Wikipedia:Reference desk/Archives/Mathematics/2008 April 25

= April 25 =

Statistical significance
I recall reading in one of my textbooks that if the samples are less than 1,500 they are not statistically representative or significant. How would I calculate the statistical significance of only 585 drivers coming from the South and turning East and only 373 drivers coming from the West and continuing straight within a 24 hour period? Also, how would I plot a histogram based on an 85 percentile speed of 33 mph for the first group and 28 mph for the second? 71.100.7.78 (talk) 00:57, 25 April 2008 (UTC)


 * I always heard 1100 was the minimum for good statistics. That's within 3% of the actual value 90% of the time, I think. StuRat (talk) 01:31, 25 April 2008 (UTC)

I don't know where you two are getting these numbers, 1500 and 1100, but unless there's some additional context, that's nonsense. The question about drivers is not clearly stated. If you're going to talk about statistical significance, you've got to say what your null hypothesis is. Your question about histograms is even vaguer. And the statement that "That's within 3% of the actual value 90% of the time" is so vague as to be meaningless. WHAT is within 3% of WHAT actual value? Michael Hardy (talk) 01:49, 25 April 2008 (UTC)


 * If we do a poll and conclude that 60% of the people prefer tweedledee over tweedledum, there is a margin of error associated with that. One way of stating the margin of error is "the poll has a 3% margin of error over a 90% confidence interval", which essentially means we will be within 3% of the percentage of the total population who prefer dee to dum, 90% of the time we do such a poll.  I believe 1100 is the number that provides that result.  Obviously, lower sample sizes mean a larger margin of error and/or a lower confidence interval. StuRat (talk) 07:20, 27 April 2008 (UTC)

PS: I think it's perfectly obvious that what's going on is that someone is attempting to post a homework question before understanding what the question is. Michael Hardy (talk) 01:50, 25 April 2008 (UTC)

No, not a homework question although i suppose my brain trys to formulate questions like a teacher would. That said and to be more specific the problem is that transportation departments are assigned the task of determining whether a roadway merits controls at intersections or speed controls or opening a cut through. What my local government transportation office has done, however is to set up what appear to be arbitrary conditions for such controls and when they do a reading violate all sorts of statistical norms. For instance, limiting data to a single 24 hour period rather than showing a histogram of speed and flow per hour of the day and day for the week, etc. for a whole week. Rather than publish or provide the citizenry with the actual data they say things like, "The control can not be installed unless the 85th percentile of the speed is 10 mph above the posted limit." What my question attempts to do is to say okay then with that information can I draw a histogram to represent a facsimilty of the data? Since the computation is based on only 585 and 373 samples then I'm also concerned about the statistical significance of the data. 71.100.7.78 (talk) 02:16, 25 April 2008 (UTC)

PS: Although I can eventually figure out what I need to do in order to create a frequency distribution or histogram for myself my concern is for ordinary folk like plumbers who are expected to understand what "The control can not be installed unless the 85th percentile of the speed is 10 mph above the posted limit." means and who would know if presented in a graphical distribution or histogram form with a line draw at the 85th percentile. Its not too much to ask that the transportation techs do this but some like to place a mystique around the work they do in order to confuse people to guarantee their job. 71.100.7.78 (talk) 03:05, 25 April 2008 (UTC)

PSS: In fact there are so many calculators on the Internet now to convert percentile and x-scores to z-scores and a mean it is not funny. Thus, I now know without the help of the reference desk that the mean for a percentile of 85, an SD of 8 and an x-value of 33 is 25 and that soon I will be able to do this in MathCad and in Excel. Thanks for the help. 71.100.7.78 (talk) 04:09, 25 April 2008 (UTC)


 * Not sure if this helps you, but Excel already has stats functions, such as NORMSINV, CONFIDENCE and NORMDIST. Thus:
 * 33-8*NORMSINV(0.85) = 24.70853
 * 24.70853+CONFIDENCE(0.3,8,1) = 33
 * NORMDIST(33,24.70853,8,TRUE) = 0.85 Gandalf61 (talk) 08:18, 26 April 2008 (UTC)


