Standard deviation and variance examples needed

#1
I am trying to calculate SS, variance and standard deviation for the following sample of n = 5 scores: 10, 3, 8, 0, 4.

Now the sum of squares would be squaring the 10, 3, 8, 0, and 4 and then dividing by 4 (take one out for sampling error) correct?

Now the mean would be 25/5 (sum of x over n) which is five.

Taking out variance it would become Sigma(25 (sum of x) - 5 (mean) squared over n-1 (5-1=4)

So the variance would be 20 squared (400) over four, leaving me with a variance of 100? That doesn't seem quite right to me. Do I have to square root it again to make the s squared into s?
 

gianmarco

TS Contributor
#2
Hi,
the variance is equal to the sum of the squared deviation from the mean, divided by the number of observation in your sample. That is, it is the average of the squared deviations from the mean.

The standard deviation is the square root of the variance.

Hope this helps.

Regards
gm
 
#3
Hi,
the variance is equal to the sum of the squared deviation from the mean, divided by the number of observation in your sample. That is, it is the average of the squared deviations from the mean.

The standard deviation is the square root of the variance.

Hope this helps.

Regards
gm
Thats what I am having problems on, finding the squared deviations. I am not quite sure how to accomplish that. And with the variance, do I square the 100 because as it stands 100 seems an unlikely varianve and its s squared = 100
 
#4
You take the difference of each data point from the mean, then square, then sum.

To get variance, you take the sum squared and divide by n-1 (since you are dealing with a sample).

Standard deviation is just square root of variance.
 
#5
You take the difference of each data point from the mean, then square, then sum.

To get variance, you take the sum squared and divide by n-1 (since you are dealing with a sample).

Standard deviation is just square root of variance.
This isn't working out. My variances are ending up in the hundreds and thousands and don't look right. Can you please check my work on the first post and see if I am getting this concept correct?

BTW, the numbers are the frequency of categorical data.
To recap let me make up some numbers:

12, 42, 87, 57, 8

Squaring = 144, 1764, 7569, 3249, 64

Adding together is 12790. Using n-1 (which equals 4) I would have a variance of 3,197.5
 
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#6
Your data set is {10,3,8,0,4}
Mean = 25/5=5

Var = {(10-5)^2 + (3-5)^2 + (8-5)^2 + (0-5)^2 + (4-5)^2}/5-1

(if this was a population instead of a sample, you would divide by 5 instead of 4)

var = 5^2 + (-2)^2 + 3^2 + (-5)^2 + (-1)^2 / 4

VAR = 25+4+9+25+1 /4 = 64/4 = 16

Std Dev = sqrt(16) = 4.
 
#7
Your data set is {10,3,8,0,4}
Mean = 25/5=5

Var = {(10-5)^2 + (3-5)^2 + (8-5)^2 + (0-5)^2 + (4-5)^2}/5-1

(if this was a population instead of a sample, you would divide by 5 instead of 4)

var = 5^2 + (-2)^2 + 3^2 + (-5)^2 + (-1)^2 / 4

VAR = 25+4+9+25+1 /4 = 64/4 = 16

Std Dev = sqrt(16) = 4.
Ok I get it now. Sum of squares if squaring the number AFTER you've subtracted the mean from it.
 

BGM

TS Contributor
#9
Of course you can compute from the statistics ∑x^2 and ∑x:

S^2 = ∑x^2/(n - 1) - (∑x)^2/[n(n-1)]

In your case
∑x^2 = 10^2 + 3^2 + 8^2 + 0^2 + 4^2 = 100 + 9 + 64 + 0 + 16 = 189
∑x = 10 + 3 + 8 + 0 + 4 = 25
n = 5
S^2 = 189/(5 - 1) - (25)^2/[5(5-1)] = 189/4 - 125/4 = 16
 
#10
Sum of squares is just what it saids... you take the squares, and what is meant by this is actually the difference between an observation and the mean squared. And then you take these squares and sum them up... to get the sum of squares.

If you devide this by n-1 you get the variance. Sometimes, when you have population data instead of a sample, you devide by n.

Then, take the square root of this variance to get the standard deviation.

It's not that hard actually...