"Average" function query

[Robin Blackwell]

Hi all,

I have an Excel function query that's really bugging me. In column "A" I
have about 80 staff annual salaries. In column "B" I have their
corresponding annual salary increase.

If I sum Column "A", and sum Column "B", and divide the resultant totals, I
get what I assume to be the average annual salary increase for the group
(3.46%).

However, if in Column "C" I calculate each individual's average annual
increase (B2/A2 etc), and then use the AVERAGE function over all of the
results in column "C" I get a different answer (3.79%).

Why is this please, and which answer is correct?

Thanks,

Robin

 

[Guest]

I believe you need to do a weighted average of the %s, not an average. To do
that use

=SUMPRODUCT(A2:A81,C2:C81)/SUM(A2:A81)

you should come up with the 3.46% figure.

 

[Guest]

It is because the denominators (the salaries in column A) of the fractions
being added (when you average the percent column) aren't all the same.
Remember back to fraction addition:

7/4 + 8/2 DOES NOT equal 15/6... but instead 8/2 = 16/4, and so 7/4+16/4 =
23/4.

 

[joeu2004]

In column "A" I have about 80 staff annual salaries.
In column "B" I have their corresponding annual salary
increase.

I assume these are in A2:A81 and B2:B81 respectively.

If I sum Column "A", and sum Column "B", and divide the
resultant totals, I get what I assume to be the average
annual salary increase for the group (3.46%).

However, if in Column "C" I calculate each individual's
average annual increase (B2/A2 etc), and then use the
AVERAGE function over all of the results in column "C"
I get a different answer (3.79%).

Why is this please, and which answer is correct?

The answer is: both(!). It depends on what you want to
compute.

The first value, SUM(B2:B81)/SUM(A2:A81), is the percentage
increase of the total salary cost for the group. (It is
not really an average.)

The second value, SUMPRODUCT(B2:B81/A2:A81)/COUNT(A2:A81),
is the average salary increase per individual.

(Oddly, SUMPRODUCT(...)/COUNT(...) is equivalent to the
way you computed the average. There might be a better
formula that is more apparent to the reader.)

The reason for the difference is because
(b2+...+b81)/(a2+...+a81) does not equal
b2/a2 +...+ b81/a81, much less (b2/a2 +...+ b81/a81)/80.

For example, consider A2:A3 = {2,3} and B2:B3 = {1,1}.
(1/2 + 1/3)/2 equals (5/6)/2 (5/12), and (1+1)/(2+3)
equals 2/5.