 * 1500 is a somewhat arbitrary number, but I believe it's the one used by the National Opinion Research Center as a guideline for the number of respondents they look for in a social survey. With a simple random sample of size 1500, a subsample of size 30 is just on the borderline of being a statistically significant result (i.e. if I calculate the standard error on the estimate of proportion - 0.002 - I get a result of about 0.0011, which means that since it's within 2 standard errors of zero it just fails the 95% significance test). On the other hand, if the proportion is around 50%, then the standard error is about 1.3%, so you can detect a swing of about 2.5 - 2.6 percentage points. Both of these may have been factors in choosing the number. Confusing Manifestation (Say hi!) 01:10, 28 April 2008 (UTC)
 * Thanks. Very helpful. 71.100.7.78 (talk) 12:21, 1 May 2008 (UTC)


 * The above is just nonsense. What is significant depends on what the hypothesis is that's being tested. Michael Hardy (talk) 11:35, 1 May 2008 (UTC)

More Quantum Theory
According to Zeno's paradox, it states that Planck length and Planck time are the smallest mesurable units of their respective dimensions. It then implies that space may be discontinuous and not infintely divisible, having the Planck length as the smallest unit of traversable space. How could this be? Where could I find more information about this? Furthermore, why is it (Planck length) the smallest mesurable unit of space, the Planck time the smallest mesurable amount of time? Why can nothing smaller be meaured? Thanks, Zrs 12 (talk) 02:50, 25 April 2008 (UTC)
 * I think you may have better luck at the Science section. Regarding the "how could this be" part - why not? Newtonian physics models the universe as a function from continuous time to continuous space, but that doesn't mean that's what it really is. Stephen Wolfram, in his infamous book A New Kind of Science, explores the idea that the universe is a cellular automaton (like Conway's Game of Life). I suppose you could find more information in any book or online resource about quantum mechanics. -- Meni Rosenfeld (talk) 09:41, 25 April 2008 (UTC)

Calculus problem
I'm solving a multipart problem for homework, and I understand all the parts except for one. A water tank holds 1,200 gallons of water at time t = 0. During the time interval 0 ≤ t ≤ 18 hours, water is pumped into the tank at the rate W(t) = 95√t sin2(t/6) gallons per hour. During the same interval, water is removed from the tank at the rate R(t) = 275sin2(t/3) gallons per hour. At what time t, 0 ≤ t ≤ 18, is the amount of water in the tank at an absolute minimum? I don't think I can just graph it and look for the solution graphically because I have to show my work. If anyone could point me in the right direction, it would be greatly appreciated. Thanks! —Preceding unsigned comment added by 70.23.84.88 (talk) 02:58, 25 April 2008 (UTC)


 * You'll want to look at the articles on maxima and minima and the antiderivative. You've been given the derivative of volume with time as well as the necessary boundary condition to find the complete function, though you'll have to think hard and ferret it out from the word problem.  --Prestidigitator (talk) 06:57, 25 April 2008 (UTC)
 * You don't really need to work out the complete function - for maxima and minima, you just need the derivative, which is what you're given (indirectly). Integrating it is just making work for yourself, since your next step would be to differentiate. --Tango (talk) 15:52, 25 April 2008 (UTC)
 * Okay, so I'm guessing I need to find the critical numbers by setting the derivative equal to 0. But should the derivative function I set equal to 0 be W(t) + R(t) or W(t) - R(t)? I'm guessing the latter, but I just want to make sure. —Preceding unsigned comment added by 71.249.156.154 (talk) 18:20, 25 April 2008 (UTC)
 * Don't guess, think it through. If you're having trouble, try making a simpler example to test your method out on (try constant flow rates - W(t)=2, R(t)=1, say - now, that won't have any critical points, but you should be able to work out the derivative pretty easily, then use the same method on the real problem). --Tango (talk) 18:34, 25 April 2008 (UTC)
 * While you don't have to integrate to find the local minima, I believe you'll have to integrate eventually and evaluate the value of the function at the endpoints and any local minima you do find in order to determine the absolute minimum within your interval. --Prestidigitator (talk) 18:46, 25 April 2008 (UTC)
 * You may have a point there. --Tango (talk) 18:47, 25 April 2008 (UTC)
 * Except that the problem statement doesn't require us to find the value of the minimum, only its location (and for matters of establishing globality, bounds are sufficient). Not to mention that the antiderivative is non-elementary. -- Meni Rosenfeld (talk) 17:12, 26 April 2008 (UTC)
 * Actually, there may be something I'm missing here, as the location of the minimum also seems to be non-elementary. -- Meni Rosenfeld (talk) 19:13, 26 April 2008 (UTC)
 * If there are multiple local minima and possibly lower values at the endpoints (which may not be local minima), then you have to figure out which of them is smallest in order to determine where the absolute minimum is. The use of half-angle identities might help a bit, but I completely missed the $$\sqrt{t}$$ in the one function the first time around.  Ouch.  That might make things a pain.  --Prestidigitator (talk) 21:45, 26 April 2008 (UTC)
 * Knowing you would say that, I have preemptively mentioned that to verify that the point is a global minimum, one does not need to actually evaluate the antiderivative\integral (which is essentially impossible in this case), but only to put a bound on its value (which should be simple enough). -- Meni Rosenfeld (talk) 21:57, 26 April 2008 (UTC)


