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August 23

Dividing by fractions

I know that to perform the following

${\frac {a}{b}}\div {\frac {c}{d}}$

you convert it to

${\frac {a}{b}}\times {\frac {d}{c}}$

However, I don't understand why inverting the second fraction and then multiplying by it works. Could someone explain? asyndeton 01:10, 23 August 2007 (UTC)[reply]

By definition of division,

{\frac {a}{b}}\div {\frac {c}{d}}=x

means

{\frac {a}{b}}={\frac {c}{d}}*x

To find x, you multiply by d and divide by c:

x={\frac {a}{b}}\times {\frac {d}{c}}

Does that help? PrimeHunter 01:20, 23 August 2007 (UTC)[reply]

Yes it does, I understand now. Thank you! asyndeton 01:34, 23 August 2007 (UTC)[reply]

Another way of looking at it is observing that x / y = x * 1/y, 1/(a/b) = b/a (turn upside down). So (a/b)/(c/d) = (a/b) * 1/(c/d) = (a/b) * (d/c). Voila.

Coordinates shown in science-fiction shows

In many science-fiction shows or movies involving a starship in space, coordinates are said in the format "123 mark 45" (one-two-three mark four-five). How does one interpret this? Does this mean at 123 degrees on the same plane as the starship (assume x-z plane), and 45 degrees in the x-y plane? It seems to be missing one more set of coordinates. Maybe it is divided into grids? Could someone kindly link me to an article that talks about this type of coordinate system? Thanks much! --Taktser 03:52, 23 August 2007 (UTC)[reply]

There's an explanation given on the Star Trek wiki [1] (though I can't vouch for its reliability), and it's explained in more detail and with an illustration on Sector (Star Trek)#Headings and Bearings. In effect, it's pretty much what you described. For a wikipedia article, try Equatorial coordinate system. Also related are Celestial coordinate system, and of course, Spherical coordinate system. nadav (talk) 04:51, 23 August 2007 (UTC)[reply]

Ah I see, so either the ship is moving in a straight line or another object is coming towards in a straight line, that's why a third set of coordinates isn't needed. Thanks a whole bunch! --Taktser 08:11, 23 August 2007 (UTC)[reply]

That's right, a "heading" is a direction only, not a location. Think of it as meaning "head in this direction". StuRat 16:06, 23 August 2007 (UTC)[reply]

The fact is that there is no canonical explanation of the system used on Star Trek. It is never explained in any of the shows or movies. The explanation given at Sector (Star Trek)#Headings and Bearings describes a system that would be sensible, and I'm guessing that it comes from some external source known to fans. But I do not believe it was used consistently; for one thing, at least on the older shows, the word "mark" was practically always followed by a single digit. I think it started as something that the writers made up so it would sound good, just like "star dates", which originally were essentially random numbers used so that the writers wouldn't have to worry about date issues and yet it would sound good. This changed later for star dates and maybe it did for bearings too. --Anonymous, August 24, 2007, 06:03 (UTC).

Hilbert Curve

Is it possible to (and if so is there an easy equation for) converting a percentage distance along a "unit square" infinite hilbert curve into an x,y coordinate. Resolution obviously plays a part in the accuracy of the result. In our article anything in the first 25% is in the top left which will extrapolate to 0% being the very top left point. Beyond this I get stuck. -- SGBailey 08:06, 23 August 2007 (UTC)[reply]

Here is a way of doing it:

function unzip (z, n):

if n ≤ 0: return (0.5, 0.5);

z := 4×z;

(q, z) := (floor(z), z mod 1);

(x, y) := unzip (z, n–1);

case q of:

0: return (y/2, x/2);

1: return (x/2, (y+1)/2);

2: return ((x+1)/2, (y+1)/2);

3: return ((2–y)/2, (1–x)/2);

The input is a number z in the range 0 ≤ z < 1, and a control parameter n, which is a natural number controlling the depth of recursion. The output is a pair of numbers (x, y), each in the range between 0 and 1, and with a precision of 2⁻⁽ⁿ⁺¹⁾. I start the curve at (0, 0) and let it go to (1, 0); other choices are easily accommodated. --Lambiam 08:57, 23 August 2007 (UTC)[reply]

Thanks -- SGBailey 11:30, 23 August 2007 (UTC)[reply]

How about the reverse? Given the coordinates (provided that both are dyadic rationals), what function gives the curve length? And while we're up, can both functions be extended to 3 or more dimensions? —Tamfang 03:36, 2 September 2007 (UTC)[reply]

Statistical representative

How many samples you need to include in an analysis to be statistical representative?

