This isn't fair guys and girls!
I try my best at giving you maths problems to solve, and making sure that the game is fair for everyone.

Please don't overreact. (Lets not ruin this thread, otherwise I'll be dumping it)

Luke.

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- Luke.

This isn't fair guys and girls!
I try my best at giving you maths problems to solve, and making sure that the game is fair for everyone.

Please don't overreact. (Lets not ruin this thread, otherwise I'll be dumping it)

Luke.

Hey, Luke

All in the best possible taste (to quote the late {great?} Kenny Everett) at this end.

Keep smiling :))

F.

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Everyone cooled down?

I don't know how a bottle can be heated up that quick. I didn't invent the question...

JD wins Q20...

Standings:
1. Jason Gee - 5 1/2 Points
2. Fred - 4 Points
3. BeefDog - 2 1/2 Points
4. JDWhale - 2 Points
5. Gas Giant - 1 Point
6. Dominique - 1 Point
7. Zach Parker - 1 Point
8. TBD...

Question 21 (1 Point): A gold mining company is testing locations for its next mine. From location A eight samples were taken of units of gold per ton of ore. The results were 1.23, 1.42, 1.41, 1.62, 1.55, 1.51, 1.60, and 1.76 . From location B six samples were taken with the following results 1.76, 1.41, 1.87, 1.49, 1.67, and 1.81 . It is assumed that the amount of gold in a sample in either location have a standard normal distribution with a fixed, yet unknown, mean and variance, and that the variance in both locations is equal. Test the hypothesis that the mean gold content of both locations is equal. Use a 10% level of significance, in other words assume that if the two means were the same the test would pass 90% of the time.

(Q22 Preview: Anyone up for a bit of calculus? ;)

Luke.

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- Luke.

Question 21 (1 Point): A gold mining company is testing locations for its next mine. From location A eight samples were taken of units of gold per ton of ore. The results were 1.23, 1.42, 1.41, 1.62, 1.55, 1.51, 1.60, and 1.76 . From location B six samples were taken with the following results 1.76, 1.41, 1.87, 1.49, 1.67, and 1.81 . It is assumed that the amount of gold in a sample in either location have a standard normal distribution with a fixed, yet unknown, mean and variance, and that the variance in both locations is equal. Test the hypothesis that the mean gold content of both locations is equal. Use a 10% level of significance, in other words assume that if the two means were the same the test would pass 90% of the time.

Darn, don't have the time to go right through the calculations for that, but it looks like it passes a t-test for equal means, of different sample size but equal variance (perhaps someone else who has time to do the calculations properly should get the point though). So yes they are considered equal gold content in both locations.

(Q22 Preview: Anyone up for a bit of calculus? ;)

Right out of practice there (~14 years) but you never know, as long as it was easy maybe ;D

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"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Well, it's 2:15 a.m. here.
I just grabbed my 5 Probability & Statistics related books.
If no one's tried it by the time I wake up, I'll give this one a shot.

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Capitalize on this good fortune, one word can bring you round ... changes.

Well, it's been a while since I've had to work out this type of problem, either a student or a teacher, but, here goes.

With calculations below.
The null hypothesis was that the two means were equal. The null hypothesis must be rejected. If I recall correctly, we can say with 90% confidence that the mean gold content of both locations is not equal.
The formulas used here come from Mathematical Statistics (5th edition), by John E. Freund. (The fact that the two random samples are small, from normal distributions with the same unknown variances.) Formulas were entered on the software package MAPLE.

> with(stats):
>
Warning, these names have been redefined: anova, describe, fit, importdata, random, statevalf, statplots, transform

> sample[1]:=[1.23, 1.42, 1.41, 1.62, 1.55, 1.51, 1.60, 1.76];
> sample[2]:=[1.76, 1.41, 1.87, 1.49, 1.67, 1.81];
>

sample[1] := [1.23, 1.42, 1.41, 1.62, 1.55, 1.51, 1.60, 1.76]


sample[2] := [1.76, 1.41, 1.87, 1.49, 1.67, 1.81]

> sample_mean[1]:=describe[mean](sample[1]);
> sample_mean[2]:=describe[mean](sample[2]);
>

sample_mean[1] := 1.512500000


sample_mean[2] := 1.668333333

> s[1]:=describe[standarddeviation](sample[1]);
> s[2]:=describe[standarddeviation](sample[2]);
>

s[1] := .1503121752


s[2] := .1671742272

> n[1]:=8;
> n[2]:=6;
>

n[1] := 8


n[2] := 6

> s[p]:=sqrt(((n[1]-1)*s[1]^2+(n[2]-1)*s[2]^2)/(n[1]+n[2]-2));
>

s[p] := .1575574925

> t:=simplify((sample_mean[1]-sample_mean[2])/(s[p]*sqrt((1/n[1])+(1/n[2]))));
>

t := -1.831377608

> alpha:=0.10;
>

alpha := .10

> alpha/2;
>

.05000000000

> degrees_of_freedom:=n[1]+n[2]-2;

degrees_of_freedom := 12

>

|t| = 1.831377608 > 1.782 = t[0.05,12] = t[alpha/2,degrees of freedom].

The null hypothesis was that the two means were equal. The null hypothesis must be rejected. If I recall correctly, we can say with 90% confidence that the mean gold content of both locations is not equal.

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Capitalize on this good fortune, one word can bring you round ... changes.