 * ...and does the local plumbing supply company really provide pumps that pump 95√t sin²(t/6) gallons of water ? I can only find pumps that pump ln(tanh-eπit³) gallons.  :-) StuRat (talk) 07:07, 27 April 2008 (UTC)

Global food meltdown dilemma: Should I hoard 10 pallets of rice or keep the money in the bank?
I'm not very good at math. How do I figure this out? Let's assume it will take my family 2 years to eat the rice. Thanks for your help.

24.130.198.167 (talk) 03:48, 25 April 2008 (UTC)


 * I'm not sure where the math part of the question is, and I don't think we can offer financial advice here, but I'll go so far as to point out that one of the key differences might be that rice doesn't generate much interest. --Prestidigitator (talk) 07:01, 25 April 2008 (UTC)


 * You need to calculate whether inflation will increase the price of the food faster than interest will increase the price of the money in the bank - without the numbers, I can't do that for you. --Tango (talk) 10:49, 25 April 2008 (UTC)


 * There are several possible ways to interpret your question: which investment will be cheaper (that is, have a lower expected cost), or how to minimize the uncertainty in gaining sufficient quantities of rice (for a particular price). But as is already stated, the solution to either question is heavily dependent on your guess of the future price of rice, as well as spoilage rates of stored rice.  You may also want to consider the ethical implications: if everyone tried to stockpile 2 years of rice simultaneously, there would not be enough even under ideal conditions.  It may be better to limit hoarding, from a purely ethical standpoint.  --TeaDrinker (talk) 19:23, 25 April 2008 (UTC)


 * There can be at least one moral reason to horde, however. If a small group of people believe, correctly, that there will be a future shortage, but the majority does not, the small group can horde the commodity, and thus drive the price up and make it financially beneficial for busineses to increase production before the shortage occurs.  If the horded commodity is then released on the market as well when the shortage occurs, this, along with the increased production, may reduce the severity of the shortage. StuRat (talk) 06:53, 27 April 2008 (UTC)
 * Knowing how to be selfish myself, I am not going to share the solution with you :) Bo Jacoby (talk) 14:10, 27 April 2008 (UTC).

x^2 + 4x^3*y^2 + y^3 = 10
I saw this curve in a textbook and out of interest wanted to know what it looked like. I know that there are vertical tangents at the x-intercepts, (+-√10, 0), that there is a local maximum stationary point at (0, 10^(1/3)), and that the curve is decreasing in the first quadrant. I also know that other possible stationary points lie somewhere on the curve 6xy^2 + 1 = 0, ie: to the left of the y-axis, but it seems impossible to solve the degree 9 polynomial to find them. It seems also that as x gets infinitely large, y gets infinitely small, and as x gets infinitely small, y gets infinitely small. It seems also that where the curve hits the y-axis is the highest point on the curve.

