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 Post subject: Age standardisation
PostPosted: Mon Jun 10, 2013 9:04 am 
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Joined: Mon Jun 10, 2013 8:15 am
Posts: 141
Hi folks,
My dd is late September-born, so I was wondering how much more dd has to get in comparison as a rough guesstimate? Can anyone help? :wink:


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 Post subject: Re: Age standardisation
PostPosted: Mon Jun 10, 2013 4:45 pm 
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Joined: Sun Sep 23, 2012 11:51 pm
Posts: 112
I don't know, but my ds is January born, so I am guessing he has to get around 95%.
When I took the 11+ there was no age standardisation, and I am June born, (46 years young yesterday :D) It made no difference to me at all...


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 Post subject: Re: Age standardisation
PostPosted: Mon Jun 10, 2013 7:35 pm 
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Joined: Mon Aug 22, 2011 8:20 pm
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Location: Warwickshire
This example standardisation table is actually from the 2012 KS2 Maths tests - but it will give you some idea of relative allowance made for age when calculating standardised scores.


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 Post subject: Re: Age standardisation
PostPosted: Mon Jun 10, 2013 8:28 pm 
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Joined: Mon Jun 10, 2013 8:15 am
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Thanks guys :lol: !

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 Post subject: Re: Age standardisation
PostPosted: Tue Sep 03, 2013 12:43 pm 
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Silly stats time

(4250/2)/12 ~ 177 dcs of a single gender e.g. Girls taking the exam born in each calendar month e.g. January.

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 Post subject: Re: Age standardisation
PostPosted: Fri Sep 06, 2013 10:18 am 
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viewtopic.php?f=5&t=32258

Ideally we need the NFER local standardized look up table for this cohort, but I doubt that will be Publically available even until the whole cohort has taken the tests, including late sitters :lol:

As I understand it, on the Lhs we have raw scores & on the top the age of the child at the test date ( in completed months). From the final bunch of marks Nfer make up a standardized score for each raw score: age axis combinations. This ensures that dcs are only compared to the same age as themselves & that the same proportion of dcs in each age group obtain the pass mark.

So we will have the raw score going in & the standardized score coming out of the "magic black box" (which has the complicated converting equation inside) :wink: .

Each 11+ year has a unique box.
But, if we wish to get a very rough idea ( of the ballpark figures), we can try & use previous years in & out scores, to very roughly guess a simple & possibly misleading version of the real equation.

If we want to do this, then we need the following information>
Raw scores in, standardized scores out & age group sub- division of child e.g. 10Y:06M.
Even better would be to get a historical Redbridge look up table as well :D . I wonder do Nfer or LBR make these publicly accessible & if not, why not, after all we ultimately pay their wages?

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 Post subject: Re: Age standardisation
PostPosted: Sun Sep 08, 2013 10:08 am 
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Joined: Mon Jun 10, 2013 8:15 am
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From the Essex forum >


This might help.

For more detailed conversion of standardised score back to precentile position try this z-table. http://www.doe.virginia.gov/testing/tes ... _table.pdf

You'd need to take your standardised score, subtract 100, and then divide by15. Then look up the result (z) in the tables. The lookup value is the fraction which will score lower up to and including this.

e.g standardised score 120 gives a z score of (120-100)/15 - 1.333 Look up 1.333 and you get 0.909 (or 90.9%) Thus 9.1% will score higher than this. If you know the number of candidates you can get an approximate rank.

You can't just add 2 or more standardised scores though! (Could explain, don't have time now). In this case your best guess will be the average of your ranks (or percentiles).

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 Post subject: Re: Age standardisation
PostPosted: Sun Sep 08, 2013 5:03 pm 
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Joined: Sat Jan 12, 2013 10:02 pm
Posts: 88
Fat pigeon wrote:
From the Essex forum >


This might help.

For more detailed conversion of standardised score back to precentile position try this z-table. http://www.doe.virginia.gov/testing/tes ... _table.pdf

You'd need to take your standardised score, subtract 100, and then divide by15. Then look up the result (z) in the tables. The lookup value is the fraction which will score lower up to and including this.

e.g standardised score 120 gives a z score of (120-100)/15 - 1.333 Look up 1.333 and you get 0.909 (or 90.9%) Thus 9.1% will score higher than this. If you know the number of candidates you can get an approximate rank.

You can't just add 2 or more standardised scores though! (Could explain, don't have time now). In this case your best guess will be the average of your ranks (or percentiles).


I hated this topic during my statistics modules in A-levels! :P But thank you, I might try this out.


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