Let's discuss Mathematics

[dutchfire]

What field did you graduate in? This is the perfect opportunity to shift the conversation to your own field of study.

 

[ParadigmShifter]

Why does this thread only get bumped on Fridays or my Birthday ?

 

[dutchfire]

What's the chance of such a thing happening....

 

[ParadigmShifter]

I think the probability of a bump occurring is p(bump).

 

[GoodGame]

I think the probability of a bump occurring is p(bump).

But do you expect that probability = # of CFC-OTers/# of people in the universe, or would you use a Poisson process to estimate via the rate of posts (posts per second) that occur on CFC?

 

[GoodGame]

Statistics question:

Hypothetical (sic): I'm testing a hypothesis (either single mean or difference of two means) and I use a confidence interval at some confidence coefficient (e.g. 1 - alpha) to support or reject the null hypothesis, and I end up supporting it by the confidence interval. Is that the same as saying I rejected or failed to reject the null hypothesis at the alpha% level of significance?

I'm thinking the answer is "no" just because the two equations (the confidence interval definition and the test statistic equation) are not identical (even though they are similar).

Am I right?

Apparently the answer is "yes" according to this:

http://davidmlane.com/hyperstat/B15183.html

I wasn't sure, since the equations are different, IIRC.

For posterity:

Confidence Intervals & Hypothesis Testing (1 of 5)
There is an extremely close relationship between confidence intervals and hypothesis testing. When a 95% confidence interval is constructed, all values in the interval are considered plausible values for the parameter being estimated. Values outside the interval are rejected as relatively implausible. If the value of the parameter specified by the null hypothesis is contained in the 95% interval then the null hypothesis cannot be rejected at the 0.05 level. If the value specified by the null hypothesis is not in the interval then the null hypothesis can be rejected at the 0.05 level. If a 99% confidence interval is constructed, then values outside the interval are rejected at the 0.01 level.

 

[ParadigmShifter]

Yep, I'd use a Poisson process (number of arrivals per time interval) for that.

 

[ThinkTank]

One of my favorites:

Two numbers m and n are chosen such that 2 <= m <= n <= 99. Mr.
S(um) is told their sum and Mr. P(roduct) is told their product. The following
dialogue ensues:

i. Mr. P: I don't know the numbers.
ii. Mr. S: I knew you didn't know. I don't know either.
iii. Mr. P: Now I know the numbers.
iv. Mr. S: Now I know them too.

In view of the above dialogue, what are the numbers?

 

[Truronian]

What field did you graduate in? This is the perfect opportunity to shift the conversation to your own field of study.

The modules I did were mainly applied... fluid dynamics, ecology, quantum physics. They were fun but there was bit too much taking-a-difficult-problem-and-simplifying-it-down-into-an-unsolvable-problem.

 

[Mise]

One of my favorites:

Two numbers m and n are chosen such that 2 <= m <= n <= 99. Mr.
S(um) is told their sum and Mr. P(roduct) is told their product. The following
dialogue ensues:

i. Mr. P: I don't know the numbers.
ii. Mr. S: I knew you didn't know. I don't know either.
iii. Mr. P: Now I know the numbers.
iv. Mr. S: Now I know them too.

In view of the above dialogue, what are the numbers?

i) implies that m and n aren't both prime
ii) implies that the sum of m and n implies (i)
iii) implies that m*n and (ii) are sufficient to know m and n
iv) implies that m+n and (ii) are insufficient to know m and n, but m+n, (ii), and (iii) are sufficient to know m and n.

If I had the wherewithal to do it, I could make a program, database or excel spreadsheet that could figure it out. I suspect, however, that such a dialogue could not happen with human beings, but only with a computer system designed to have such a dialogue

(Obviously I'm not smart enough to work it out analytically )


EDIT: No, I'm missing something... those statements imply something that I haven't thought of yet.

 

[ParadigmShifter]

Well the product must be the product of 3 or more prime numbers. Hmm, I was gonna say that the factors would be 3 or more from 2, 3, 5, 7, 11, 13 (since 6*17 > 99) but I see that the individual numbers, not the product are <= 99, so there are loads of possibilities.

 

[ParadigmShifter]

The numbers have to be 5 and 6.

So the sum is 11 and the product is 30.

30 is a product of 3 primes, and numbers 2<=x<=99 that sum to 11 are {2,9}, {3,8}, {4,7} and {5,6}.

Dots denote multiplication here:

2.9 = 18 = 2.3.3
3.8 = 24 = 2.2.2.3
4.7 = 28 = 2.2.7
5.6 = 30 = 2.3.5

And all those possibilities are products of 3 or more primes.

Therefore, for the question to make sense, this has to be the only solution.

It's now a tedious process to verify that all other possibilities have exactly one additive representation that is a product of 2 primes only. I'll leave that as an exercise

 

[dutchfire]

S is not the sum of any two primes. S is not 198.

