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Standard Deviation and TI-84 Questions
Zeromus
Registered User regular
Registered User regular
First off: God damn required math courses at college. 
Now. I'm doing some statistics work. In this problem, I need to calculate the standard deviation and variance of a small set of numbers by hand. The numbers are:
5.6 , 5.2 , 4.6 , 4.9 , 5.7 , 6.4
Using instructions I pulled from the book, I learned that in order to calculate the variance, I should subtract the mean from each value of interest, square the results, and then divide them by the amount of values (6) minus 1 (so, 5). Thus:
4.6 - 5.4 = -.8 ^2 = -.64
4.9 - 5.4 = -.5 ^2 = -.25
5.2 - 5.4 = -.2 ^2 = -.04
5.6 - 5.4 = .2 ^2 = .04
5.7 - 5.4 = .3 ^2 = .09
6.4 - 5.4 = 1 ^2 = 1
(-.64 + -.25 + -.04 + .04 + .09 + 1) / 5 = .04
Then, to get the standard deviation, I am to square root .04, which gives me .2
So that all seems to check out, but when I put the info into my TI-84 and use it to calculate the standard deviation using the 1-Var Statistics option, it tells me the standard deviation is (.586).
Does anyone know what might be going wrong? Did I calculate it wrong by hand? Is there some setting off in my calculator?
Now. I'm doing some statistics work. In this problem, I need to calculate the standard deviation and variance of a small set of numbers by hand. The numbers are:
5.6 , 5.2 , 4.6 , 4.9 , 5.7 , 6.4
Using instructions I pulled from the book, I learned that in order to calculate the variance, I should subtract the mean from each value of interest, square the results, and then divide them by the amount of values (6) minus 1 (so, 5). Thus:
4.6 - 5.4 = -.8 ^2 = -.64
4.9 - 5.4 = -.5 ^2 = -.25
5.2 - 5.4 = -.2 ^2 = -.04
5.6 - 5.4 = .2 ^2 = .04
5.7 - 5.4 = .3 ^2 = .09
6.4 - 5.4 = 1 ^2 = 1
(-.64 + -.25 + -.04 + .04 + .09 + 1) / 5 = .04
Then, to get the standard deviation, I am to square root .04, which gives me .2
So that all seems to check out, but when I put the info into my TI-84 and use it to calculate the standard deviation using the 1-Var Statistics option, it tells me the standard deviation is (.586).
Does anyone know what might be going wrong? Did I calculate it wrong by hand? Is there some setting off in my calculator?
Zeromus on
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See the first example here: http://en.wikipedia.org/wiki/Standard_deviation
Edit: yeah... all your squared numbers are right except for the sign... also not sure why your book says to divide by the number of samples-1, because it should just be by the number of samples (so 6).
Huh. I knew that. Yet, when I type them into my calculator, it gives me a negative result...
Edit: Ah, need to put the numbers in parentheses before I square them. Forgive me, I'm not a math major.
Hmm... what exact steps/buttons are you pressing to do the square on the calculator?
Edit: ah... apparently the TI-8x series is somewhat odd about this, though I don't recall this from my TI-82... http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv016.cgi?read=98825 So if you type in -2^2 it gives you -4, because it squares then applies the negative. If you put in (-2)^2 it will give the correct 4... weird.
So play around with how your calculator is interpreting how you're entering it in. If you're entering it into a Solver or similar, it may not be looking in the order you want. If you're just punching it in and then getting Ans^2, it *should* be doing it right.
It's a difference between exact standard deviation and estimating the population standard deviation. If these six numbers are your entire population than you can accurately and precisely calculate the exact standard deviation (.586). However, if this is just a sample of your population, you need to calculate it differently (http://en.wikipedia.org/wiki/Standard_deviation#Estimating_population_standard_deviation_from_sample_standard_deviation)
Edit: This is probably where the confusion of 6 vs 6-1 came in. Because in estimating it you multiply your sum of squares by 1/N-1 before taking the square root.
(And like, really, I knew that stuff about exponentials, Eggy. Honest.)
:...:
n-1 degrees of freedom and what not
As already mentioned, Sx is the sample standard deviation, and sigma x is the population standard deviation. They're calculated differently, and here's why (CAUTION: Maybe more of an explanation than you really want)
Let's say your variable is height. In the population (let's say the US population) there will be some people with extreme height, high or low. Shaq and midgets and such. By definition, they're very rare. These people tend to increase the standard deviation by...well, existing.
If you take a sample out of the population (let's say 1000 people), there's a very strong chance that you won't get any people who are extremely tall or short, because there arent' many. Most likely the people in your sample will tend to be near the average height, because that's much more common. By excluding those extreme values, the standard deviation of the sample will be artificially low compared to the population. To help equalize the two, you divide the sample standard deviation by a smaller amount (n-1 instead of n) in order to raise it up slightly, closer to the population standard deviation. It's basically a mathematical hack.
Of course, your calculator doesn't know whether your set of data represents a sample or a population, so it gives you both options.
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