Reply With QuoteWhy not just compute the standard deviation (S) for all the data as:Originally Posted by PabloZzzz
S = Sqrt[ Total Sums of Squares / df ]
where df = N - 1. (N is the total sample size).
Finding Standard Deviation
I have been working through a problem in which I am given 4 means from four different conditions, and I am also given the between, within, and total SS, df, MS and F.
I used this set of on-line directions to try to figure it out. The answer was in response to this question:
Q: The sum of 50 observations in the sample is 400. The sum of the squares is 3544. That is all I know.
How do I find variance and Standard Deviation? Do I even have enough information?
A:
The Standard Deviation (σ) is a measure of how spread out numbers are.
σ = the square root of the Variance
so Variance (v) is the square of the standard deviation, ie: σ^2
To solve the problem, you first calculate v which will then give σ
v = The average of the squared differences from the Mean (m)
= (1/m)(sum (Xi - m)^2)
= (1/m)(sum (s)^2)
You have the number of samples (N) and their sum (T)
So you have m=(T/N) = (400/50) = 8
You have the sum of the squares(S),
S = 3544
but not the sum of the squared differences (sum (s)^2)
If you do the algebra, you can calculate σ^2 anyway by using
σ^2 = (S - (sum (s))^2/N) )/N
= ( 3544 -(160000/50) )/50
= ( 3544 - 3200 )/50 = 6.88
and the standard deviation is the square root.
________
Since used the average score of the four means and multiplied by the total number of subject to get an estimate of the sum of scores. I then used my sum of squares total and subtracted it from my sum of scores squared divided by my total number of subjects. After the subtraction I divided the number again by my total number of subjects and then sq rooted that number. So is this right? When I graph it, it seems as if the standard dev is way too large. Help, please.
Why not just compute the standard deviation (S) for all the data as:
S = Sqrt[ Total Sums of Squares / df ]
where df = N - 1. (N is the total sample size).
I am not given the data, the question asks that I guess about the stand dev, but I was thinking that I could actually work backwards from the data at hand to find a general estimate. I am assuming homogeniety of variance. Is there a way to work backwards in order to make a guess?
Ok, I went back and read your response again. I think I get it. The results surprise me a bit, again, however. Since I am getting an F of 6 with an F critical of 2.7 and the distance between the lowest and highest mean is 7 points, and st dv is comes out to be around 5--meaning the .05 line would be drawn 15 points out. Does this mean I have fairly steep curves? or am I even reading this correctly?
The overall S is not a good statistic to use if you're wondering why the F statistic is 6. A better statistic to consider at would be the MS_within...it's the average of the variances for each group. Also, if you take the
Sqrt[MS_within] it is the "standard error of the estimate". In other words, if you conducted this ANOVA via Regression (with dummy codes) it would be the standard of the regression.
I am not familiar with what it is you're looking at here.--meaning the .05 line would be drawn 15 points out. Does this mean I have fairly steep curves? or am I even reading this correctly?
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