studiot

Both the full population and the sample have a mean and a variance (or standard deviation). There is no reason for these two parameters to be the same in both the population and the sample or between two samples unless the sample size equals the whole population. If we take the variance of the sample to be [math]{T^2} = \sum {\frac{{{{\left( {{X_i} - \overline X } \right)}^2}}}{n}} [/math] We would wish it to be equal to the variance of the population, [math]{\sigma ^2}[/math] However some several lines of algebra shows it to be actually equal to [math]{\sigma ^2} - \frac{{{\sigma ^2}}}{n} = \frac{{n - 1}}{n}{\sigma ^2}[/math] So if we 'correct' this deficiency by multiplying this by [math]\frac{n}{{n - 1}}[/math] we obtain the wanted equality. You can see that Bessel's correction is equivalent to using (n-1) instead of n in the calculation of sample variance. Do you really want the algebra proof?

Thank you , SJ, That is what I usually do. But someone told me that this is the 21st century, and I naively thought that computers might have advanced at least to the stage of pencil and paper so I really ought to be able to type it straight in by now. Oh well, back to the chalk and slate.

What is the 5.1mm measurement?

I'm sure I remember seeing a thread somewhen that someone posted how to do it. But I couldn't find it.

Thanks John

Test2 flick toggle switch before posting. This time I copied the table in white under your code box. Last time I copied the code box. n/(n-1) - 2 2.0 5 1.25 10 1.11111 100 1.01010 1000 1.00100 test3 finally got it thanks Acme +1 Number in sample,n Bessel's Correction, n/(n-1) 2 2.0 5 1.25 10 1.11111 100 1.01010 1000 1.00100

test [ta=left][tr][td]n/(n-1)[/td][td]-[/td] [/tr] [tr][td]2[/td][td]2.0[/td][/tr] [tr][td]5[/td][td]1.25[/td][/tr] [tr][td]10[/td][td]1.11111[/td][/tr] [tr][td]100[/td][td]1.01010[/td][/tr] [tr][td]1000[/td][td]1.00100[/td][/tr][/ta]

So what did you do that I didn't?

[ta=left][tr][td] Number in sample, n[/td][td] Bessel’s Correction n/(n-1)[/td] [/tr] [tr][td]2[/td][td]2.0[/td][/tr] [tr][td]5[/td][td]1.25[/td][/tr] [tr][td]10[/td][td]1.11111[/td][/tr] [tr][td]100[/td][td]1.01010[/td][/tr] [tr][td]1000[/td][td]1.00100[/td][/tr][/ta] Well thanks for trying anyway.

So, once again, explain your point. This constant provocative denial of others is beginning to smack of troll.

Here is a word document of what I want to post, if it's any help. bessel's correction.doc

Here is a nice safety oriented video for those playing with Mike's Capacitors asnd Coils. http://www.youtube.com/watch?v=AGXQNLq19FQ

So assume I don't understand, and explain how the link supports your claims. As a matter of fact John Cuthber (and I) referred to carbon dioxide, which is 'heavier' than air. You referred to carbon monoxide which is marginally 'lighter'.

Thank you Acme I see them, can someone tell my how to post a 2 column x 6 row table and how to enter the data?

Additional to the Wiki article I was hoping to post a table of Bessel's correction, but I have had to ask how to post a table (http://www.scienceforums.net/topic/86509-posting-a-table/ Bessels correction and the (n-1) is also associated with statistical 'degrees of freedom'. The ultimate for this is Goset's "Student's t distribution' Edit I now have the table (thanks Acme) and it is interesting how quickly the correction approaches 1 as the number of samples increases. Number in sample,n Bessel's Correction, n/(n-1) 2 2.0 5 1.25 10 1.11111 100 1.01010 1000 1.00100

Statement issued with no proof.

Searching doesn't find any useful help on this so how do I post a table here please?

So, Talos, your University link says that you were wrong and everybody else was right.

@CasualKilla How did your brother's exam go? I would appreciate it if you posted the exact question as written by the exam board or whoever since there may be something in their wording we have missed, that leads to the solution you say they sent in a memo? Textbooks are rarely wrong in their published answers, Exam boards even more rarely wrong.

If you have anything to say, say it here. Or do you deny the rules of Science Forums as well?

I am aware of the laws of hydrostatics. But that doesn't matter to the question I asked you or to Newton's laws of motion or to the Laws of thermodynamics. Particularly as you denied their truth

You would have to ask a botanist for an explanation, but my Chambers Scientific Dictionary has: Anodal or anodic (Bot) In the upward direction on the genetic spiral. There is, however, no impasse. It's just that some engineers seem to have forgotten what we regard as the internal cicuit and what the external circuit and mix them up. It is important to understand which is which because directions are reversed for these two parts of a complete circuit. So they cheerfully take a battery, write + on one terminal and draw conventional current passing out from this terminal round the 'external' circuit and back into the battery at the other one. Yest these same engineers then take a vacuum tube, mark one terminal + and cheerfully show conventional current entering the valve at this terminal passing through the valve to the other terminal. Strangely they want to call both these terminals marked + the anode!

You are so nearly there it is a shame if you don't want to listen. Or are you, perhaps, just repeating something you half heard sometime ago? You have a balloon sitting limply on the ground, showing no signs of going anywhere. Newtons First Law : A body continues in its state of rest or steady motion unless acted on by a (net) Force. So the limp balloon is not being acted on by a force. You come along and inflate the balloon so that it rises off the ground. So Newton requires there to be a force acting on the balloon. Can you name this force and describe it please?

No I do not agree with your chain of 'reasoning' and I have already stated why. You state that the balloon causes the atmosphere to rise. Why so? Suppose I removed some air adjacent to the balloon (perhaps by freezing it) and equal in volume to the balloon, Would the atmosphere now rise? Would the balloon still rise when released? What would happen if I filled the balloon with carbon dioxide instead of helium by mistake? Would the balloon skin not still push back the atmosphere by the same amount? The point is that the apparatus that inflates the balloon causes the balloon volume to increase against the pressure of the atmosphere around the balloon skin. Work is done against this pressure and is equal to the volume change times the pressure. Since this work is done against the atmospheric pressure energy is transferred to the atmosphere as pressure energy, not gravitational potential. That is what I mean by saying that you need to distinguish types of potential energy. I also mentioned bouyancy force. This is independent of the balloon (once inflated) and is the same upward force regardless of the weight of the balloon. So if the weight of the balloon is less than the BF the balloon will rise, and Newton's First Law is not violated. Work is then done by the bouyancy force minus gravity on the balloon. This work appears as increased gravitational potential energy of the balloon.

Which is exactly double the correct answer. Not quite. positions (2) and (3) could be vowels or consonants. position (4) is specified as a vowel. imatfaal What would happen if you only had an alphabet with two vowels, but otherwise the same question? it is because the fourth one is specified as a vowel that you cannot reorder. By this reordering you can separate the 6 character word into three independent segments. But you cannot do this when the final letter is specified as a vowel. In your reordering you definitely have 4 vowels available to choose from in position (2) In the original you may have 4 or 3 or 2 This difference must be reflected in different probabilities for the second specified vowel.