Let's discuss Mathematics

[azzaman333]

Not certain, but I would guess it's because the confidence interval is taken from a random sample of the population at a whole. So depending on the specific sample you get, you could get vastly different means and standard deviations and therefore confidence intervals. You can't strictly say that the specific CI you've calculated has a 95% chance of containing the true mean, as that may not be a property that is strictly true to the calculated interval.

I'm sure I remember being given a good reason for it at some point during my undergrad, whether or not this is right I can't recall though.

 

[Atticus]

Thanks for the input!

To use your deck of cards example, if you know that the deck is a normal 52 card deck well shuffled you could say that P(red card) = 0.5 with zero confidence interval.

I'm not sure we're talking about the same thing here. That's an a priori probability for the card being red, not a result of a statistical investigations.

The whole card example wasn't meant to be about the CI either, but to make the point about the power of statistics and probabilites: while the card is unknown, it makes sense to assign a probability to it's colour, although strictly speaking it either is or is not red. Similarly, while the paratmeter you're trying to estimate by statistical inference is unknown, it makes sense to say that it has 95% probability on being in the interval.

 

[Rashiminos]

Not certain, but I would guess it's because the confidence interval is taken from a random sample of the population at a whole. So depending on the specific sample you get, you could get vastly different means and standard deviations and therefore confidence intervals. You can't strictly say that the specific CI you've calculated has a 95% chance of containing the true mean, as that may not be a property that is strictly true to the calculated interval.

I'm sure I remember being given a good reason for it at some point during my undergrad, whether or not this is right I can't recall though.

In a sense you could consider the population parameter to be a random variable** with a probability density function, and then construct an interval for which the integral of that function over that interval is not necessarily .95, but the process which generates that interval has at least a 95% chance of generating an interval that contains the value of the variable for this specific population.

With X as a random variable; constants aci, bci
Compare:
P( aci < X < bci), *

versus

P{Generate an interval (a,b) containing X}

_______________________________________________

* or P (aci < X < bci | sample data)

**Something more often done in the context of Bayesian inference where the variable represents uncertainty in the knowledge of the parameter. I wouldn't consider myself sufficiently versed in credible intervals to discuss them here.

Now, I accept that the interval either contains or doesn't the right value. But if the value is unknown, isn't it "philosophical nitpicking" to say that it's incorrect to say that there's a 95% chance it does?

If you don't consider a random variable approach like the above, then the parameter itself is not random, so saying the probability of it lying in the observed confidence interval is 95% {P(in the interval)=.95} is problematic. Saying it is in the interval or not seems to imply P(in the interval)=1 XOR P(in the interval)=0 . The 95% probability of a confidence interval has to do with the sample estimate and not the parameter.

it makes sense to say that it has 95% probability on being in the interval.

More succinctly, it shouldn't make sense:

Couldn't you similarly say that Brazil either winds or doesn't win the world cup 2014? So assigning a probability value to it would be incorrect.

How would you model the probability of Brazil winning the cup? Would it be like flipping a (biased) coin?

In a typical coin model, the probability of getting heads on a fair coin is 50%. We can simulate flipping a coin a number of times to estimate the probability. Suppose we get a 95% confidence interval from a sample that is between 35% and 43%. What sense does it make to you to say that the true probability of getting heads (assumed to be 50%) has a 95% chance of lying within our interval between 35% and 43%?

 

[Mise]

This is cool: http://plus.maths.org/content/non-transitiv-dice

So there's a set of 3 dice, A, B and C, but the numbers on the dice are set up such that, even though they still average out to 3.5, dice A beats dice B, B beats C, and C beats A.

 

[dutchfire]

If you want to play a dice game, but don't like chance, you could do the following:

Every player gets an empty dice and is allowed to put 21 dots on there. You calculate which die will win according to probability theory, and this player wins the round.

(I assume for every die A with 21 points, there is a die B with 21 points with B>A.)

 

[timtofly]

How would that work in playing Civ games?

 

[PlutonianEmpire]

I have a trigonometry question about space. If I know the distance in light years and orbital separation in astronomical units of a binary star, how do I find out the angular separation between the two in arcseconds?

