Total average of averages not same as the average of total values

averagemeansrounding errorstatistics

I am having a strange problem while computing the overall percentage. Let me demonstrate my problem using an example.

Assume that I am receiving multiple batches of apple from the vendor and the number of good apple received in each batch are computed to display as percentage as below:

If I find the average of the last column (Good Apple Percentage),

I get $\frac{(98.5+98+99)}{3} = 98.5 $%

However, if I find the total no of apples and total number of bad apples first (column 2 and 3),

Total No of Apples $= 200+100+300 = 600$

Total No of Bad Apples $= 3+2+3= 8$

Then compute the total percentage, $\frac{600-8}{600}= 98.67$%

Why is the total average of the average not the same as the average of total values? There is no rounding up or rounding down necessary for individual percentages (Column 4), so shouldn't the two percentage be the same? What am I missing?

Best Answer

The reason you are getting different numbers is because if you look at just the averages, it completely obliterates any sense of the size of the original data. Let me give you a more extreme example to illustrate:

Batch A: 2 apples total, 1 bad apple, 50% good apples Batch B: 100 total apples, 1 bad apple, 99% good apples

Totaling the two we have 102 apples, 2 of which are bad, 100 good. But if we average just the percentages, we are at $74.5%$.

The moral of the story is, you can't just average the averages because it loses all sense of how big the individual samples are.

Example

Let say you compute the two indicators for summer:

strength_sum summer = n_june + n_july + n_august = 62
strength_avg summer = (n_june + n_july + n_august) / 3 = 20.67

And for winter:

strength_sum winter = n_december + n_january + n_february = 70
strength_avg winter = (n_december + n_january + n_february) / 3 = 23.33

Then strength_sum summer < strength_sum winter is equivalent to strength_avg summer < strength_avg winter (just by multiplying by 3). In both cases, the summer had less animals than the winter.

But what if...

If your seasons has a different number of months, or if you want to generalize to other time periods, I think that using the average of averages is more meaningful, regarding your issue.

Imagine you were in the Alps mountains, where the summer and autumn are way shorter than winter. It makes sense to favor the average of averages in order to correctly compare a 2-month long summer with a 5-month long winter.

For example, if you use the sum of average criterion: if strength_sum summer = 20 * 2 = 40 (n = 20 for each month of summer) and strength_sum winter = 10 * 5 = 50 (n = 10 for each month of winter), you don't want the winter to "win" simply because it's more than twice as long as summer : strength_sum summer = 40 < 50 = strength_sum winter, but strength_avg summer = 20 > 10 = strength_avg winter. It seems to me than strength_avg makes more sense here.

Calculate a Weighted GPA

You calculated an unweighted GPA. To weight it, you must factor in the course credit variable.

> sum=45+30+45+60
> (86*45+74*30+88*45+77*60)/sum
[1] 81.5

As the original units are in a 0-100 scale, your weighted gpa is about a B-.

To convert to a 4.0 scale, simply use the appropriate unit conversion

> 81.5*4/100
[1] 3.26

Your original calculation was missing a multiplication by 4. The procedure for your unweighted gpa should be

> (86+74+88+77)/4*4/100
[1] 3.25

Best Answer

Related Solutions

[Math] average of averages vs sum of averages

Example

But what if...

Calculate a Weighted GPA

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