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

= 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? &#x2013; 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 $$n\equiv m\pmod2$$. 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, &hellip;, An are abelian groups then the direct sum A1 ⊕ &hellip; ⊕ An is the set of n-tuplues (a1, &hellip;, an), where a1 ∈ A1, &hellip;, an ∈ An, 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 ⊕ &hellip; ⊕ An into an abelian group. — Fly by Night  ( talk )  13:43, 5 January 2011 (UTC)