09-06-2004 21:38
-
- Follow
- 2
- 09-06-2004 21:52
There is unfortunately no shortcut. You have to work out (sum over i) d(i)^2, where d(i) is the ith difference in ranks. The PMCC doesn't involve ranks.
-
Sang
- Follow
- 0 followers
- 0 badges
- Send a private message to Sang
- Thread Starter
Offline0ReputationRep:- Follow
- 3
- 09-06-2004 21:59
(Original post by Jonny W)
There is unfortunately no shortcut. You have to work out (sum over i) d(i)^2, where d(i) is the ith difference in ranks. The PMCC doesn't involve ranks. -
- Follow
- 4
- 09-06-2004 22:24
This question is OK provided you are happy taking constants out of sums. For example, if you know
(sum over i from 1 to 100) [y(i)^3 + 7] = 1000000
then the sum counts the '7' 100 times. So
(sum over i from 1 to 100) y(i)^3 = 1000000 - 700.
(a)
sum(x - 200) = 17000
sum(x) - 200n = 17000
sum(x) = 17000 + 200n = 18600
mean = sum(x) / n = 18600 / 8 = 2325
(b)
sum[(x - 200)^2] = 86000000
sum[x^2 - 400x + 40000] = 86000000
sum(x^2) - 400sum(x) + 40000n = 86000000
sum(x^2)
= 86000000 - 40000n + 400sum(x)
= 86000000 - 40000*8 + 400*18600
= 93120000
variance = (1/n)sum(x^2) - mean^2 = 6234375
standard deviation = sqrt(variance) = 2496.87
(c)
sum(x - 300)
= sum(x) - 300n
= 18600 - 300*8
= 16200 -
Sang
- Follow
- 0 followers
- 0 badges
- Send a private message to Sang
- Thread Starter
Offline0ReputationRep:- Follow
- 5
- 10-06-2004 17:51
ok im having some troubles with part 3, im not totally sure of what you mean
I found the mean by 17000/8 + 200
I found the SD by [root]86000000/8 - (mean-200)^2
but with the last part I dont understand how you relate {(x-300) with {(x-200)
the answer claims {(x-200) - 800 how does it get the 800? Is it only possible by doing Jonny's method? Surely its easier somehow? with relation to the number of marks available.
Any help would be asap would be great as exam is tomorrow morn!
Thanks -
- Follow
- 6
- 10-06-2004 18:53
Let y(i) = x(i) - 200 (ie, subtract 200 from each of the data values).
Then
mean(y) = mean(x) - 200 . . . shifting the data shifts the mean,
sd(y) = sd(x) . . . shifting the data doesn't change the sd.
The given sums imply
mean(y) = (1/8)sum(y) = (1/8)sum(x - 200) = 17000/8 = 2125,
sd(y)
= sqrt[(1/8)sum(y^2) - mean(y)^2]
= sqrt[86000000/8 - (17000/8)^2]
= 2496.87.
So
mean(x) = 2125 + 200 = 2325,
sd(x) = 2496.87.
In the last part we want to relate sum(y) with sum(z), where y(i) = x(i) - 200 and z(i) = x(i) - 300. Each item in the second sum is 100 less than the corresponding item in the first sum. Since there are eight items, it follows that sum(z) = sum(y) - 8*100 = sum(y) - 800. We know that sum(y) = 17000, so sum(z) = 17000 - 800 = 16200. -
Sang
- Follow
- 0 followers
- 0 badges
- Send a private message to Sang
- Thread Starter
Offline0ReputationRep:- Follow
- 7
- 10-06-2004 19:06
ok so if i wanted to find {(x-400)
id notice a difference of 200
=> 200n
=>200*8 = 1600
=> {(x-400) = {(x-200) - 1600
=> 17000 - 1600
= 15400
Correct?
[Cheers Jonny you are a huge help] -
Sang
- Follow
- 0 followers
- 0 badges
- Send a private message to Sang
- Thread Starter
Offline0ReputationRep:- Follow
- 8
- 10-06-2004 19:16
Just furthering this,
now I find {(x-100) difference is = +100
=> 100*8
= 800
But, its + 100
=> 17000+800
= 17800
Yes? I think ive got it now -
- Follow
- 9
- 10-06-2004 19:36
Your last two posts are correct.
- Your exam discussion threads for 1st June - 5th June 2015
- AS and A-Level maths and further maths resources thread
- AS and A-Level maths and further maths resources thread
- Edexcel S1 A Level Maths
- AS Edexcel S1 - (5th June 2015) Exam Thread
- Edexcel Mathematics: S1 6683 7th June 2017 [Exam ...
- AQA A2 Mathematics MS2B Statistics 2 ? Friday 12th June ...
- AQA A2 Mathematics MPC3 Core 3 - Friday 5th June 2015 ...
- Anybody else finding s1 really hard?
- Edexcel Applied ICT Unit 3 June 2013
-
University of St Andrews
-
University of Surrey
-
University of Sheffield
-
Royal Holloway University of London
-
Mathematics with High Performance Computing
University of Plymouth
-
Nottingham Trent University
-
Brunel University London
-
Durham University
-
Mathematics with Physics (Including Placement Year)
University of Essex
-
Secondary Mathematics with Science with QTS (2 years)
University of South Wales
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
- SherlockHolmes
- Notnek
- charco
- Mr M
- Changing Skies
- F1's Finest
- rayquaza17
- RDKGames
- davros
- Gingerbread101
- Kvothe the Arcane
- TeeEff
- The Empire Odyssey
- Protostar
- TheConfusedMedic
- nisha.sri
- claireestelle
- Doonesbury
- furryface12
- Amefish
- harryleavey
- Lemur14
- brainzistheword
- Rexar
- Sonechka
- TheAnxiousSloth
- EstelOfTheEyrie
- CoffeeAndPolitics
- an_atheist
- Labrador99
- EmilySarah00