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CSCI A113
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We will "prove" today in lab that the sum of the squared deviations of all the scores about their mean is a minimum. In other words, the formula below, in which zeta is an unknown (or variable)
admits a minimum for
Let's prove that (and in the process calculate other things as well).
Here are the steps:
That's our data (the scores).
=average(a1:a5)For me that is 2.4 (the arithmetical mean of these 5 numbers).
In B1 write the formula for the first deviation:
=A1-$E$1Notice that the second element has an absolute reference to column E row 1.
When we paste this formula that will become relevant.
The first deviation is -1.4 so the formula works fine.
The deviations are calculated.
=A5-$E$1Excel updated only the relative components of the formula.
=sum(b1:b5)The value should be 0 (zero).
We have just verified that the sum of all the deviations about the mean is zero.
In cell C1 write this formula
=b1^2The value is 1.96 for me, the square of -1.4 as expected.
Select C1. Drag lower right corner to C5. Release mouse button.
Squared deviations have been calculated.
In cell C6 enter the following formula
=sum(c1:c5)That's the sum of the squared deviations about the mean, and it's 5.2 for me.
=count(a1:a5)That's 5. I leave it to you to label your spreadsheet nicely.
=sqrt(c6/e2)
=sqrt(sum(c1:c5)/count(c1:c5))
=stdevp(a1:a5)
=stdev(a1:a5)That should clarify the difference between STDEVP and STDEV.
In A9 through F9 enter 0, 1, 2, 3, 4, 5.
We'll calculate the sum of squared deviations around each of these numbers.
=($A1-A$9)^2
The cells from A10 to F14 should now contain squared deviations.
In cell A15 enter this formula
=sum(a10:a14)
Choose "Line" and click "Finish".
Still another way would be to
There will be a new mean, but that's where the minimum will also be.
Please let me know if you have any questions or if you need help.
A113