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So far, I have learnt how to carry out a hypothesis test using the binomial expansion in the following way:
A company believes that 80% of the lightbulbs it makes are not faulty. They test a sample of 15, and find that six are faulty. Can their initial claim be justified?
Hence, X~B (15, p) and H0: p = 0.8, where p is P (Not faulty). H1: p < 0.8.
Or something along those lines. Anyway, then you use cumulative binomial tables to figure out whether H0 can be accepted or rejected at a given percentage tail. That's all well and good, but how would one go about doing this backwards, as it were? Id est, if one began by carrying out that test and finding that p = 0.6, how would one find out the range of H0s that one could reasonably fob off on people (without changing the result of the test, of course)? It Is Me Here t / c 09:09, 7 June 2009 (UTC)
What the original poster asks for is a confidence interval for p. Certainly there are reasonable approximate confidence intervals based on the normal approximation to the binomial distribution. I'm not sure if we have an account of exact confidence intervals for this situation somewhere on Wikipedia. More later.... Michael Hardy ( talk) 03:34, 8 June 2009 (UTC)
Following my question on June 6, there's a bigger problem I need help with. I'm trying to find a space that contains two non-empty closed subsets A and B such that they are disjoint and A is disconnected, and their union is connected. I have a possible model:
Let C1 be the circle with radius 1 centered at the origin, C2 be the circle of radius 1 centered at x=1.5 y=0 and C3 be the circle with radius 1 centered at x=3 y=0.
Let X=union of these circles, and its topology be generated by the 3 circles. Let A=complement of C2, B=complement of ((C1)U(C3))
Clearly they are disjoint, and closed because they are complements of open sets, and also A is disconnected because it can be dissected by C1 and C3. But is AUB connected? It seems impossible to check with all the possible open sets... I really need an answer... And please say it's correct, I've spent 2 days working on this problem now... Standard Oil ( talk) 09:55, 7 June 2009 (UTC)
But how can you get A and B from just performing union and intersection on the circles? Maybe I just don't see it... Can you write the process to attain A and B? —Preceding unsigned comment added by Standard Oil ( talk • contribs) 10:24, 7 June 2009 (UTC)
Ohhhhh I see what you mean... There's a misunderstanding between us here, my mistake. Your talking about connected space. I should have said connected set, meaning a set A such that you can't find two open disjoint sets V1 V2 in the embedding space that dissects A (V1 intersection A and V2 intersection A both non-empty, and A is a subset of (V1)U(V2)). What I'm asking here is, is it possible to find two disjoint open sets in X that dissects AUB, so I'm not asking whether AUB is a connected space in its own right, but rather is AUB a connected set in X. Standard Oil ( talk) 10:39, 7 June 2009 (UTC)
Consider X':=X U {*}, with * a clopen point, and A':=AU{*} and B. Is it what you want?
Thanks for your help, maybe I'm just too confusing (a beginner can't be too lucid). Let's say that for a set A, you can find two open disjoint sets V1,V2 such that V1 intersection A and V2 intersection A are both not empty, and furthermore A is a subset of (V1)U(V2). There are 4 conditions here, and we'll just call A dissectable. So, what I want is two non-empty disjoint closed sets A and B satisfying: A is dissectable, AUB is NOT dissectable. Standard Oil ( talk) 12:20, 7 June 2009 (UTC)
I'm having difficulty trying to show that the unit square, quotiented out by the relation that identifies with , is homeomorphic to S2. I've had no luck simply trying to stretch the space around and gluing the relevant points together (I can't find a way to carry this out in a finite number of steps, there are always points left unglued) - the closest I came was to put a half twist in, then join one pair of opposing edges together (like a Moebius strip), but that just gets me into a mess. I've noticed the square (with identifications) has half-turn symmetry, but I can't see how to use that fact. It seems at first that any line drawn on the square, passing through the centre, is a great circle on the sphere (since the ends are identified), which seems like a good avenue to head down, but then you have to perform subsequent gluings to pairs of points on the great circle, which gets into a bit of a mess.
I've tried thinking about it purely in terms of open sets of each space and the like, but I'm never really sure how to do that generally. Any help would be much appreciated, thanks! Icthyos ( talk) 12:48, 7 June 2009 (UTC)
I'm not following this so far (maybe I will after I've thought it through....). If all this is correct, there should be a two-to-one continuous mapping from the torus to the sphere that's locally one-to-one everywhere and whose local inverse is continuous. As if you had a torus-shaped map of the earth on which every location appears twice. Can you specify that mapping? Michael Hardy ( talk) 20:53, 8 June 2009 (UTC)
Thank you, Gandalf.
