Wikipedia:Reference desk/Archives/Mathematics/2016 July 5
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July 5
editSampling voters
editDoes the result of a referendum (or any vote between two candidates) represent a Binomial distribution, a Hypergeometric distribution or something else? If a poll is taken before hand, to estimate the outcome, is the size of sample required influenced in any way by the predicted result? And if a poll is taken some time afterwards, to estimate any shift in preference, is the size of sample required influenced in any way by the actual result? (Apparently, to achieve a margin of error of 3%, at the 95% level of confidence, a sample of "about 1000" is required, even for infinite populations - so perhaps those last two questions are redundant.) Martinevans123 (talk) 18:29, 5 July 2016 (UTC)
- If you assume that both the people who turn up to place a vote, and the people polled, are random samples of the population, then you can estimate the margins of the vote pretty easily, although you might want to look into methods for sampling a finite population - the main thing is that you introduce a finite population correction into your estimate of variance, which relates to the fraction of the population you sample (with the important result that if you sample everyone, the variance becomes zero because you know the values for the population with certainty). Other than that, though, you can use something fairly close to the binomial/multinomial distribution formulas, noting in particular that the margin of error on your estimate will relate to the population proportions themselves, which then affects the sample size you need to measure the result to within a particular tolerance. Confusing Manifestation(Say hi!) 02:14, 6 July 2016 (UTC)
- Thanks. And is is true that "to achieve a margin of error of 3%, at the 95% level of confidence, a sample of "about 1000" is needed", regardless of the size of the population? Because this is what most polling companies appear to use to estimate the results of elections and referenda. Martinevans123 (talk) 07:54, 6 July 2016 (UTC)
See [[1]]. Bo Jacoby (talk) 10:35, 6 July 2016 (UTC).
Equation of plane
edit(moved here from science desk)
I want to define a plane for which i know only one vertex.the normal passing through the centroid of the triangle intersects xy plane at 0,0.additionally i know area of triangle and length of sides.please help how to find other two vertex of triangle.i would be extremely grateful to anyone answering... — Preceding unsigned comment added by 103.206.113.41 (talk) 18:31, 5 July 2016 (UTC) I could add one more constraint or freedom..the said normal may intersect in 10mm dia on xy plane. — Preceding unsigned comment added by 103.206.113.78 (talk) 20:38, 5 July 2016 (UTC)
- @103.206.113.78:
- A plane is infinite, so that's not what you're asking for.
- What does "10mm dia" mean?
- Your constraints do not uniquely determine the solution. If you give me a solution of the problem, I can find uncountably many others by rotating the triangle about the line extending from the origin to your known vertex.--Jasper Deng (talk) 22:01, 5 July 2016 (UTC)
- This seems like a question asked here recently. And the answer has not changed either, as seen from that given above. —Quondum 03:51, 6 July 2016 (UTC)
- Imagine a ray (a half-line, Line (geometry)#Ray) beginning at the origin. You have two degrees of freedom choosing the ray direction (it can be defined with two coordinates φ,λ on the unit sphere). Next, choose a point Q on the line; the distance from the ray's starting point is a third degree of freedom. Now, make a plane through the Q point, normal to the ray. In that plane put your triangle, so that the triangle's centroid coincides with Q. Of course you can rotate the triangle in the plane, keeping the centroid at Q – so the rotation angle is the fourth degree of freedom.
Now, you have four degrees of freedom in the solution you seek. Put it another way, constraints you defined allow for a 4-dimensional space of solutions. Nobody knows which specific solution from that space you seek. --CiaPan (talk) 12:14, 6 July 2016 (UTC)- The way I interpret the question, the absolute position of one of the vertices is known. This constrains the solution space to one degree of freedom (plus a flip of the triangle). —Quondum 14:59, 6 July 2016 (UTC)
- Imagine a ray (a half-line, Line (geometry)#Ray) beginning at the origin. You have two degrees of freedom choosing the ray direction (it can be defined with two coordinates φ,λ on the unit sphere). Next, choose a point Q on the line; the distance from the ray's starting point is a third degree of freedom. Now, make a plane through the Q point, normal to the ray. In that plane put your triangle, so that the triangle's centroid coincides with Q. Of course you can rotate the triangle in the plane, keeping the centroid at Q – so the rotation angle is the fourth degree of freedom.