what is the argument for not using the average of an average?

I want to disprove someone's calculation of percentage of cash sales for a year by taking summing percentage of cash sales by month and dividing by 12. I sense the correct way is to take total cash sales for the year and divide by total sales for year to arrive at percentage of sales but need correct reasoning why it is wrong to use an average of an average.

• What is 'percentage of cash sales'? How is it calculated? Jan 4 '13 at 21:54

You need a counter example. E.g.

Mon     Sales   Cash    Percent

Jan     100     70      70%
Feb      10      1      10%
Mar      10      1      10%
Apr      10      1      10%
May      10      1      10%
Jun      10      1      10%
Jul      10      1      10%
Aug      10      1      10%
Sep      10      1      10%
Oct      10      1      10%
Nov      10      1      10%
Dec     100     70      70%

Total sales are $300$, cash sales are $150$ so the overall percentage is $50\%$. But the average of the monthly percentages is $20\%$.

• Thank you very much. I will be back to this site I'm sure. Jan 5 '13 at 14:42

You are correct, here's a mathematical example:

Let $s_1, s_2, s_3 \cdots s_{12}$ be the sales of the months.

Let $c_1, c_2, c_3 \cdots c_{12}$ be the cash sales of the months.

$$\frac{1}{12} \sum_{k=1}^{12} \frac{c_k}{s_k}$$

Is the same as: $$\frac{\sum_{k=1}^{12} c_k}{\sum_{k=1}^{12} s_k}$$

You can show it is false by plugging in some numbers.

• Thanks for taking the time to respond. Very helpful. Jan 5 '13 at 14:41

The average of the averages leads to the right average only when the samples have the same size.

If each month has exactly the same number of sales, then taking the average of the averages would be right. But if the number of sales per month is different, then the average of averages leads in general to the wrong answer.

Simple situation: First month 100 sales of 1 each. Average 1. Second month 1 sale of 100. Average 100 .

The average of the averages is 50.50, but there were 101 sales and only 200 income....

• Right, otherwise you need a weighted average. A simple unweighted arithmetic mean would be wrong. Jan 5 '13 at 1:27
• Thank you very much. Language very helpful. Jan 5 '13 at 14:40

Suppose in January I sell \$100 worth of product, and all of it is sold for cash. Every other month I sell \$1 worth of product, and none of it is sold for cash. I have then sold \$111 dollars worth of product, of which \$100 was sold for cash, so my percentage of cash sales is $\frac{100}{111}\approx90\%$. Calculated the other way, we would get that for $1$ month $100\%$ of my sales were cash, while for the other $11$, $0\%$ of my sales were cash, giving $\frac{100\%}{12}\approx 8.3\%$. This is clearly wrong.

• Thanks so much for responding. Examples are so helpful! Jan 5 '13 at 14:41

It is clear to you that taking cash sales for a year, and dividing by total sales for the year, will give you the correct proportion of cash sales for that year.

The problem with taking monthly proportions and averaging them (average of averages) is that that procedure could produce a quite different number.

Let's take a simple numerical example. January to October: total sales each month $\$1000$, cash sales each month$\$500$. November and December (Christmas season, people are buying a lot, mostly on credit): total sales each month $\$10000$, cash sales each month$\$500$.

So total sales for the year, 30000, cash sales 6000. Thus the yearly proportion of cash sales is $\dfrac{6000}{30000}=20\%$. This is the correct percentage.

Now let's compute the monthly averages. For January through October, they are $50\%$. For each of November and December, they are $5\%$.

To find the average of the monthly proportions, as a percent, we take $\frac{1}{12}(50+50+50+50+50+50+50+50+50+50 +5+5)$. This is approximately $42.5\%$, which is wildly different from the true average of $20\%$.

For many businesses, sales exhibit a strong seasonality. If the pattern of cash sales versus total sales also exhibits seasonality, averaging monthly averages may give answers that are quite far from the truth.

• Exactly what I needed! Thanks very much. Jan 5 '13 at 14:39