(first, don't forget to sign your posts) Short formally correct answer: 1, if your randomization is acceptable. In the sense that the (unconditional) sampling distribution of this one is the same as the underlying distribution of what you are sampling from.

Longer more useful answer: Since the properties of the underlying distribution cannot be inferred at all (unless you do something Bayesian), it is hard to justify a sample of 1. What do you mean by "statistical representative" exactly, in terms of usefulness? Baccyak4H (Yak!) 14:25, 23 August 2007 (UTC)[reply]

Whether a sample is representative of the whole population is not so much an issue of the size of the sample, but of the way its individuals have been selected. If the selection method is biased, so that certain aspects are overrepresented or underrepresented, then the sample is, statistically, unrepresentative. Of course, if the sample is very small, things tend to get underrepresented or overrepresented. In a sample consisting of one person, the genders are not evenly represented.

However, the main reason for having a large sample size is that it is hard to obtain significant conclusions from a small sample. Suppose that you have a sample of 10 people, you ask how many have read Finnegans Wake cover to cover, and the answer is: nobody. What can you conclude for the whole population? At the 95% confidence level, not much better than that three out of four people haven't read the book cover to cover: the probability of getting "no" then 10 times in a row is about 5%. That is not a very strong conclusion, but given the small sample size you can't do better. --Lambiam 15:07, 23 August 2007 (UTC)[reply]

The figure I look for is 1100. That's enough to provide a 3% margin of error over a 90% confidence interval. If a sample size is at least that large, and the sample appears to be blind and unbiased, I actually pay attention to it. Otherwise, I consider the results to be crap. StuRat 16:00, 23 August 2007 (UTC)[reply]

When was the last time you saw a poll or study that you considered valid? I'm not sure I've ever seen a poll with over 1000 respondents. Tesseran 19:57, 23 August 2007 (UTC)[reply]

I've seen them quite often; in medical studies, election polls, etc. The fact that many newspapers and other unprofessional pollsters fail to use a large enough sample size to provide accurate results doesn't mean we should accept such invalid samples. StuRat 01:23, 24 August 2007 (UTC)[reply]

It all depends on whether proper account is taken of the confidence intervals. If 34 respondents in a representative sample of 48 voters prefer candidate A over candidate B, then it is a fair conclusion that a majority of voters prefer A over B. It is wrong to state that the poll has shown that over 70% prefer A over B. Some journalists misrepresent such results out of innumeracy; that makes their reports invalid, not the polls they report on. --Lambiam 04:39, 24 August 2007 (UTC)[reply]

First, I doubt if it is possible for such a small sample to be representative, in that some ethnic, religious, class, employment, income, geographic or language groups are sure to be entirely excluded from the sample. For example, if some subgroup represents 1% of the population, it would either not be represented at all or be over-represented as over 2% of the sample. There are likely to be many such small groups which collectively represent a substantial portion of the population. Second, even if the sample was representative, mathematically it would still be possible for 34 of 48 to say they prefer A even though the majority of the population actually prefers B. StuRat 13:31, 24 August 2007 (UTC)[reply]

What you say is still true if you replace 34 and 48 by 34,000,000 and 48,000,000, so your conclusion should be that no sample ever is representative. The subgroup issue is a red herring, in that it does not invalidate the conclusion: (with high confidence) a majority of voters prefer A over B. The possibility of absence or underrepresentation of small minorities does not alter the probabilities involved. --Lambiam 16:17, 24 August 2007 (UTC)[reply]

If 34 million people of a 48 million sample size in an unbiased, fair poll say they prefer one candidate, the chances that the majority of the population prefers the other candidate is incredibly small. This is not true if 34 of 48 say they prefer one candidate. This is because random effects tend to even out in larger samples. Also, note that the minorities, when viewed collectively, may very well form the majority of the population (Blacks + Hispanics + Asians + Catholics + Jews + homosexuals, etc.), so representing their views accurately is important. StuRat 04:22, 25 August 2007 (UTC)[reply]

