ATAR/UMAT for a place in UNSW Medicine/Arts?

[ardria]

Does anyone have an idea of what I'd need to score for a safe shot at a place? Would it be roughly the same or higher than Monash Medicine?

Just to be clear, I'm referring to the double degree for Medicine (MD aka MBBS) and Arts. I'm doing VCE in Melbourne this year. http://www.handbook.unsw.edu.au/undergraduate/programs/2015/3855.html

Thanks  !

[A1P]

The handbook says This program is only open to students who have been accepted for entry into UNSW Medicine, which means the entrance score is the same as for Medicine itself. Score required for Unsw (both NSW & interstate students) is quite higher than for Monash.

It's hard to call what is a safe score, know of a guy with 99.x + GPA 7 + 100%ile Umat who didn't get in (he got in two other unis though). I'd say low 99s + low 90s %ile gives a 50/50 shot, high 99s + high 90s %ile about 80% chance i.e. if your interview score is in the bottom 20% of the interview cohort you still miss out.

[RuiAce]

Essentially (so just building on) the important thing to realise that both the ATAR and UMAT contribute to your placement, as well as the interview.

It is unlikely that UNSW should disclose the required balance between the three criteria that would be sufficient to label you 'safe'.

[A1P]

Essentially (so just building on) the important thing to realise that both the ATAR and UMAT contribute to your placement, as well as the interview.

What we know is, after using z-score on the interview cohort (to standardise the scales between ATAR UMAT Interview) UNSW adds each applicant's three z-scores together to give a final ranking score, and the top 1/3rd are successful for a place.

Since you are studying Advanced Maths, can you help figure out a way to estimate the required interview score (relative to the interview cohort) to be in the top third overall, for certain ATAR/UMAT combos like top in one bottom in the other, or median in both, or one top + one median etc? Thanks

[RuiAce]

What we know is, after using z-score on the interview cohort (to standardise the scales between ATAR UMAT Interview) UNSW adds each applicant's three z-scores together to give a final ranking score, and the top 1/3rd are successful for a place.

Since you are studying Advanced Maths, can you help figure out a way to estimate the required interview score (relative to the interview cohort) to be in the top third overall, for certain ATAR/UMAT combos like top in one bottom in the other, or median in both, or one top + one median etc? Thanks

It's probably within my abilities but of course to infer results we're going to need actual data.

Under realistic assumptions (i.e. no excessive outliers) z-scores take between -3 and 3. The sum of three z-scores is therefore between -9 and 9. If it's true that just the sum of z-scores are used, then an applicant who has achieved a cumulative z-score of between 3 and 9 will be accepted in.
__________________________

Given a set of data, the z-score of an observation x is

where mu denotes the mean, and sigma denotes the standard deviation.

We may, of course, use this formula to our advantage with the statement given above. We should have the following data obtained to deduce the required performance in the interview
- Mean and standard deviation of each of the following. (Variance can be used to replace standard deviation.)
  - ATARs of students applying to take medicine
  - UMAT scores
  - Interview scores

Then, given an observed value for ATARs and UMAT scores, i.e. the actual score the candidate received, we can deduce the interview score necessary.

Edit: This model is quite limited in what it has to offer. I will address some limitations after I complete a course review I'm writing up.

[RuiAce]

I'm scrapping some bits of the model above. The model has a slight flaw in that the condition 3 <= z <= 9 will require z to follow a uniform distribution. This is most likely not the case.

Note that the required values are still as listed above. It is what goes on behind the scene that has to be adjusted.

This would be a standard application of the central limit theorem (as I believe we are safe to assume that there's always going to be more than 50 candidates applying to take medicine). I will first make assumptions based off the CLT of the distribution of the sample sum or sample mean of scores (my brain's not working properly today so I'll figure out which exactly later). Then, I can use properties of the normal distribution to find the distribution of 3 z-scores added together.

From there, everything will be just be first year computations for me. Basically I'll just look at quantiles.

(Intentionally split-posted for now. May revert it later)

[A1P]

Under realistic assumptions (i.e. no excessive outliers) z-scores take between -3 and 3. The sum of three z-scores is therefore between -9 and 9. If it's true that just the sum of z-scores are used, then an applicant who has achieved a cumulative z-score of between 3 and 9 will be accepted in.

The interviewees have been selected based on their combination of ATAR+UMAT, ranging from 96.0 to 99.95 and say 80%ile to 100%ile. It's commonly known if lowest in one they need a top in the other to get into the interview round. So while there may be ones with ATAR+UMAT= +6 I think the lowest is zero not -6, true?

It then follows the final score possible range is -3 to +9, would this change the top third line to be not +3?

[RuiAce]

The interviewees have been selected based on their combination of ATAR+UMAT, ranging from 96.0 to 99.95 and say 80%ile to 100%ile. It's commonly known if lowest in one they need a top in the other to get into the interview round. So while there may be ones with ATAR+UMAT= +6 I think the lowest is zero not -6, true?

It then follows the final score possible range is -3 to +9, would this change the top third line to be not +3?

That second post I made is intended to essentially scrap that assumption.