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### Arithmetic vs. Geometric Mean

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Date: 03/27/2001 at 19:04:27
From: Mark
Subject: Arithmetic vs. geometric mean

When would it be appropriate to use the arithmetic mean over the
geometric, or vice versa? I am not trying be vague, but the geometric
mean will always be less than the arithmetic. Are there specific
circumstances where one is used over the other?
```

```
Date: 04/17/2001 at 10:51:58
From: Doctor Floor
Subject: Re: Arithmetic vs. geometric mean

Hi, Mark - thanks for writing.

Let me try to give examples for when you use one, and when you use the
other.

The profit of Company A, SYZO Ltd., has grown over the last three
years by 10 million, 12 million, and 14 million dollars. It is
appropriate to say that it has grown by an average of 12 million
dollars yearly, for which we use the arithmetic mean.

The profit of Company B, OZYS Ltd., has grown the over last three
years by 2.5%, 3%, and 3.5%. Here we cannot use the arithmetic mean
and say that the average growth was 3%. Why not?

Suppose that Company B, OZYS Ltd., started with a 100-million-dollar
profit. Three years later it will have become:

\$100,000,000 * 1.025 * 1.03 * 1.035 = \$109,270,125

This is less than a yearly increase of 3% would yield, since:

\$100,000,000 * 1.03 * 1.03 * 1.03   = \$109,272,700

Here we see that we should use the geometric mean of the growth
factors 1.025, 1.03, and 1.035 to find the average percentage. That is
always less than the arithmetic mean would yield.

I hope this helps.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 06/11/2001 at 21:49:57
From: Wesley A. Payne
Subject: Geometric Mean vs. Arithmetic Mean

Dr. Math:

I am attempting to determine why the geometric mean is more
appropriate to use when dealing with percentages and rates of change.
However, I am not sure why it works better. I know that the result of
applying the geometric mean will be the same as or lower than the
arithemtic mean, and that the arithmetic mean is used when finding the
average of numbers that are added to find the total, and that the
geometric mean is used when the items of interest are multiplied to
gain the total (found by looking at your Web site). I still do not
know why it works this way.

I looked in some older texts that were in print prior to the use of
scientific calculators and saw that there is a formula for the
geometric mean that uses logs. This leads me to believe that in doing
so, the data was being normalized, and maybe this would be the reason
that the geometric mean would be better.

Wes Payne
```

```
Date: 06/12/2001 at 11:35:25
From: Doctor Rick
Subject: Re: Geometric Mean vs. Arithmetic Mean

Hi, Wes.

I think Dr. Floor did a fine job of briefly explaining the benefits of
each mean. I will just say the same thing with a little expansion.

The big idea behind means is this. You have a bunch of different
numbers. You want to replace each of these different numbers by the
*same* number, in such a way that the net effect (the result of
combining the numbers) is unchanged.

Different means are used depending on what we mean by "combining" the
numbers. In Dr. Floor's example of a company growing annually by 10,
12, and 14 million dollars, the "combined" growth is the sum of the
three numbers. We want the total growth (the sum of the annual
growths) to equal the total if the company grew by N million dollars
each year. The mean we seek, therefore, is the number N such that

10 + 12 + 14 = N + N + N

The number that works is the arithmetic mean:

N = (10 + 12 + 14)/3

In Dr. Floor's second example, of a company growing annually by 2.5%,
3%, and 3.5%, the growth rate over the three years is computed
differently:

Profit in year 1 = 1.025 times profit in year 0

Profit in year 2 = 1.03 times profit in year 1
= 1.03 * 1.025 times profit in year 0

Profit in year 3 = 1.035 times profit in year 2
= 1.035 * 1.03 * 1.025 times profit in year 0

Thus the ratio of the profit in the third year to the profit in the
base year is

1.025 * 1.03 * 1.035

If the growth rate had been the same each year, the ratio would be

N * N * N

For the growth over 3 years to be the same in both cases, we must have

1.025 * 1.03 * 1.035 = N * N * N

The solution is

N = (1.025 * 1.03 * 1.035)^(1/3)
= 1.029992

That is, the cube root of the product of the annual growth factors.
This is the geometric mean.

Note: The mean growth rate comes out to

(1.029992 - 1) * 100% = 2.9992%

which is very close to the arithmetic mean (as seen in Dr. Floor's
answer: compare \$109,270,125 to \$109,272,700). The difference would be
greater if the growth rates were greater.

Your comment about logs prompts me to point out that there is a close
connection between the two means. The log of the geometric mean of a
set of numbers is the arithmetic mean of the logs of the numbers:

log((abc)^(1/3)) = (1/3)log(abc)
= (1/3)(log(a)+log(b)+log(c))

Thus you can find the geometric mean by taking the logs of your data,
finding the arithmetic mean, then taking the antilog (exponential) of
the mean. This is presumably the formula to which you refer.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 06/12/2001 at 17:40:46
From: Payne, Wesley
Subject: Re: Geometric Mean vs. Arithmetic Mean

Doctor Rick,

That was the formula (logs) that I was referring and after further
examination, I realized it was not a transformation but simply the
non-electronic way to find the answer.

response.

Wes
```
Associated Topics:
High School Statistics
Middle School Statistics

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