 

[Timsup2nothin]

I have a trigonometry question about space. If I know the distance in light years and orbital separation in astronomical units of a binary star, how do I find out the angular separation between the two in arcseconds?

You don't have enough information. You need to know the angle between the line connecting the two stars and the line of sight from Earth. If they are lined up they will have no angular separation at all, while if that angle is ninety degrees the angular separation will be maximized.

 

[Samson]

I have a trigonometry question about space. If I know the distance in light years and orbital separation in astronomical units of a binary star, how do I find out the angular separation between the two in arcseconds?

WARNING: I am not an expert (or even close).

I do not think it matters, if you are asking what the maximum seperation a star system will have. I think that whatever the angle of the plane of rotation is, there will be a point that they are seperated by that distance relative to earth. So it becomes simple trignometry, angle = atan(diameter of rotation/distance from earth). For example, from here Eta Cass is 19 light years (1.8e17m) away, and seperated by 70 AU (1.0e13m), diameter 35 AU (5e12). So the angle between them is atan(5e12/1.8e17) = 2.9e-05 radians = 5.7 arcseconds.

I will re-interate that I could be wrong at any point in this working. I have left more significant figures in my calculation than I have shown.

 

[Timsup2nothin]

WARNING: I am not an expert (or even close).

I do not think it matters, if you are asking what the maximum seperation a star system will have. I think that whatever the angle of the plane of rotation is, there will be a point that they are seperated by that distance relative to earth. So it becomes simple trignometry, angle = atan(diameter of rotation/distance from earth). For example, from here Eta Cass is 19 light years (1.8e17m) away, and seperated by 70 AU (1.0e13m), diameter 35 AU (5e12). So the angle between them is atan(5e12/1.8e17) = 2.9e-05 radians = 5.7 arcseconds.

I will re-interate that I could be wrong at any point in this working. I have left more significant figures in my calculation than I have shown.

If you are going for a maximum value this seems correct. It does seem like there will be that point no matter what the plane of rotation, though I'd hate to have to prove it.

Edit: Okay, proving that would not be that bad. So if you are looking for the maximum potential angular separation you can find it.

 

[Harv]

I do not think it matters, if you are asking what the maximum seperation a star system will have. I think that whatever the angle of the plane of rotation is, there will be a point that they are seperated by that distance relative to earth. So it becomes simple trignometry, angle = atan(diameter of rotation/distance from earth). For example, from here Eta Cass is 19 light years (1.8e17m) away, and seperated by 70 AU (1.0e13m), diameter 35 AU (5e12). So the angle between them is atan(5e12/1.8e17) = 2.9e-05 radians = 5.7 arcseconds.

From the same source:

Eta Cass 11.9" 70 AU 478 yrs 19 LY

The short answer to the question is:

a = 3.26 x / D, where:

a is the angle between the stars, in arcseconds
x is the distance between the stars, in AU
D is the distance from the Earth to the binary system, in Light Years

So if I take 3.26 times 70, divided by 19, I get 12.0 arcseconds. The reference says 11.9.

Here is my derivation:

According to the Wikipedia article on "light-year," a light-year is 63241 AU.

If I assume the distance to the binary star system is much greater than the separation between them, then the tangent of the angle a is approximately equal to the sine of the angle, which is approximately equal to the angle in radians.

A ~= sin(a) ~= tan(a) = x / D / 63241, where A is the angle between the stars in radians.

a = 60 * 60 * 180 * A / pi = 206265 A

a = 206265x / D / 63241 = 3.26x / D

Also note that 3.26 is also the number of light-years in a parsec.

 

[Robert FIN]

I have a maths test tomorrow and I can't figure out these two examples. I don't know math terms in english but I think the task is to divide the thing in 'factors' which basically means that there shall not be any + or - outside ( ) signs. Let's say e=power, for example "3 e2" is "3 power 2" which is 3×3=9.

1) Divide to factors: (1 - x)e2 + (1 - x)e3

2) Same task: (&#960; - 4)e2 + &#960; - 4

Thanks

 

[Harv]

I don't know math terms in english but I think the task is to divide the thing in 'factors' which basically means that there shall not be any + or - outside ( ) signs.

Is this what you mean? It does not directly answer your question, but it might make the question more clear.

