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April 5

Cubic capacity conversion

I'm working on an article about airships and tried to convert 20,000 cubic feet into SI Units. I couldn't remember how to work it out myself and used an on-line converter, which came up with 566 cubic metres. Is that right? It sounds rather on the low side to me. Alansplodge (talk) 19:00, 5 April 2013 (UTC)[reply]

Remember 1 m = 3.28 feet. Now just use dimensional analysis:

20000ft^{3}\cdot ({\frac {1m}{3.28ft}})^{3}=566.77m^{3}

. 72.128.82.131 (talk) 19:04, 5 April 2013 (UTC)[reply]

Hmmmm, there are two different feet. The international one is 30.48 cm, but the US one is 30.4800609601 cm. Not much difference but still not exactly the same. Anyway, I prefer to do the conversion by substitution. So, 20,000 ft^3 = 20,000 (30.48 cm)^3 = 20,000 (0.3048)^3 m^3 = 566.34 m^3. Count Iblis (talk) 19:18, 5 April 2013 (UTC)[reply]

Thank you both - I must have been asleep in that lesson. Thanks Count Iblis; however I'm not sure that the figure of 20,000 was that accurate in the first place, as it describes an airship constructed in a do-it-yourself shed on the edge of a sportsfield in 1902. Probably 566 m^3 is going to be good enough. Alansplodge (talk) 01:02, 6 April 2013 (UTC)[reply]

Also note that the volume of an airship may not be constant. In the case of soft-sided airships, like blimps, the interior space itself changes depending on the level of inflation. However, even in a rigid airship, like a dirigible, the amount of gas inside isn't constant, as more gas will squeeze into the same volume under higher pressure. So, if you wanted to know "what would the volume of gas inside the dirigible be, if it were to be removed and brought to STP", you might get different answers depending on the initial temperature and pressure inside. StuRat (talk) 05:22, 6 April 2013 (UTC)[reply]

April 7

Confidence interval

The textbook Elementary Statistics by Larson and Farber, 2nd ed., states in the chapter on confidence intervals:

"After constructing a confidence interval, it is important that you interpret the results correctly. Consider [the above example]. Because µ already exists, it is either in the interval or not. It is not correct to say 'There is a 90% probability that the actual mean is in the interval (22.35,23,45).' The correct way to interpret your confidence interval is 'There is a 90% probability that the confidence interval you described contains µ.' [...]" [emphasis in original]

There are a number of differences between these two sentences, but I don't see which of these illustrates an important point. Is it the change from "actual mean" to "µ"? If so, then a better place to say so would be when introducing µ. Or is it "is in" vs. "contains"? That seems to be merely a grammatical change. — Sebastian 04:42, 7 April 2013 (UTC)[reply]

I think the key is Because µ already exists. As we know µ and know the interval it either lies in the interval or not, if it does then there is 100% chance, if it does not then there is a 0% chance.--Salix (talk): 05:58, 7 April 2013 (UTC)[reply]

In this case, we don't know µ yet. But if we did - as the IP editor below writes in their thought experiment, this would indeed be a possible distinction. — Sebastian 06:44, 7 April 2013 (UTC)[reply]

I'm not a specialist in statistics, but it seems to me that what the authors are trying to get across is poorly stated. I agree that both sentences appear to say the same thing. (I've taught classes from Larson's algebra and calculus books - though not statistics - and I don't have a great deal of confidence in the accuracy of their mathematical content, particularly on subtler points.) Here is my understanding of what is meant by a confidence interval.

You might say that, for any particular value of the actual mean µ, there is a 90% chance that if you take a sample and construct an interval in the indicated way, then µ will be in the given interval. The 90% probability is viewed as being prior to taking the sample and determining the confidence interval to be (22.35,23.45). This is different from saying, once the sample has been taken and the interval constructed, that there is a 90% probability that µ is in that interval.

For example, imagine that some purely theoretical considerations, or perhaps earlier reliable studies, make it highly unrealistic that µ actually lies in the calculated confidence interval. Even excluding the possibility of a biased sample or other methodological problems, you might conclude in those circumstances that a rare event probably occurred with your sample. (This makes sense mathematically at least. As to whether this is proper scientific methodology, that's another question.) So you can no longer say at that point that there's a 90% chance that µ is in the confidence interval, since outside factors have convinced you this is unlikely. But before taking the sample, you could have said that there would be a 90% chance that µ would eventually lie in the confidence interval, once the sample was taken and the interval calculated.

