How many ii until...

[llamasarefuzzy]

---

Members don't see this ad.

How many ii do you think it takes to feel fairly comfortable that you'll get in somewhere provided you are a relatively socially capable person?

 

[Goro]

There's no magic formula. If you're a bad interviewee, you're not getting accepted.

So basically, as many IIs as you can get until you land the actual acceptance.

How many ii do you think it takes to feel fairly comfortable that you'll get in somewhere provided you are a relatively socially capable person?

 

[llamasarefuzzy]

There's no magic formula. If you're a bad interviewee, you're not getting accepted.

So basically, as many IIs as you can get until you land the actual acceptance.

I figured lol. Just can't help myself from trying to predict the future. Apologies to all for my neuroticism

 

[baxt1412]

yeah too hard to tell. especially because some schools interview more than others even if they have the same number of spots. so it's hard to really try to estimate any given probability

 

[moop]

(nvm)

 

[demystifie]

Well if you want a specific number, I would say 4. My reasoning is that since most schools have about 25% acceptance after an interview, you have a decent chance of getting into at least 1 out of the 4 (given you have basic social skills).

 

[FictiousForce]

Go to US news, get the number of interviewees, and the number of acceptances.

Let P_i be the ratio of acceptance to number of interviewees for each school.

The probability that you will get accepted somewhere is equal to

A = 1 - (1-P_1) x (1-P_2) x (1-P_3) x ...(1-P_i)

So if you had 3 IIs, where the fraction accepted of interviewees are 0.25, 0.3, and 0.5, the probability that you will get accepted SOMEWHERE is 1-(1-.25)(1-.3)(1-.5) = 0.7375 ~ 73% chance.

Well if you want a specific number, I would say 4. My reasoning is that since most schools have about 25% acceptance after an interview, you have a decent chance of getting into at least 1 out of the 4 (given you have basic social skills).

Four schools at 25% each actually gives a 57% chance of getting in at least one place. So, yeah, pretty good chance, but still just slightly better than a coin flip.
EDIT: nevermind, it's 68%

 

[pluto101]

Four schools at 25% each actually gives a 57% chance of getting in at least one place. So, yeah, pretty good chance, but still just slightly better than a coin flip.

The chance of not getting into any of four schools, if each school accepts 25% of interviewees, would be .75^4, or 32%. So you'd have a 68% chance of getting into at least one school, not 57%.

Of course, that assumes your chances of getting into each school are independent of one another. Being a good or poor interviewer would drastically change those numbers.

There are no guarantees, get as many interviews as lets you sleep at night. Personally, I think that'd be around 3, but I think I'm a good interviewer.

 

[llamasarefuzzy]

Loving the stats guys. October 15 made me a bit crazy... Went out and bought the us news thing (seriously, why couldn't the msar have that info). Did the calculations which came up at at 96.45% chance that I'll get in somewhere, given no more interviews. Feeling much less anxious☺

 

[llamasarefuzzy]

Edit: 95% Wald Confidence interval of .76-1.16. However, this assumes a binomial distribution with no bounds. Currently scouring my stats textbook to remember the proper test for bound binomial data. Finally came in handy, those statistics

 

[moop]

People calm down damn

 

[llamasarefuzzy]

People calm down damn

Physically impossible at this point lol

 

[llamasarefuzzy]

Edit: 95% Wald Confidence interval of .76-1.16. However, this assumes a binomial distribution with no bounds. Currently scouring my stats textbook to remember the proper test for bound binomial data. Finally came in handy, those statistics

I think it involves a log transformation because log has ian outpu of (0-1) and input of (-infinity-+infinity) but it could also be an exp transformation and I guess I don't care enough to find the conidence interval. Point estimate will have to do it for me. Besides, once I find the 95%CI, I'll also have to do a odds ratio, relative risk, etc, and I just don't have the time for that. Ain't no one have time for that.😱

 

[llamasarefuzzy]

Omg what has grad school done to my brain???

 

[moop]

I think it involves a log transformation because log has ian outpu of (0-1) and input of (-infinity-+infinity) but it could also be an exp transformation and I guess I don't care enough to find the conidence interval. Point estimate will have to do it for me. Besides, once I find the 95%CI, I'll also have to do a odds ratio, relative risk, etc, and I just don't have the time for that. Ain't no one have time for that.😱

Odds ratio...you went there..

As a quant junkie (see sig), I approve. As an SDNer, I disapprove and snub you at your attempts at quantitative rigor.

 

[Make Or Break]

I think it involves a log transformation because log has ian outpu of (0-1) and input of (-infinity-+infinity) but it could also be an exp transformation and I guess I don't care enough to find the conidence interval. Point estimate will have to do it for me. Besides, once I find the 95%CI, I'll also have to do a odds ratio, relative risk, etc, and I just don't have the time for that. Ain't no one have time for that.😱

I don't even....