Wikipedia:Reference desk/Archives/Mathematics/2008 April 5

= April 5 =

Statistical significance
Dear Wikipedians:

How do I calculate the statistical significance level of a multiple regression model?

Thanks.

76.65.13.132 (talk) 19:55, 5 April 2008 (UTC)


 * Do you have a test statistic? What is your null hypothesis? The concept of significance level is not associated with models but with hypothesis testing. --Lambiam 20:18, 5 April 2008 (UTC)


 * I interpret the poster is referring to the global significance of the regresion. There is an F-test (Analysis of variance) to study this signification. (In fact, most statistic packages give the corresponding F-statistic.) The section Linear regression explains the underlying tenets behind the test construction.
 * The null hypothesis of the test is informally stated as "the coefficients &beta;i are all zero"; in more formal terms, the base hypothesis is that the variance in the dependent variable explained by the regression is insignificant in comparison with the residual (i. e. not explained) variation of the variable. Pallida  Mors  00:55, 7 April 2008 (UTC)

Derivative of an integral
Well I don’t really know if or how the fundamental theorem of calculus would apply to this case, but it came up recently and I realized I have no idea how to approach the problem if $$f$$ is not separable. That is, to find the derivative of an integral of a multivariate function. That is, something like
 * $$\frac{d}{dx} \int_{a}^{x} f(x,y)\,dy $$

Any ideas or know of any pages that address this? GromXXVII (talk) 20:08, 5 April 2008 (UTC)


 * Can you do
 * $$\frac{d}{dx} \int_{a}^{u} f(v,y)\,dy, $$
 * where u and v depend on x?
 * In general,
 * $$\frac{d}{dx} E(u,v) = \frac{du}{dx} \frac{\partial}{\partial u} E(u,v) + \frac{dv}{dx} \frac{\partial}{\partial v} E(u,v) = \frac{du}{dx} E_u(u,v) + \frac{dv}{dx} E_v(u,v).\,$$
 * After you have obtained the functions Eu and Ev, simply substitute u := x and v := x, giving
 * $$\frac{d}{dx} E(x,x) = \frac{dx}{dx} E_u(x,x) + \frac{dx}{dx} E_v(x,x) = E_u(x,x) + E_v(x,x).\,$$
 * --Lambiam 20:38, 5 April 2008 (UTC)
 * That looks right to me. Assuming x and y are independent, I would myself think of it like this:
 * $$g(x) = \int_{a}^{x} f(x,y)\,dy$$
 * The value of g(x) at any value of x is equal to the integral of the function f(x,y) between the paths y=a and y=x. For a total derivative in x, you have to ask both how f(x,y) changes with x, and how the change in the upper bound of the integral (the path y=x) changes.
 * $$\frac{d}{dx}g(x) = \frac{d}{dx}\int_{a}^{x} f(x,y)\,dy = \int_{a}^{x}\frac{\partial}{\partial x}f(x,y)\,dy + f(x,x)$$
 * where the second term is due to the change in the upper bound (y=x) of the integral with a change in x. This is equivalent to Lambiam's answer above, but it helps me to think of what is actually happening and why the answer turns out as it does.  Hope it proves helpful to you too.  --Prestidigitator (talk) 05:19, 10 April 2008 (UTC)

Projectile motion and tetrahedra
Consider this question: A projectile is launched at some initial angle $$\theta$$ above the horizontal, in a uniform gravitational field, and with no obstacles to collide with. What is the largest value of $$\theta$$ such that the overall distance between the projectile and the launch point never decreases?

I already obtained the solution $$\theta = \arctan(\sqrt{8})$$ by analytic geometry. My question is, why is this exactly the dihedral angle of a regular tetrahedron? There are no coincidences in math, so what does this have to do with tetrahedra? Is there some way to solve it by synthetic geometry that actually gives you a tetrahedron? —Keenan Pepper 21:06, 5 April 2008 (UTC)


 * Well, I can confirm your answer, more than that, I'm not sure. I think there are coincidences in maths, although it's a little difficult to define "coincidence". It's probably possible to find a connection, but it could an extremely convoluted one. Of course, there may be a really elegant one that I'm missing. --Tango (talk) 00:23, 6 April 2008 (UTC)
 * The same equations have the same solutions. The two problems must lead to the same equations. It is easier to work with equations than with solutions. Bo Jacoby (talk) 14:41, 6 April 2008 (UTC).
 * The same equations will have the same solutions, sure, but the converse isn't true. It's perfectly plausible for 2 unrelated equations to have the same solutions. f(x)=x and g(x)=ex-1 have the same zeros, but any relationship between them would be pretty contrived. (A first order Taylor approximation of g gives you f, I suppose, but I wouldn't say that meant much.) --Tango (talk) 15:26, 6 April 2008 (UTC)