a crucial maths problem (solved... thank god)

[Tuomas]

I'm so close to killing myself right now...

Serisously, I just chewed my headphones broken because I hate everything that moves or doesn't move at the moment. why? because I can't solve this bloody thing! So I thought maybe someone could help a bit. So, I need to count the standard error (of the mean)... WEll, here's the function or what'sit.
Soo... well, the baisc thing is, that I have 672 pupils, all got notes in between 0 and 5 as following

noteÃ,  Ã,  amount of notes(x)
0Ã,  Ã,  Ã,  Ã,  Ã, 100
1Ã,  Ã,  Ã,  Ã,  Ã, 168
2Ã,  Ã,  Ã,  Ã,  Ã, 180
3Ã,  Ã,  Ã,  Ã,  Ã, 164
4Ã,  Ã,  Ã,  Ã,  Ã,  41
5Ã,  Ã,  Ã,  Ã,  Ã,  19
---------------------
Ã,  Ã,  Ã,  Ã,  Ã,  Ã, 672

So here... well,

the figure n stands for the amount of the observations, that is 672 if I'm not wrong.

the X_i is a single observation, (a note, 0,1,2...,5)

and the X with the ¯ is the average note, that is 1,90.... I have count it already and it was correct, so that I'll give to you.

Well, I don't get it, should I really give x the values from 1 to 672? I really don't get anything...

Well, basically I can't finish the course if I don't get this right!

[fred]

Try this, although be warned, I might be completely wrong

It's basically a loop:
You add allÃ,  the squares of the differences between achieved and average scores and divide the result by the number of observations. Then you get the square-root of the result...

I think... someone please help us out here...

EDIT: ehm, right

[Fabiano]

Im a bit rusty at math, but as far as I can see, teh E(xi-x)2 means:

(note from pupil 1 - average) 2
(note from pupil 2 - average) 2
(...)
(note from pupil 672) - average) 2

Guess that it you wanted, right?

rememberin that the X with a - means the statistic average note (that you get with a statistic calc)

[Kweepa]

Since there are only 6 different "notes", you can calculate the summation very easily as follows.
x_i = 0 (x_i - x_bar) = -1.9 (x_i - x_bar)^2 = 3.61 sum of 0s = 100*3.61 = 361
x_i = 1 (x_i - x_bar) = -0.9 (x_i - x_bar)^2 = 0.81 sum of 1s = 168*0.81 = 136.08
x_i = 2 sum of 2s = 180*(2 - 1.9)^2 = 1.8
x_i = 3 sum of 3s = 164*(3 - 1.9)^2 = 198.44
sum of 4s = 41*(4 - 1.9)^2 = 180.81
5s = 182.59
total sum = 878.13
sample variance = sum/(n-1) = 878.13/(672-1) = 1.308
standard deviation (SD) = sqrt(variance) = 1.144
standard error = SD/sqrt(n) = 0.0441
So basically you are giving x_i all 672 values, but shortcutting the calculations, since most of them are the same.

[EDIT] Whoops!

[Ashen]

I think you missed the Sum of the 5s out of the total sum, Steve.

So the last bit should be:
Total Sum = 1060.72
Sample Variance = sum/(n-1) = 1.581
SD                                           = 1.257
Standard Error                        = 0.0485

[Darth Mandarb]

Anybody else want to join the "man I suddenly feel dumb" club with me??

[Tuomas]

So the moment I wrote this I got desparate, called my friend and we went out for a drink. Then I came home and checked this out, and hey, those guys have the correct answer. I can't even say how thankful I am right now, I was so frustrated. Well, thanks to you , I'll be passing this course... that is if I manage to get through the test, but that's later.

THanks... I'll now take my time and figure out what exactly it was that you did there

PS. Darth, do I need to join anymore?

[Nikolas]

Anybody else want to join the "man I suddenly feel dumb" club with me??

Yes! ME!

And to think that once I was studying physics in the uni :p Now all I can think about it boobies...

[MarVelo]

Wow, and i feel special for getting an A in Advanced Algebra 2....