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Old 10-28-2009, 11:52 AM   #16
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Interesting. Well, one thing I've learned in my classes is that if you set out with a set of data and try to come up with some sort of statistic that fits, you can slap a bunch of logical (or random) coefficients on your prediction formula and get stuff that is kind of legit for some data points and stuff that is extremely, EXTREMELY inaccurate for other data points. While your formula is cute, I'd urge you to use some real-life examples and try to alter the coefficients (or the relationships between the variables) if you're actually interested in creating something that will predict admission or rejection. Otherwise, it is a cute project, but I'm unconvinced of its accuracy.
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Old 10-28-2009, 11:55 AM   #17
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Oh I completely understand that. I just made this up late one night last week and then posted it the next day. I plan on modifying it, but I think that right now, for a first draft, it works out pretty well. I will have to do some more research on the topic to further develop my model.

EDIT: What I could do is ask people who are applying to MIT on this board to test drive the formula so that I can see if there is a correlation between higher scores and admissions. They would just have to provide the ratings on essays/recs and I would do the rest of the work for calculating their scores.
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Old 10-28-2009, 06:36 PM   #18
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Quote:
I originally watched an MIT admissions video and went to a Splash class at MIT on admissions.
My EC actually did explain something similar to me, so yes.

I don't think it's very accurate regardless. If I'm reading this correctly three 5s would mean a chance of 100%, which is obviously unrealistic. Also, fix those parentheses. Is it

e^((ln(%accepted)+(zscore)ln(100/%accepted))/3)

or e^(ln(%accepted)+(zscore)ln(100/%accepted)/3)?
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Old 10-28-2009, 08:02 PM   #19
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e^(ln(%accepted) + ((z-.01)(ln(100/%accepted)))/3)

(See I added that little -.01 to the z so you never get to 100). Dividing by three is only for the last ln part, not for the whole thing, so my original one was correct (I think). The two people who got into MIT who used it reported high 40s for chances. I do agree that I need more data to work with, which I will start looking for after I submit my own app to MIT and after I finish my linear algebra course (ends a few days after Nov 1st).
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Old 10-29-2009, 01:18 AM   #20
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Quote:
and went to a Splash class at MIT on admissions
Were you in my class?!

LOL.
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Old 10-29-2009, 05:11 PM   #21
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Quote:
(See I added that little -.01 to the z so you never get to 100).
99.24% is not exactly a huge improvement. Adding coefficients to the equation doesn't necessarily make it more precise anyway.

Quote:
Dividing by three is only for the last ln part, not for the whole thing, so my original one was correct (I think).
I figured it was that one, but I wanted to make sure. The other one gave numbers that tended to hover around 10-20%, which seemed a little more realistic either way (even if you were a stellar student).

Quote:
The two people who got into MIT who used it reported high 40s for chances.
But isn't it largely subjective anyway? I think this could be improved upon if you had access to detailed admissions statistics from elite colleges as well as an intimate understanding about how the process works.

Quote:
I do agree that I need more data to work with, which I will start looking for after I submit my own app to MIT and after I finish my linear algebra course
I'm pretty much done with my app; I have yet to submit Part 2 yet, which I'll hold on until tomorrow or so. And I'm quite familiar with linear algebra but my school doesn't offer anything more advanced than Calc AB (and even local college courses cost money and time).
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Old 10-29-2009, 05:12 PM   #22
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It was two years ago when I was a sophomore. Not sure if you taught it then. It was an asian guy; his roommate added all these extra things to his powerpoint that showed off certain things about MIT.
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