Yossarian wrote:

Is zero an invented number? Is zero a number? Is zero? Is?

https://www.youtube.com/watch?v=9Y7gAzTMdMA

For me zero was all about notation - marking an empty space more than anything else. Just drawing a circle around it. An advance on from roman numerals for example.

But I know there's more to it than that. You have to start somewhere and have a point of reference.

davyK wrote:

I was amazed to hear about complex numbers being able to solve real world problems.

Yeah, it's pretty much the cornerstone of quantum mechanics, and nobody knows why. All maths has application I suppose because the universe is mathematical. An electron for example is nothing more than a set of equations and is completely defined as such.

But that's it , isn't it?

How close is pure maths to the actual truth?  

We use it to model reality. It's a description of what we observe given a set of conditions , not how actually something works.

But how far down the Russian doll stack will we ever be able to go?

Is it the only tool we have to punch through our layers of perception?

If it isn't we are blind.

And who knows if that would be a good thing or not?

Numbers are all invented, it's a pure abstraction. Just it's one so old and so versatlie that it's been used to describe almost everything in (and out) the universe

edit: imo and all that, also didn;t see this page

Ignoring the maths, you’re not going to be able to accurately interpret any of your results.

Ok. I'm basically aiming to calculate a confidence interval... I thought there had to be a term for the above.

I'm not 100% sure what the question is, but addressing the easy "what's the probability 12 people get menu A" - it's a simple binomial calculation:

40C12 x 0.3^12 x 0.7^28 = 13.66% - assuming that the distribution of both the menus and the questionnaires was random

If you want confidence intervals you can use the general properties of the binomial distribution. The variance of the binomial distribution is 

N x P x (1-P)

So 40 x 0.3 x 0.7 = 8.4 in this case. 

A 95% probability confidence interval would then broadly be 1.96 x 8.4^0.5 plus or minus the mean, which is something like 6 to 17.

There's all sorts of things you can do with this, but the questions a bit vague so I'm not sure what else you want doing. I'm not going to start getting into censoring and extracting information involving peoples views on either menu because I'd have to actually think about that!

voices wrote:

I'm not 100% sure what the question is, but addressing the easy "what's the probability 12 people get menu A" - it's a simple binomial calculation: 40C12 x 0.3^12 x 0.7^28 = 13.66% - assuming that the distribution of both the menus and the questionnaires was random If you want confidence intervals you can use the general properties of the binomial distribution. The variance of the binomial distribution is  N x P x (1-P) So 40 x 0.3 x 0.7 = 8.4 in this case.  A 95% probability confidence interval would then broadly be 1.96 x 8.4^0.5 plus or minus the mean, which is something like 6 to 17. There's all sorts of things you can do with this, but the questions a bit vague so I'm not sure what else you want doing. I'm not going to start getting into censoring and extracting information involving peoples views on either menu because I'd have to actually think about that!

That's interesting. That first idea is pretty much completely what I'd originally asked.
So I get what you mean by how muddled the question is.

This though:

voices wrote:

...A 95% probability confidence interval would then broadly be 1.96 x 8.4^0.5 plus or minus the mean, which is something like 6 to 17...

Is what I've found since asking the question, and is what I think I'd like to do.

I've been using a tool to calculate it so far, but the formula doesn't look too complicated, so the next step for me, is to break it down, and understand it.

But from what I've learnt so far, if I have 100 people, and feed 30 of them set menu A, and 70 set menu B, if I pick 40 people at random, and 12 of them claim to like the food, then I can be 95% sure that between 18.94 and 41.06(!!?) of the 100 people liked the new menu (A). So even taking the lower number as the estimate on which to base my decision, I can be pretty sure that the new menu was fairly well received.

At least I think that's what I'm looking for.

I'm still open to ideas though.

Cheers for the replies. :]

In regards to the above, the only thing I think was lost in translation (my maths is that bad - :])
is the fact that I meant that the 12 people in that 40 guest sample did in fact both receive and like menu A (the one that I gave to 30 people in the 100). I fucked that bit up.

The tool that I used to give the figures above ("I can be 95% sure that between 18.94 and 41.06(!!?) of the 100 people liked the new menu (A)"), was this one:

https://www.surveysystem.com/sscalc.htm#one

It's a bit ropey, but I'd been using it to get a rough idea of how the concept works.
I'd been using the second of the two calculators ('sample size', population, percentage), and what I basically did was:

Sample Size = 40
Population = 100
Percentage = 30

Doing that gave me a CI of 11.06

I've since learnt that the calculator is based on some instance of the first equation on this page - https://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval - but the result of that equation doesn't take the original population size into account. The result I got for calculating it by hand is a CI of 14.2 (I think).

So the sentence above changes to:
("I can be 95% sure that between 15.8 and 44.2(!!?) of the 100 people liked the new menu (A)")

I mean, it's a tiny population, and a tiny sample... And to try and understand the idea, I've fudged the perfect result of 12 people in 40 both happening to have received and liked menu A (which I suspect will never happen).

It was your description of the problem as a binomial one that's got me to this point so far (if it's anywhere close to correct, lol!... I can't even post about fucking tofu with 'perfect' competence).

This is essentially a Bernoulli trial, right?

https://slate.com/technology/2021/06/golden-ratio-phi-irrational-number-ellenberg-shape.html

interesting article about the golden ratio, or at least interesting for maths thickos like me.

I would definitely say 'fie' not 'fee' despite not being in a fraternity

Here's what i assume is a simple maths odds question, but I can't decide the right answer.

the lottery. let's say you play the same numbers each week.
one week, you get a couple of numbers - forget that currently that would get you an extra lucky dip for the next draw.
does that reduce your chances of getting all 6 numbers the following week with the same numbers? and would that carry on effecting your odds into the future?
I've been playing the lottery basically since it's started and using the same 6 numbers, and never had more than 3 numbers...are my chances of winning getting worse each time some of my numbers do come up? or does it being random mean each draw is a completely fresh chance?

Yeah each draw is independent of what's gone before, so having got stuff one week doesn't make it more or less likely the next.

It's like tossing a coin. You get a head, the next toss is still 50/50.

(The exceptions to this are eg loaded dice, or eg FA cup draws where balls aren't replaced, etc)