Sometimes a page that seems not accessible by google books becomes accessible by cleverly altering the search terms. But you might have to think of some words that occur together only on that one page. Any ideas along those lines? Michael Hardy ( talk) 05:00, 9 June 2009 (UTC)
I'd forgotten this thing I should have remembered: doubly periodic holomorphic functions from C to C ∪ {∞} assume every value twice. That means that is a double covering of the sphere by the torus, and it's conformal. Michael Hardy ( talk) 02:10, 10 June 2009 (UTC)
Could it turn out to be the case that no axiomatic system of mathematics is consistent? If that happens, would Mathematics as a discipline then turn into a branch of Physics, or what? 94.27.225.206 ( talk) 15:57, 7 June 2009 (UTC)
We also have proofs that Peano arithmetic is consistent (Gödel's incompleteness theorem only proves that systems like PA can't prove their own consistency, not that their consistency can't be proven from "outside"). The first one, Gentzen's consistency proof, uses an induction principle on trees that turns out to be equivalent to PA but is still "intuitive"; another, in Gödel's Dialectica interpretation, uses higher-order functions, subject to the same problem.
Presberger arithmetic and Peano arithmetic both contain a "successor axiom" that says that every integer has a successor, i.e. there are infinitely many integers. There are actually a few mathematicians (see ultrafinitism) who reject that axiom and say there is a largest integer, which means all proofs relying on the successor axiom are wrong. Doron Zeilberger has written some interesting articles about this. Zeilberger is of the view that we are approaching an era when mathematics indeed is becoming like physics, where we have a lot of observations that we can experimentally seem to confirm, but which are beyond our capability to prove. Gregory Chaitin says similar things but from a different philosophical standpoint. 67.122.209.126 ( talk) 03:12, 8 June 2009 (UTC)
For every natural number k there is a natural number G such that if each member of a set C is a set of size at most k and any two members of C have a non-empty intersection, then there exists a set A of size at most G such that the intersection of A and any two members of C is still non-empty.
This statement is not hard to prove. What I'd like to know is whether it is known, and where it appears in literature. I'd be interested about anything similar too.
The draft of the proof is this. Given C, construct a set Ap by induction on p such that for every element X of C either X and Ap share at least p elements or X intersects every member of C even in Ap. Once you have Ap, choose one element X of C for each possible intersection of X with Ap, and then choose Ap+1 as the union of these sets. The size of these Ap can be bounded by a number depending only on k and p.
This also proves a slightly stronger variant of the above statement, namely that you can have X and Y and A intersect for any member X of C and any set Y that intersects all members of C.
– b_jonas 20:13, 7 June 2009 (UTC)
Mathematics desk | ||
---|---|---|
< June 6 | << May | June | Jul >> | June 8 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
So far, I have learnt how to carry out a hypothesis test using the binomial expansion in the following way:
A company believes that 80% of the lightbulbs it makes are not faulty. They test a sample of 15, and find that six are faulty. Can their initial claim be justified?
Hence, X~B (15, p) and H0: p = 0.8, where p is P (Not faulty). H1: p < 0.8.
Or something along those lines. Anyway, then you use cumulative binomial tables to figure out whether H0 can be accepted or rejected at a given percentage tail. That's all well and good, but how would one go about doing this backwards, as it were? Id est, if one began by carrying out that test and finding that p = 0.6, how would one find out the range of H0s that one could reasonably fob off on people (without changing the result of the test, of course)? It Is Me Here t / c 09:09, 7 June 2009 (UTC)
What the original poster asks for is a confidence interval for p. Certainly there are reasonable approximate confidence intervals based on the normal approximation to the binomial distribution. I'm not sure if we have an account of exact confidence intervals for this situation somewhere on Wikipedia. More later.... Michael Hardy ( talk) 03:34, 8 June 2009 (UTC)
Following my question on June 6, there's a bigger problem I need help with. I'm trying to find a space that contains two non-empty closed subsets A and B such that they are disjoint and A is disconnected, and their union is connected. I have a possible model:
Let C1 be the circle with radius 1 centered at the origin, C2 be the circle of radius 1 centered at x=1.5 y=0 and C3 be the circle with radius 1 centered at x=3 y=0.
Let X=union of these circles, and its topology be generated by the 3 circles. Let A=complement of C2, B=complement of ((C1)U(C3))
Clearly they are disjoint, and closed because they are complements of open sets, and also A is disconnected because it can be dissected by C1 and C3. But is AUB connected? It seems impossible to check with all the possible open sets... I really need an answer... And please say it's correct, I've spent 2 days working on this problem now... Standard Oil ( talk) 09:55, 7 June 2009 (UTC)
But how can you get A and B from just performing union and intersection on the circles? Maybe I just don't see it... Can you write the process to attain A and B? —Preceding unsigned comment added by Standard Oil ( talk • contribs) 10:24, 7 June 2009 (UTC)
Ohhhhh I see what you mean... There's a misunderstanding between us here, my mistake. Your talking about connected space. I should have said connected set, meaning a set A such that you can't find two open disjoint sets V1 V2 in the embedding space that dissects A (V1 intersection A and V2 intersection A both non-empty, and A is a subset of (V1)U(V2)). What I'm asking here is, is it possible to find two disjoint open sets in X that dissects AUB, so I'm not asking whether AUB is a connected space in its own right, but rather is AUB a connected set in X. Standard Oil ( talk) 10:39, 7 June 2009 (UTC)
Consider X':=X U {*}, with * a clopen point, and A':=AU{*} and B. Is it what you want?