I believe the distinction I'm drawing here is typical of the objection made by Bayesian statisticians to some statistical techniques, but I'm getting beyond my level of expertise on this. Other than that, it's hard for me to see what the authors might have in mind. 64.140.121.87 (talk) 06:10, 7 April 2013 (UTC)[reply]

Thank you, I think you captured what they meant. It is quite a subtle point though, especially since the book isn't particularly rigorous otherwise. They seem to have no qualms, e.g., (in section 4.3 on discrete probability distributions) to "use the fact that the variance of a Poisson distribution is σ² = µ" to solve the following question:

20. Snowfall. The mean snowfall in January for Evansville, Indiana, is 4.0 inches.
(a) Find the variance and standard deviation. Interpret the results.
(b) Find the probability that the snowfall in January for Evansville, Indiana, will exceed seven inches.

(In this example, the SD would have the dimension √inch, which doesn’t make sense. That this is absurd can be illustrated if one ignores the unit: One would get √4 = 2 — or, if one uses 100 mm instead of 4 inches, √100 = 10.) — Sebastian 06:44, 7 April 2013 (UTC)[reply]

That example has worse problems! Why should the snowfall be a discrete variable? And even if it is, you have to know how small the quanta are in order to apply a discrete analysis. (This is the same issue as your units issue, fundamentally.) It strikes me that it would be fascinating, if the measurements could be made sensitive enough, to measure something like the fundamental electric charge by observing the standard deviation of the total charge of many small boxes snatched out of an ionized, rarefied gas. --Tardis (talk) 03:32, 12 April 2013 (UTC)[reply]

It's because in the frequentist world, you're not supposed to make probability statements about parameters, as they're not random variates - see the first bullet point in that frequentist link. But you can make prob statements about CI's, because they are random variates. I think some statisticians get quite insistent on this point, but I think they're being pointlessly pedantic :-) Also it confuses non-specialists for no benefit - practically there is a 90% chance µ is in the CI. Mmitchell10 (talk) 15:10, 7 April 2013 (UTC)[reply]

64's analysis is quite correct (as is everyone else here). An analogy might help. Imagine you are about to roll a die. Suppose the numbers have worn off slightly, and you are looking at it through a pair of glasses that have been splashed with mud. Before you roll it, you have a 1 in 6 chance of throwing a 2. Afterwards, you look at it, and decide you can't make out whether you are looking at spots on the die, or mud on your glasses. You try to discern the dots, but eventually you give up. Can you say "the chance of a 2 is 1 in 6"? Well, sort of. Because you are looking at the actual die, you would rather say it either is or isn't a 2. There is no longer any real randomness. You just can't see properly. If you could prove beyond doubt that your glasses are caked in mud, and you can see nothing at all, then it is a 1 in 6 chance. But if you can see through the glasses at all, you might be gleaning something from the die. Then you can't talk like that anymore. So it is better to talk more precisely, and talk of confidence intervals as being, a priori, 90% (or 95%) likely to contain the parameter of interest, but not a posteriori, that is, after the data are gathered. IBE (talk) 22:57, 7 April 2013 (UTC)[reply]

Thanks a lot, everyone; these were really excellent answers! BTW, the link to Frequentist inference led me to Fiducial inference, which under the bullet "A confidence interval, in frequentist inference, ..." expresses the same idea in a way I find even clearer than the first bullet in Frequentist inference: "The probability concerned is not the probability that the true value is in the particular interval that has been calculated since at that stage both the true value and the calculated are fixed and are not random." I find it very interesting that this question turned out to be a cornerstone of different schools of how to draw conclusions from samples of data. 03:13, 8 April 2013 (UTC)

(In response to Mmitchell10's unindented comment:) This is not an issue of frequentist obstinacy (the Bayesian approach has no issue with credence in the value of

\mu

) nor of the mire of Bayesian priors (how much did we really know about

\mu

, anyway?): if we construct enough (say, 90%) confidence intervals from one population, we will eventually find that two of them are disjoint, at which point it is simply nonsense to say that, in any sense, there is a

>50\%

probability of

\mu

lying in each. --Tardis (talk) 14:23, 12 April 2013 (UTC)[reply]

April 8

MS Windows calculator: Tanh?

I have the opposite and adjacent lengths and I'm trying to use the 'tan' function on the Windows built-in calculator. How do I do this?

I could probably do it on my calculator from muscle memory (but it's at work). This isn't homework. I'm 28 and got an A in A level Maths before going on to forget everything.