Hmmm,
That's odd, I get t = -1.75129 which passes the test against 1.782 (Just), meaning they'd be equal. Oh well, I'm probably using the wrong formula(usual story).
Specifically I'm using the test of equal means one to get my t:
t = (Xa-Xb)/ SQRT( ( Na*Sa +Nb*Sb)/(Na+Nb))) , X being the mean, N being the count, and S representing the variance (normally written as s-squared)

Jason

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"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Hmmm,
That's odd, I get t = -1.75129 which passes the test against 1.782 (Just), meaning they'd be equal. Oh well, I'm probably using the wrong formula(usual story).
Specifically I'm using the test of equal means one to get my t:
t = (Xa-Xb)/ SQRT( ( Na*Sa +Nb*Sb)/(Na+Nb))) , X being the mean, N being the count, and S representing the variance (normally written as s-squared)

Jason

The difference is in where we summed the total from the two samples. I have this, but 2 is subtracted from it. Like I said, it's been awhile, but notice how that is the same as the degrees of freedom. Why should these be the same, and why use n[1] + n[2] - 2 for the degrees of freedom? I don't recall. I'm just guessing that it is 1 off from each sample size.

[EDIT 1] Na*Sa +Nb*Sb vs. Na * Sa + Nb * Sb - 2.
[EDIT 2] There's another difference. sqrt (1 / Na + Nb) vs. sqrt((1/Na) + (1/Nb)).

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Capitalize on this good fortune, one word can bring you round ... changes.

Ah, well It'll be interesting to see, been spending time with simultaneous equations for circuit analysis of late, so haven't used much probability.

---

"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Sorry Sarge, Jason Gee is correct. The tests are the equal...

Standings:
1. Jason Gee - 6 1/2 Points
2. Fred - 4 Points
3. BeefDog - 2 1/2 Points
4. JDWhale - 2 Points
5. Gas Giant - 1 Point
6. Dominique - 1 Point
7. Zach Parker - 1 Point
8. TBD...

Question 22 (1 Point) : Calculus required. At dark is thrown at a dart board of radius 1. The dart can hit anywhere on the board with equal probability. What is the mean distance between where the dart hits and the center?

Luke.

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- Luke.

The probability to hit a circle with radius r is 2*pi*r*dr/(pi*r0^2) = 2xdx, where r0=1 radius of the board, x = r/r0.
So the mean distance is R = 2*r0/3 = 2/3

---

Looks right to me.

---

"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Sorry Sarge, Jason Gee is correct. The tests are the equal...
....

Hi Luke, I've spotted the problem with that gold sampling question and think Sarge should get at least a 1/2 point. It states:"It is assumed that the amount of gold in a sample in either location have a standard normal distribution with a fixed, yet unknown, mean and variance, and that the variance in both locations is equal..."

To get the 'right' answer, I calculated the variance independently in both locations, for the 'test for equal means':

Location A variance = 0.018626667
Location B variance = 0.033536667

clearly different (well depending on if there's a 'test for equal variance' I'm unaware of... LoL.)

Assuming equal variance implies selecting a different test formula, which is probably what Sarge has done. I was lucky enough to choose to ignore the bolded part of the question in favour of the 'simpler' test.

---

"Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions.

Sorry Sarge, Jason Gee is correct. The tests are the equal...
....

Hi Luke, I've spotted the problem with that gold sampling question and think Sarge should get at least a 1/2 point. It states:"It is assumed that the amount of gold in a sample in either location have a standard normal distribution with a fixed, yet unknown, mean and variance, and that the variance in both locations is equal..."

To get the 'right' answer, I calculated the variance independently in both locations, for the 'test for equal means':

Location A variance = 0.018626667
Location B variance = 0.033536667

clearly different (well depending on if there's a 'test for equal variance' I'm unaware of... LoL.)

Assuming equal variance implies selecting a different test formula, which is probably what Sarge has done. I was lucky enough to choose to ignore the bolded part of the question in favour of the 'simpler' test.

Even if the means and variances were the same, the sample means and sample variances could differ. But, I did use a formula that made the assumption of equal but unknown variances as stated in the problem conditions.

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Capitalize on this good fortune, one word can bring you round ... changes.

Worms Do Calculus to Find Food
Like humans with a nose for the best restaurants, roundworms also use their senses of taste and smell to navigate. And now, researchers may have found how a worm's brain does this: It performs calculus.

Worms calculate how much the strength of different tastes is changing - equivalent to the process of taking a derivative in calculus - to figure out if they are on their way toward food or should change direction and look elsewhere, says University of Oregon biologist Shawn Lockery, who thinks humans and other animals do the same thing.

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The probability to hit a circle with radius r is 2*pi*r*dr/(pi*r0^2) = 2xdx, where r0=1 radius of the board, x = r/r0.
So the mean distance is R = 2*r0/3 = 2/3

Luke, your response? :)

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Capitalize on this good fortune, one word can bring you round ... changes.

The probability to hit a circle with radius r is 2*pi*r*dr/(pi*r0^2) = 2xdx, where r0=1 radius of the board, x = r/r0.
So the mean distance is R = 2*r0/3 = 2/3

Luke, your response? :)

Luke?

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Capitalize on this good fortune, one word can bring you round ... changes.

The probability to hit a circle with radius r is 2*pi*r*dr/(pi*r0^2) = 2xdx, where r0=1 radius of the board, x = r/r0.
So the mean distance is R = 2*r0/3 = 2/3

Luke, your response? :)

Luke?

Sob!!! I think he has deserted us...

Or, maybe, he takes a break sometimes?

F.

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Hello Everyone! I do not want to desert you at all! But, I have family issues that take priority over SETI and they need sorting out. Fingers crossed I'll have a bit of time in a few days...

Cya All!
Luke.

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- Luke.

Hello Everyone! I do not want to desert you at all! But, I have family issues that take priority over SETI and they need sorting out. Fingers crossed I'll have a bit of time in a few days...

Cya All!
Luke.

Just like Boinc, we run at lowest priority - life has to be lived (outside the boards).

Whenever other things ease up, Luke. This is a fun thread and we'll be waiting.

F.

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