Arithmetic vs. Geometric MeanDate: 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. Thank you for your help, 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. I appreciate your answer and I appreciate the quickness of your response. Wes |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/