What is this curve called? What is its class? Is it special? —Preceding unsigned comment added by 124.191.116.62 (talk) 04:53, 25 April 2008 (UTC)


 * Here's a plot of it. &mdash; Kieff | Talk 08:10, 25 April 2008 (UTC)

Very interesting. How did you come to draw that? Not many calculators can graph this because it's implicit. From your image, it seems as if I was wrong about the y-intercept being a local maximum. How did you approach the graphing of this? What are the asymptotes? I assume you consider the behaviour for extreme values of x. How do you calculate the limits of this behaviour? —Preceding unsigned comment added by 124.191.116.62 (talk) 08:29, 25 April 2008 (UTC)


 * It looks like he just plugged it into an appropriate piece of Mathematical software (Mathematica, Maple, or similar). --Tango (talk) 10:47, 25 April 2008 (UTC)


 * Nope. PHP+GD. Also, I wonder if Wikipedia could have a graphing system built-in... &mdash; Kieff | Talk 16:26, 25 April 2008 (UTC)
 * Ok, so you wrote your own appropriate piece of mathematical software! That works too. --Tango (talk) 18:35, 25 April 2008 (UTC)
 * Pacific Tech makes a good program called Graphing Calculator (or NuCalc) that handles implicit, and a variety of other types of plots. It's about $40 if you show them a student ID. -GTBacchus(talk) 23:03, 25 April 2008 (UTC)

I've found a neat graph program inbuilt on the Mac. Thanks for your help. —Preceding unsigned comment added by 124.191.116.62 (talk) 02:12, 26 April 2008 (UTC)

Simple verification of rectangles and "content zero"
A subset A of Rn has content zero if for every $$\epsilon > 0$$ there is a finite cover $$\{U_1,\ldots,U_m\}$$ of A by closed rectangles such that $$\lambda(U_1) + \cdots + \lambda(U_m) < \epsilon$$, where &lambda; is the elementary rectangle volume. Assume the following theorem:
 * If $$a < b$$ and $$\{U_1,\ldots,U_m\}$$ is a finite cover of $$[a,b] \subset \mathbf{R}$$ by closed intervals, then $$\lambda(U_1) + \cdots + \lambda(U_m) \geq b - a$$. In particular, $$[a,b]$$ does not have content zero.

I want to show that $$[a_1,b_1] \times \cdots \times [a_n,b_n]$$ does not have content zero if $$a_i < b_i$$ for each i.

Let $$A = [a_1,b_1] \times \cdots \times [a_n,b_n]$$ and suppose the closed rectangles $$\{U_1,\ldots,U_m\}$$ cover A. Write $$U_i = [u_{i1},v_{i1}] \times \cdots \times [u_{in},v_{in}]$$ and $$U_{ij} = [u_{ij},v_{ij}]\,$$, so that $$U_i = U_{i1} \times \cdots \times U_{in}$$. Then $$\{U_{1j},\ldots,U_{mj}\}$$ is a sequence of closed rectangles that cover $$[a_j,b_j]$$. Consequently,
 * $$\sum_{i=1}^m \lambda(U_{ij}) \geq b_j-a_j,$$

so that
 * $$\sum_{i=1}^m \lambda(U_i) = \sum_{i=1}^m \sum_{j=1}^n v_{ij}-u_{ij} = \sum_{j=1}^n \sum_{i=1}^m v_{ij}-u_{ij} \geq \lambda(A).$$

I saw this result shown differently, so I'm just wondering if the above is correct or if there's something I'm missing. &mdash; merge 15:24, 25 April 2008 (UTC)
 * Surely $$\sum_{i=1}^m \lambda(U_i) = \sum_{i=1}^m \prod_{j=1}^n v_{ij}-u_{ij}$$? Algebraist 16:03, 25 April 2008 (UTC)


 * Er, sorry. I obviously have some wires crossed today.  :)  Thanks.  &mdash; merge 16:32, 25 April 2008 (UTC)
 * Out of interest, how does the proof you've seen go? Algebraist 17:13, 25 April 2008 (UTC)

I'll describe it, but first, how about this, since we already happen to have linearity and monotonicity of the integral and we can assume each Ui is contained in A:
 * $$\lambda(A) = \int_A 1 \leq \int_A \sum_{i=1}^m \chi_{U_i} = \sum_{i=1}^m \int_A \chi_{U_i} = \sum_{i=1}^m \lambda(U_i).$$

&mdash; merge 18:27, 25 April 2008 (UTC)
 * Sure, that works if you've got integration. I'd got the impression you were after a first-principles proof; otherwise this result comes out for free from the existence of Lebesgue measure. Algebraist 10:56, 26 April 2008 (UTC)

Well, what's going on here is that this question comes from a book which happens to develop Riemann integration on Rn using the notion of Jordan measure. It shows up in the section just after the integral has been defined and its basic properties proved (as problems, by the reader). The funny thing is that after you have these properties the statement that $$\lambda(A) \leq \sum \lambda(U_i)$$ becomes trivial via the above proof, yet the author doesn't use this to prove the theorem mentioned at the beginning (i.e. the case n=1), and instead proceeds directly. Presumably he also intends the reader to proceed directly here as well, though this seems unnecessary to me.