X-post

 

[Mise]

I don't understand, why can't it be 15 and 2, rather than 5 and 6? 17 is not the sum of any two primes either (just like 11), so S would be satisfied knowing that P can't know it. But P wouldn't know whether it's 15 and 2 or 5 and 6, even knowing that S knew P didn't know.

 

[ParadigmShifter]

I say the question is wrong!

2, 15 case:

S = 17, P = 30

Possibles: {15, 2}, {14, 3}, {13, 4}, {12, 5}, {11, 6}, {10, 7}, {9, 8}

15.2 = 30 = 2.3.5
14.3 = 42 = 2.3.7
13.4 = 52 = 2.2.13
12.5 = 60 = 2.2.3.5
11.6 = 66 = 2.3.11
10.7 = 70 = 2.5.7
9.8 = 72 = 2.2.2.3.3

And those possibilities would mean Mr. S would not know either.

 

[Mise]

Oh, it must simply mean that P can't possibly have been told the number 30.

So the next step is to exclude all possibilities like 30, where there are two sets of numbers that multiply to give P's number and sum to a number that is not also the sum of two primes.

I will add this to my ever expanding spreadsheet....

 

[ParadigmShifter]

I looked it up on the internet and have confused myself now

 

[Truronian]

The numbers have to be 5 and 6.

So the sum is 11 and the product is 30.

30 is a product of 3 primes, and numbers 2<=x<=99 that sum to 11 are {2,9}, {3,8}, {4,7} and {5,6}.

Dots denote multiplication here:

2.9 = 18 = 2.3.3
3.8 = 24 = 2.2.2.3
4.7 = 28 = 2.2.7
5.6 = 30 = 2.3.5

And all those possibilities are products of 3 or more primes.

Therefore, for the question to make sense, this has to be the only solution.

It's now a tedious process to verify that all other possibilities have exactly one additive representation that is a product of 2 primes only. I'll leave that as an exercise

P = 30
S = 11

P considers possibilities: {2,15},{3,10},{5,6}
S considers possibilities: {2,9},{3,8},{4,7},{5,6}

i)P doesn't know.

ii)S knew P wouldn't know as all of S's possibilities have three plus factors, as you've shown. In addition, S doesn't know.

iii)This is where I lose it... how does P discount {2,15}?

EDIT: I see we've moved on. 30 is wrong.

 

[nc-1701]

One of my favorites:

Two numbers m and n are chosen such that 2 <= m <= n <= 99. Mr.
S(um) is told their sum and Mr. P(roduct) is told their product. The following
dialogue ensues:

i. Mr. P: I don't know the numbers.
ii. Mr. S: I knew you didn't know. I don't know either.
iii. Mr. P: Now I know the numbers.
iv. Mr. S: Now I know them too.

In view of the above dialogue, what are the numbers?

Well... Repeating some of your work here but...
i=>m, n not both prime
ii=>m+n is odd. Since every even number can be a sum of primes (Goldbach's conjecture is certainly valid for integers <100) and if the number could be a sum of primes then Mr. P might have known what it was. And we can get m+n=/=5,197.
iii: Here I start to lose it... I think we have to do some pretty in depth analysis on the sets...
m+n odd => m*n is even and exactly one of m, n is even with the other odd
further m+n-2 cannot be prime since n=2 gives a sum of two primes which would be invalid by ii.

That's as far as I can get easily, running analysis on m+n...
m+n={11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 83, 87, 89, 93, 95, 97, 101, 107, 113, 117, 119, 121, 123, 125, 127, 131, 135, 137,143, 145, 147, 149, 153, 155, 157, 161, 163, 165,167, 171, 173, 177, 179, 185, 187, 189, 191}
This cuts down the possibilities somewhat...
In order for the problem to be solved on iii) we need a a number where we can eliminate all but 1 pair f additive factors.
m+n=11 =>m*n={18, 24, 28, 30} Mr. P. can solve the problem if all of the possible factor representations of of his numbers do not appear on Mr. S's list except the on matching 11.
18=2*9, 3*6, which give 11 and 9
24=2*12, 3*8, 4*6, which give 14, 11, and 10. Now 11 has been created twice so it can no longer work.
Moving on to 17...
m+n=17=>m*n={30, 42, 65, 70, 66, 70, 72} I believe since 30 is shared with the decomposition of 11, it eliminates 17 as well...

Further exhaustive analysis by hand will be difficult, but I'm sure I could write a Maple program to do this next time I'm at school. Or a few few more observations could help reduce the search space some?

 

[Mise]

Here's another thing: If P gets 722, for example, he knows that the numbers are 19 and 38, because those are the only two m,n that multiply to 722.

So n is such that m is not a factor of n?????

Another thing: P gets 212 -- he knows it's 4 and 53, because although 106,2 works as well, 106 is > 99.

So.... how do you discount this one?????????