How do I find sample standard deviations?
The data set I have is 14.2, 15.6, 16.3, 17.1, 17.9, 18.6, 20.0
+128707 The data set I have is 14.2, 15.6, 16.3, 17.1, 17.9, 18.6, 20.0
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Let's see how much of statistics I remember!!
First, let's find the mean (average) of this data set. We just sum the numbers and divide by the number of them that we have, (7)........
This gives: ( 14.2+ 15.6 +16.3 + 17.1 +17.9 + 18.6 + 20.0 ) / 7 = 17.1
The, we subtract each of the values from the mean and square the result....note that we can disregard 17.1 itself !! (Subtracting it from the mean = 0)
So we have
(17.1 - 14.2)2 = (2.9)2 = 8.41
(17.1 - 15.6)2 = (1.5)2 = 2.25
(17.1 - 16.3)2 = (.8)2 = .64
(17.1 - 17.9)2 = (-.8)2 = .64
(17.1 - 18.6)2 = (-1.1)2 = 1.21
(17.1 - 20.0)2 = (2.9)2 = 8.41
Now, sum the values on the right hand side
We have (8.41 + 2.25 + .64 + .64 + 1.21 + 8.41) = 21.56
Now, since this is a "sample," we need to divide this by (the number of values - 1) = (7-1) = (6)
This gives us (21.56/6) ≈ 3.59
Now, take the square root of this SQRT(21.56/6) = 1.895
And, assuming I didn't make any math errors, that's the standard deviation ... (S.D.)
Now....you might have the same question that I had when I first saw S.D......So what???
Well....what this tells us is how far (on average) each score is from the mean. A "high" S.D. indicates a "dispersed" data set and a "low" S.D. indicates a data set that is bunched closely around the mean. This data set seems to be fairly closely bunched around the mean.......we don't have any "large" outlier values that would tend to "skew" the S.D. very much. Of course, the terms "high," low" and "closely bunched" are all relative to the situation.
Note....because this process is lengthy, it's usually better to use some sort of application to find S.D.
Here's an app to help you with that:
http://www.wolframalpha.com/input/?i=standard+deviation+98.17%2C+112.3%2C+102.6%2C+94.3%2C+108.1
This will find S.D. for samples or populations
+128707 The data set I have is 14.2, 15.6, 16.3, 17.1, 17.9, 18.6, 20.0
-----------------------------------------------------------------------------------------------------------------------------
Let's see how much of statistics I remember!!
First, let's find the mean (average) of this data set. We just sum the numbers and divide by the number of them that we have, (7)........
This gives: ( 14.2+ 15.6 +16.3 + 17.1 +17.9 + 18.6 + 20.0 ) / 7 = 17.1
The, we subtract each of the values from the mean and square the result....note that we can disregard 17.1 itself !! (Subtracting it from the mean = 0)
So we have
(17.1 - 14.2)2 = (2.9)2 = 8.41
(17.1 - 15.6)2 = (1.5)2 = 2.25
(17.1 - 16.3)2 = (.8)2 = .64
(17.1 - 17.9)2 = (-.8)2 = .64
(17.1 - 18.6)2 = (-1.1)2 = 1.21
(17.1 - 20.0)2 = (2.9)2 = 8.41
Now, sum the values on the right hand side
We have (8.41 + 2.25 + .64 + .64 + 1.21 + 8.41) = 21.56
Now, since this is a "sample," we need to divide this by (the number of values - 1) = (7-1) = (6)
This gives us (21.56/6) ≈ 3.59
Now, take the square root of this SQRT(21.56/6) = 1.895
And, assuming I didn't make any math errors, that's the standard deviation ... (S.D.)
Now....you might have the same question that I had when I first saw S.D......So what???
Well....what this tells us is how far (on average) each score is from the mean. A "high" S.D. indicates a "dispersed" data set and a "low" S.D. indicates a data set that is bunched closely around the mean. This data set seems to be fairly closely bunched around the mean.......we don't have any "large" outlier values that would tend to "skew" the S.D. very much. Of course, the terms "high," low" and "closely bunched" are all relative to the situation.
Note....because this process is lengthy, it's usually better to use some sort of application to find S.D.
Here's an app to help you with that:
http://www.wolframalpha.com/input/?i=standard+deviation+98.17%2C+112.3%2C+102.6%2C+94.3%2C+108.1
This will find S.D. for samples or populations