Thanks for your help, maybe I'm just too confusing (a beginner can't be too lucid). Let's say that for a set A, you can find two open disjoint sets V1,V2 such that V1 intersection A and V2 intersection A are both not empty, and furthermore A is a subset of (V1)U(V2). There are 4 conditions here, and we'll just call A dissectable. So, what I want is two non-empty disjoint closed sets A and B satisfying: A is dissectable, AUB is NOT dissectable. Standard Oil ( talk) 12:20, 7 June 2009 (UTC)
I'm having difficulty trying to show that the unit square, quotiented out by the relation that identifies with , is homeomorphic to S2. I've had no luck simply trying to stretch the space around and gluing the relevant points together (I can't find a way to carry this out in a finite number of steps, there are always points left unglued) - the closest I came was to put a half twist in, then join one pair of opposing edges together (like a Moebius strip), but that just gets me into a mess. I've noticed the square (with identifications) has half-turn symmetry, but I can't see how to use that fact. It seems at first that any line drawn on the square, passing through the centre, is a great circle on the sphere (since the ends are identified), which seems like a good avenue to head down, but then you have to perform subsequent gluings to pairs of points on the great circle, which gets into a bit of a mess.
I've tried thinking about it purely in terms of open sets of each space and the like, but I'm never really sure how to do that generally. Any help would be much appreciated, thanks! Icthyos ( talk) 12:48, 7 June 2009 (UTC)
I'm not following this so far (maybe I will after I've thought it through....). If all this is correct, there should be a two-to-one continuous mapping from the torus to the sphere that's locally one-to-one everywhere and whose local inverse is continuous. As if you had a torus-shaped map of the earth on which every location appears twice. Can you specify that mapping? Michael Hardy ( talk) 20:53, 8 June 2009 (UTC)
Thank you, Gandalf.
Sometimes a page that seems not accessible by google books becomes accessible by cleverly altering the search terms. But you might have to think of some words that occur together only on that one page. Any ideas along those lines? Michael Hardy ( talk) 05:00, 9 June 2009 (UTC)
I'd forgotten this thing I should have remembered: doubly periodic holomorphic functions from C to C ∪ {∞} assume every value twice. That means that is a double covering of the sphere by the torus, and it's conformal. Michael Hardy ( talk) 02:10, 10 June 2009 (UTC)
Could it turn out to be the case that no axiomatic system of mathematics is consistent? If that happens, would Mathematics as a discipline then turn into a branch of Physics, or what? 94.27.225.206 ( talk) 15:57, 7 June 2009 (UTC)
We also have proofs that Peano arithmetic is consistent (Gödel's incompleteness theorem only proves that systems like PA can't prove their own consistency, not that their consistency can't be proven from "outside"). The first one, Gentzen's consistency proof, uses an induction principle on trees that turns out to be equivalent to PA but is still "intuitive"; another, in Gödel's Dialectica interpretation, uses higher-order functions, subject to the same problem.
Presberger arithmetic and Peano arithmetic both contain a "successor axiom" that says that every integer has a successor, i.e. there are infinitely many integers. There are actually a few mathematicians (see ultrafinitism) who reject that axiom and say there is a largest integer, which means all proofs relying on the successor axiom are wrong. Doron Zeilberger has written some interesting articles about this. Zeilberger is of the view that we are approaching an era when mathematics indeed is becoming like physics, where we have a lot of observations that we can experimentally seem to confirm, but which are beyond our capability to prove. Gregory Chaitin says similar things but from a different philosophical standpoint. 67.122.209.126 ( talk) 03:12, 8 June 2009 (UTC)
For every natural number k there is a natural number G such that if each member of a set C is a set of size at most k and any two members of C have a non-empty intersection, then there exists a set A of size at most G such that the intersection of A and any two members of C is still non-empty.
This statement is not hard to prove. What I'd like to know is whether it is known, and where it appears in literature. I'd be interested about anything similar too.
The draft of the proof is this. Given C, construct a set Ap by induction on p such that for every element X of C either X and Ap share at least p elements or X intersects every member of C even in Ap. Once you have Ap, choose one element X of C for each possible intersection of X with Ap, and then choose Ap+1 as the union of these sets. The size of these Ap can be bounded by a number depending only on k and p.
This also proves a slightly stronger variant of the above statement, namely that you can have X and Y and A intersect for any member X of C and any set Y that intersects all members of C.
– b_jonas 20:13, 7 June 2009 (UTC)