The probability is indeed small, but mathematically it would still be possible for 34,000,000 of 48,000,000 to say they prefer A even though the majority of the population actually prefers B. If the majority of the population does not prefer A over B, then the likelihood of finding a preference as strong as 34 for A out of a randomly selected sample of 48 is less than 0.3%. While not vanishingly small, it is small, small enough to allow one to make certain pronouncements with fairly high confidence. The only criterion for the validity of "less than 0.3%" is that there be no selection bias with regard to preference for a particular candidate; as long as that is fulfilled, the selector may have preferentially selected Hispanic Jewish homosexuals at their heart's content: for it is immaterial to the conclusion. This is equally true if the whole population consists of nothing but minorities of one. In this context "representative" means "not having a selection bias within the population with regard to the aspect(s) under investigation", neither more nor less, and the size of the sample is only relevant for the confidence levels with which findings can be extrapolated to the population. --Lambiam 06:47, 25 August 2007 (UTC)[reply]

0.3% seems a bit low, to me, may I see the math ? I also have a hard time imagining any situation where such a small sample is fair/unbiased. StuRat 18:26, 25 August 2007 (UTC)[reply]

[killing indents] StuRat, it seems like you're confused about what "unbiased" means. Say you want to poll a precinct which has exactly 100 voters, but you only have the resources to talk to 10 of them. Despite the fact that this precinct is packed chock-full of minorities (in fact there isn't a single straight white male in town), here's a way to guarantee a statistically unbiased sample. Number the voters from 1 to 100. Choose ten numbers from that range at random. Poll the voters whose numbers you chose. (Lambiam has addressed your questions about minority groups, so I'll just note that if minorities really do make up the majority of your sample, then you don't have to worry that a representative sample will exclude them, right?) Tesseran 20:46, 25 August 2007 (UTC)[reply]

Perhaps "bias" isn't the term for it, but if you had only 10 in a sample you would have a certain "lumpiness" to your sample. Let's say the population breaks down as follows:

47 white      52 female  61 Protestant
23 black      48 male    18 Catholic
18 Hispanic              13 atheist
 3 Asian                  8 other
 9 other

If, by some astounding luck, you ended up representing each group as closely as possible in your sample of 10, you would end up with:

50 white      50 female  60 Protestant
20 black      50 male    20 Catholic
20 Hispanic              10 atheist
 0 Asian                 10 other
10 other

Now, if those different groups had significantly different opinions, you would have significantly miscalculated the "average opinion", even if you managed to get the "typical answers" from each member of each group you polled. The reality, though, is that in randomly selecting such a small number of people, you are likely to severely overrepresent or underrepresent just about every subgroup, which will have a profound effect on your poll results. StuRat 23:22, 26 August 2007 (UTC)[reply]

Going at it the other way; my experience has been that statistically knowledgable folks will not pay much attention to any sample with less than 20-30 as a rule of thumb even if everything about it appears purely correct and unbiased. This lumps "no information available or sample size less than 20 or more than 20% of observations with missing information" all together as unknown. Here's a guy who recommends 12 but that includes real world things like cost. Gzuckier 17:14, 23 August 2007 (UTC)[reply]

Exponential function

So I know that ${\frac {a^{x}}{a^{y}}}=a^{x-y}$

But what do I do with the following?

${\frac {e^{-3x}}{2e^{-x}}}$

That factor of 2 in the denominator has completely ruined my day. I can't see how it simplifies to ${\frac {e^{-2x}}{2}}$ . Break it down one time!!—Preceding unsigned comment added by 86.144.56.62 (talk • contribs) 11:25, August 23, 2007

This can be expressed as

\left({\frac {1}{2}}\right)\left({\frac {e^{-3x}}{e^{-x}}}\right)

. Simplify the exponents as usual and recombine the fractions. Strad 18:42, 23 August 2007 (UTC)[reply]

Ah yase, because

{\frac {a}{b}}\times {\frac {c}{d}}={\frac {ac}{bd}}

, thus you can have a 1 in front of

e^{-3x}

and factorise! Simple! I wish I was dead

Sorry, the Reference Desk cannot help with that. :-) --Anonymous, August 24, 2007, 06:05 (UTC).

Oh, oh, I know! Cause it would be considered medical advice, right? —Bromskloss 09:40, 24 August 2007 (UTC)[reply]