EDIT: I am editing in solutions with the assumption that I correctly understood the question:

Problem 1: Find the factors for (1-x)e2 + (1-x)e3

A. I determined f(x) = (1-x)e2 + (1-x)e3 to be equal to -xe3 + 4xe2 - 5x + 2 which we are solving to equal 0. We already know x=1 works, so:

B. Using long division, divide (-xe3 + 4xe2 - 5x + 2) by (x-1) to get -xe2 + 3x - 2. We can solve this for zero with the quadratic formula.

C. The quadratic formula gives solutions for x = 1 and x = 2.

D. There are three roots for this formula. x = 1, x = 1, and x = 2. Note that x = 1 is repeated. This means the root x = 1 has a multiplicity of 2, or that it is used two times.

E. The function (1-x)e2 + (1-x)e3 can therefore be expressed as: (x-1)(x-1)(x-2)
EDIT: This should be -(x-1)(x-1)(x-2) or (1-x)(1-x)(2-x)

F. Check the work. The function above multiplies out to xe3 - 4xe2 + 5x -2.
EDIT: The edit will work out to -xe3 + 4xe2 -5x +2.

Problem 2: Same task: (&#960; - 4)e2 + &#960; - 4

A. This function expands out to ne2 - 7n + 12. We can solve this for zero with the quadratic formula.

B. The roots of this function are 3 and 4. You can use n = 3 and n = 4 to get 0. Therefore:

C. We can rewrite the function as (n-3)(n-4).

D. Check work: This expands out to ne2 - 7n + 12.

 

[CKS]

I would solve this differently. I'm using * for times.

Problem 1:
We have (1-x)e2 + (1-x)e3, which is (1-x)e2 + {(1-x)e2 * (1-x)}. Each term has (1-x)e2 in it, so I factor it out to get (1-x)e2 * {1 + (1-x)} or (1-x)e2 * (2-x). (Notice that Harv dropped a minus sign - F and A are opposites of each other.)

Problem 2:
(&#960; - 4)e2 + &#960; - 4 has n-4 in each term. Factor it out to get (n-4) * {(n-4) + 1}, so we have (n-4)* (n-3).

 

[Harv]

I would solve this differently. I'm using * for times.

Yes, this is simpler. My mentality has been to find a brute-force solution first, and a simpler solution second.

(Notice that Harv dropped a minus sign - F and A are opposites of each other.)

This is correct. I re-constructed the formula based on the roots and I was only thinking of rewriting the formula f(x)=-f(x)=0. I will edit in a correction.

 

[sanabas]

Is this what you mean? It does not directly answer your question, but it might make the question more clear.

Not factor theorem. Just basic finding the common factor. i.e. x^2y + xy^2, common factor is xy, so you get xy(x+y). Or 8x^2 + 6x, common factor is 2x, so 2x(4x+3).

These questions are the next step from that, because the common factor isn't just xy or 2x, but is in the form (a+b), which makes it a bit trickier to divide each term by the common factor.

So (1-x)^2 + (1-x)^3, common factor is (1-x)^2. So you get [(1-x)^2](1 + (1-x)) = (2-x)(1-x)^2

(n-4)^2 + (n-4), common factor is (n-4), so (n-4)((n-4)+1) = (n-4)(n-3)

*edit* and already answered over the page. oops. */edit*

 

[Robert FIN]

^^ Whoah, thanks to 3 of you! Makes sense now. Those example "problems" were really difficult to my level. Thanks.

And those are absolutely correct, checked the answers. I will send you more if there will be more weird examples in the future. Thanks.

 

[Robert FIN]

Hey, here's a new problem:

Have a pair of equations (I hope this is the right term...) but I don't know how to solve it the smart way (this one is easy to count without any smart ways too).

x + y = 20
x * y = 91

If I multiply another one with negative number, it makes everything to go zero... Is this even possible to do? And is it even right to have * marks in that kinda problem?

 

[dutchfire]

x = 20-y
(20-y)*y = 91
-y^2+ 20y - 91 =0
y=7 or y=13
x=13 or x=7

 

[Robert FIN]

^ Yes, cool.

The first move was the thing I was missing. Thanks.

And the answer is right.