Opp is 15 and Adj is 142. I know that the angle is about 6 (from messing about) but I still want to know how calculate it properly with Windows calculator. --89.241.233.102 (talk) 04:56, 8 April 2013 (UTC)[reply]

It seems you forgot that you need arctan, not tan or tanh. So, entering 15 / 142 = Inv tan should do the trick. — Sebastian 05:20, 8 April 2013 (UTC)[reply]

Thanks. I don't remember ever addressing something called tanh. I knew I wanted the inverse. I didn't realise click Inv gave me those options. Thanks a bunch. <3 89.241.233.102 (talk) 06:41, 8 April 2013 (UTC)[reply]

"tanh" is the hyperbolic tangent. Looie496 (talk) 16:01, 8 April 2013 (UTC)[reply]

... It was on the Further Maths syllabus, but not the ordinary A-level. Dbfirs 07:06, 9 April 2013 (UTC)[reply]

April 9

April 10

Largest regular hexagon inscribed in a square

What is the size of the largest regular hexagon that can be inscribed in a unit square? This is not a homework question, I'm trying to machine a hex bolt out of a square bolt[1]. Google got me nothing.Dncsky (talk) 02:02, 10 April 2013 (UTC)[reply]

I think its going to be one at 15° this touches an axis-aligned square at 4 points. A vertex of a hexagon with side length r will be at (r cos(15°),r sin(15°)). Now

\cos {\frac {\pi }{12}}=\cos 15^{\circ }={\tfrac {1}{4}}{\sqrt {2}}({\sqrt {3}}+1)\approx 0.9659258263\,

.Exact trigonometric constants So the max side length fitting inside a unit square will be 0.5/0.9659258263≈0.5176380902. The vertices will be (0.5,0.134),(0.134,0.5),(-0.366,0.366),(-0.5,-0.134),(-0.134,-0.5),(0.366,-0.366).--Salix (talk): 04:45, 10 April 2013 (UTC)[reply]

Compare this to a hexagon at 0° which has side length 0.5.--Salix (talk): 05:19, 10 April 2013 (UTC)[reply]

Thanks a lot!Dncsky (talk) 08:04, 10 April 2013 (UTC)[reply]

I agree with Salix - it's the one rotated by 15 degrees. 96.46.201.254 (talk) 04:28, 11 April 2013 (UTC)[reply]

Convergent or Divergent ?

$\lim _{n\to \infty }{\int _{0}^{\infty }{\underbrace {x^{{\color {Red}-}x^{\cdot ^{\cdot ^{x}}}}} _{{\color {Red}2}n}\ dx}\ =\ ?}$

For 2n = 2 we have 1.995⁺. For large values of 2n (up to about 50) we have 1.91⁺. But whether it actually converges (either to 0, or to a value close to 2) or diverges (towards infinity, or because of perhaps possessing an oscilant nature) is beyond me... — 79.113.210.163 (talk) 10:29, 10 April 2013 (UTC)[reply]

Unless I'm misunderstanding your notation, this doesn't converge for n > 1. I assume that for n = 2 the integrand would be x^(-x)^x^(-x). 70.162.4.242 (talk) 16:48, 10 April 2013 (UTC)[reply]

You are misunderstanding my notation. There's a single minus sign there. — 79.113.210.163 (talk) 17:38, 10 April 2013 (UTC)[reply]

In that case it still doesn't converge. 70.162.4.242 (talk) 17:52, 10 April 2013 (UTC)[reply]

The minus sign is not inside any paratheses. — 79.113.210.163 (talk) 18:15, 10 April 2013 (UTC)[reply]

This takes me back to old times. The function $x^{\cdot ^{\cdot ^{x}}}$ is known as tetration, and it has quite remarkable properties. As the number of x's goes to infinity:

If $x<e^{-e}$ , in the limit it oscillates between two asymptotes.
If $e^{-e}<x<e^{1/e}$ , it converges.
If $e^{1/e}<x$ , it diverges.

The result is that if you have an even number of x's, your integral actually does converge. Looie496 (talk) 18:25, 10 April 2013 (UTC)[reply]

I knew that too, I just can't see the connection. — 79.113.210.163 (talk) 19:03, 10 April 2013 (UTC)[reply]

EXPONENTS AND NEGATIVE NUMBERS, Why is (-6)^2=36 but -6^2=-36 ??

