Wikipedia:Reference desk/Archives/Mathematics/2011 January 4

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January 4

statistics

I'm having difficulty with my thesis. can "knowledge or awareness of people to statistics" be subjected to study? what statistical tool can be used? — Preceding unsigned comment added by Rionsgeo (talk • contribs) 02:06, 4 January 2011 (UTC)

If I understand you, you want to measure people's knowledge and awareness of statistics, using statistical methods. I suppose you could do a poll/quiz where you ask people how often they use statistics, then ask them to solve some statistics problems, and use that data to calculate standard deviations, confidence intervals, etc.

One suggestion for a refinement of your thesis: "Resolved, that people who are ignorant or distrustful of statistics tend to engage in statistically unhealthy habits and thus shorten their lives." You could design a way to test this assertion and either prove or disprove it. StuRat (talk) 05:43, 4 January 2011 (UTC)

Why does this remind me to Correlation from xkcd? – b_jonas 10:12, 7 January 2011 (UTC)

Permutations

How can I compute the number of ways to choose n elements in sets of size k (with replacement), so that no element occurs in each set more than x times? 70.162.9.144 (talk) 07:30, 4 January 2011 (UTC)

I don't know if this helps, but it should be equal to the coefficient of $y^{k}$ in $(y^{x+1}-1)^{n}/(y-1)^{n}$, if I followed your notation correctly. I'd guess there isn't any nice closed-form solution. Are you looking for a way to efficiently compute it? Eric. 82.139.80.114 (talk) 01:39, 5 January 2011 (UTC)

Sorry, could you explain what you mean by coefficient and how it is derived from $(y^{x+1}-1)^{n}/(y-1)^{n}$? 70.162.9.144 (talk) 04:37, 5 January 2011 (UTC)

${\frac {(y^{x+1}-1)^{n}}{(y-1)^{n}}}=\left(\sum _{k=0}^{x}y^{k}\right)^{n}$ is expanded by the Multinomial theorem. Bo Jacoby (talk) 12:56, 5 January 2011 (UTC).

Identifying a quotient

Let A be free abelian on 3 generators a,b,c and K the subgroup (n+m)a+(n-m)b+(m-n)c for all integers n,m. Is A/K just Z x Z/2Z? This seems like a very trivial task a computer should be able to do, is there any software to identify stuff like this? Money is tight (talk) 08:26, 4 January 2011 (UTC)

K is generated by the elements a + b − c and 2a, it thus consists of elements of the form na + mb + kc where k = −m and $n\equiv m{\pmod {2}}$. From this it follows easily that K = Ker(f), where f: A → Z × (Z/2Z) is defined by f(na + mb + kc) = (m + k, n + m mod 2). Since f is clearly onto, A/K is indeed isomorphic to Z × (Z/2Z).—Emil J. 12:41, 4 January 2011 (UTC)

Thanks, I'm used to doing "show/prove" questions and this question wasn't one of those, just needed some confirmation to check my understanding is correct. Money is tight (talk) 00:56, 5 January 2011 (UTC)

I think the convention is to use the symbol ⊕ to denote the direct sum of abelian groups. (Instead of using the direct product, which is often used with non-abelian groups.) So it would be normal to write Z ⊕ Z/2Z. If A₁, …, A_n are abelian groups then the direct sum A₁ ⊕ … ⊕ A_n is the set of n-tuplues (a₁, …, a_n), where a₁ ∈ A₁, …, a_n ∈ A_n, under the binary operation

$(a_{1},\ldots ,a_{n})+(a_{1}',\ldots ,a_{n}')=(a_{1}+a_{1}',\ldots ,a_{n}+a_{n}').$

This turns A₁ ⊕ … ⊕ A_n into an abelian group. — Fly by Night (talk) 13:43, 5 January 2011 (UTC)