STTS vs standard distributions
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STTS vs standard distributions
In an attempt to understand how the Bucks STTS works, I have produced a spreadsheet to roughly compare Bucks STTS with a 'standard' standardised score used in other 11+ areas. It also shows (roughly) how a score might translate to a centile rank.
https://docs.google.com/spreadsheets/d/ ... sp=sharing
Please note this is no official information, just my own way of trying to compare the numbers.
I have used these assumptions:
- Bucks average STTS is 100 (known fact, I think)
- Bucks STTS score of 121 qualifies 30% of the candidates sitting the test (rough estimate but historic figures are in that region)
- Bucks distribution of results follows a standard distribution around the mean (reasonable assumption)
- Bucks scores are calculated using the standard formula but a different multiplier (reasonable assumption I think)
Most areas/exam providers seem to age standardise setting the average at 100 and the standard deviation at 15, giving a bell curve distribution of results around the average. Bucks seem to do it similarly but not with '15' being the magic number but something else. I've been trying to figure out what it is, and from my spreadsheet it seems to be something in the region of 36-37.
Age-standardisation geeks like me will already know that you calculate a standardised score amongst any group by using this formula:
S = x(b — a)/sd + 100
where S is the pupil’s standardised score, b is the pupil’s raw score, a is the average raw score of all the pupils, and sd is the standard deviation of the raw scores.
x is the multiplier - and by most exam providers this is set at 15. However, I think in Bucks it must be set somewhere around 36-37 to make the numbers work.
None of this can tell you anything about the underlying raw score btw. It does help (I hope) to make results comparable cross-county.
https://docs.google.com/spreadsheets/d/ ... sp=sharing
Please note this is no official information, just my own way of trying to compare the numbers.
I have used these assumptions:
- Bucks average STTS is 100 (known fact, I think)
- Bucks STTS score of 121 qualifies 30% of the candidates sitting the test (rough estimate but historic figures are in that region)
- Bucks distribution of results follows a standard distribution around the mean (reasonable assumption)
- Bucks scores are calculated using the standard formula but a different multiplier (reasonable assumption I think)
Most areas/exam providers seem to age standardise setting the average at 100 and the standard deviation at 15, giving a bell curve distribution of results around the average. Bucks seem to do it similarly but not with '15' being the magic number but something else. I've been trying to figure out what it is, and from my spreadsheet it seems to be something in the region of 36-37.
Age-standardisation geeks like me will already know that you calculate a standardised score amongst any group by using this formula:
S = x(b — a)/sd + 100
where S is the pupil’s standardised score, b is the pupil’s raw score, a is the average raw score of all the pupils, and sd is the standard deviation of the raw scores.
x is the multiplier - and by most exam providers this is set at 15. However, I think in Bucks it must be set somewhere around 36-37 to make the numbers work.
None of this can tell you anything about the underlying raw score btw. It does help (I hope) to make results comparable cross-county.
Re: STTS vs standard distributions
I think there are far too many assumptions to make this useful.
assume - it can make an 'ass' of 'u' and 'me'
Scores, once qualified, are irrelevant in Bucks.
assume - it can make an 'ass' of 'u' and 'me'
Scores, once qualified, are irrelevant in Bucks.
Re: STTS vs standard distributions
Ha, ha - fair enough.Guest55 wrote: assume - it can make an 'ass' of 'u' and 'me'
Absolutely. Agreed.Guest55 wrote:Scores, once qualified, are irrelevant in Bucks.