I really don't understand. Aren't they both just -6x-6x-6x-6x-6x-6x ? And if so aren't both answers going to be positive for their being an even number of factors? How does the parenthesis change the answer? Please help me understand this, thank you. — Preceding unsigned comment added by 71.142.71.205 (talk) 17:51, 10 April 2013 (UTC)[reply]

See Order of operations in -6^2 the minus sign is applied last so -6^2 is interpreted as -(6^2)=-(36).--Salix (talk): 17:59, 10 April 2013 (UTC)[reply]

... also, you've misunderstood the power of 2 (squared). (-6)^2 is just "(-6) times (-6)". You are correct that any negative number raised to any even power will give a positive answer. Dbfirs 08:33, 11 April 2013 (UTC)[reply]

April 11

Probability

If I have a large number of black and white balls in a container, in a known ratio, then the probability of drawing a (say) black ball is directly related to that ratio. My question is - given that ratio (and a pot of balls sufficiently large) how do I calculate the probability of ending up with at least one black ball after a given number of draws. Or putting it another way, how many draws do I have to make for there to be an 80%, 90%, or other probability of seeing a black ball? Apologies for what is probably either a very basic question, or incorrectly framed - I lack the terminology to search for the answer!

Question relates to the probability of capturing rare events on a digital storage oscilloscope given a timebase and a waveforms/second update rate.

Ic0nwiki (talk) 11:01, 11 April 2013 (UTC)[reply]

If your rare event will not influence future occurrences (equivalent to a very large number of black and white balls) then a simple Binomial distribution is appropriate. For calculations (to save lots of arithmetic), this is often approximated using a Normal distribution with mean np and variance np(1-p) where p is the proportion of black balls, or the probability that the event will occur, and n is the number of draws or trials. (Drawing balls from a limited container without replacement requires a hypergeometric distribution, but this doesn't seem appropriate here.) For your question, it will be easier to calculate the probability that the event will not occur, then subtract from one to get the probablility that it will occur. See Binomial_distribution#Normal_approximation for detail, or ask again here. Dbfirs 11:33, 11 April 2013 (UTC)[reply]

That's tantalisingly close - I've been looking at the binomial theorem page and doing some sums. I started with some simple n, p, k combos - n=some suitably large number, p=0.5, k=1. I couldn't work out where I was going wrong at first - it was giving me implausibly small numbers - until I tried n=1, n=2, ... - I realised that the formula is telling the probability of exactly one success (yes, obvious to you all!), whereas I don't care *how* many successes, so long as n>=1 (n=1, n=2... n=p). Does this make it easier or harder to do? Ic0nwiki (talk) 14:24, 11 April 2013 (UTC)[reply]

I think you missed the part of Dbfirs' answer that said "it will be easier to calculate the probability that the event will not occur, then subtract from one to get the probablility that it will occur." -- if you don't do it that way, then you have to add together a lot of probabilities, e.g. the chance of exactly one success, exactly two,... up to exactly n. So, instead use the formulae to compute the probability of no black balls, then P(at least one black)=1-P(no black). Does that make sense? SemanticMantis (talk) 15:28, 11 April 2013 (UTC)[reply]

Ranking that best satisfies set of orderings

If I have a set of orderings of the form (for example) {a > b, c > b, b > d, ...}, what techniques could I use to determine a ranking of the elements involved that maximises the number of constraints satisfied? Are there any better objective functions I could seek to maximise other than number of constraints satisfied? Thanks in advance for any pointers, I'm finding this quite hard to search for because I'm not sure what terms to use. --81.101.105.36 (talk) 14:39, 11 April 2013 (UTC)[reply]

If you're not careful about how you pose such problems, it's easy to run afoul of Arrow's paradox. Sławomir Biały (talk) 19:20, 11 April 2013 (UTC)[reply]

Your first question is the maximum acyclic subgraph problem. -- BenRG 22:44, 11 April 2013 (UTC)

Thanks, I would never have found that. As it happens, I solved this with a soft-margin SVM by determining ratings for each item that minimise the error of unsatisfied constraints. --81.101.105.36 (talk) 01:22, 12 April 2013 (UTC)[reply]

April 12

Kendall's tau and confidence intervals

What's the best way to calculate confidence intervals for Kendall's tau (specifically tau-b)? I've seen a lot of pages on the web saying it's possible, but none of them necessarily say how. (I guess that might mean that it much more complex than calculating tau itself.) I'm interested in the general formula, but if there's an R package which can calculate it automatically, that would be great. (I did find the "Kendall" package in R, which apparently gives the variance of the numerator for Kendall's tau, but provides little assistance in transforming that into a confidence interval on tau itself.) -- 205.175.124.30 (talk) 03:51, 12 April 2013 (UTC)[reply]