Anyway, the direct approach is as follows: each $$U_i$$ induces a partition on A with itself as one of the subrectangles. Let P be the common refinement of these partitions. Then P induces a partition on each $$U_i$$, the subrectangles of which are exactly the subrectangles of P that contain an interior point of $$U_i$$. The volume of $$U_i$$ is then the sum of the volumes of the subrectangles in this induced partition, while the volume of A is the sum of the volumes of the subrectangles in P. Because every subrectangle of P also appears in at least one of these induced partitions, the volume of A cannot exceed the sum of the volumes of the $$U_i.$$  &mdash; merge 11:23, 26 April 2008 (UTC)

Controllability Grammian
Can anyone point me to a clear derivation of the controllability grammian of a linear system?

Given a system:

$$\dot{\mathbf{x}}=\mathbf{Ax}+\mathbf{Bu}$$

$$\mathbf{y}=\mathbf{Cx}$$

the controllability Grammian is:

$$W_c=\int_0^\infty e^{\mathbf{A}t}\mathbf{BB}^T e^{\mathbf{A^T}t} dt$$

The input energy required to get from the state $$\mathbf{x}=\mathbf{0}$$ to $$\mathbf{x}=\mathbf{x}_0$$ is $$\mathbf{x}_0^T W_c \mathbf{x}_0$$.

Thanks, moink (talk) 17:43, 25 April 2008 (UTC)


 * Oh, and note that I didn't need to give the output equation at all, since this is just the controllability grammian. I find the observability grammian significantly easier to derive.  moink (talk) 17:47, 25 April 2008 (UTC)

Dont understand this puzzle
Could somebody solve the puzzle on the far right and give an explanation cos im stumped --Hadseys Chat Contribs 21:31, 25 April 2008 (UTC)
 * Look how red and green move. If you only see one of them then maybe it's covering the other. PrimeHunter (talk) 21:36, 25 April 2008 (UTC)
 * Does red move 1 space clockwise and green move 2? --Hadseys Chat Contribs 21:54, 25 April 2008 (UTC)
 * You tell me. Another method you can use to get to the answer a bit quicker: There are only certain corners the green is ever in, so you can rule out any choice which has the green anywhere else. --Tango (talk) 22:27, 25 April 2008 (UTC)
 * PS Or, yet another method: Just spot the repeating pattern in the tiles as a whole (don't worry about separating what's happening in the tiles, just worry about whether certain tiles are identical). --Tango (talk) 22:29, 25 April 2008 (UTC)


 * Fascinating. With the first hint, it can be solved by studying the red and ignoring the green, or studying the green and ignoring the red.  Fortunately, one gets the same answer both ways!
 * -- Danh 63.226.145.214 (talk) 02:56, 26 April 2008 (UTC)
 * So the answer's A, right? --Hadseys Chat Contribs 14:39, 26 April 2008 (UTC)
 * The green dot oscillates between two of the corners and the red dot rotates clockwise, from picture to picture. So only A fits - Adrian Pingstone (talk) 16:10, 26 April 2008 (UTC)


 * Just a quick off-topic question, but did you create that image yourself, or did you just take a screenshot of wherever you found it? If the latter, it shouldn't be hosted on wikipedia as the ref desk doesn't really qualify for fair use. But yes, just follow the pattern of the green one for instance, and see what corner that must end up in, and then follow the red. This sort of thing can usually be solved by looking at the individual elements. -mattbuck (Talk) 18:58, 26 April 2008 (UTC)


 * In this case, though, you don't need to follow them around, just notice that all the up-left to down-right diagonals are the same all the way along. Black Carrot (talk) 19:25, 28 April 2008 (UTC)