# Weighted average vs “unweighted” average in probability

[UPDATE] Terms I used: - Weighted average: weighted arithmetic mean - "Unweighted" average: arithmetic mean

I went through some of the forums here and looked for explanations online but I couldn't figure out the answer for this:

If percentages represent probabilities, what does weighted and "unweighted" average of a list of values represent?

Example: There are two shops that sell vegetables and two that don't sell vegetables. All of them ask their customers for feedback on their shopping. They say either that they were happy or they were not. Let's assume that every customers answers.

Veg Shop 1 has 10 customers and 5 are happy. (50% average) Veg Shop 2 has 20 customers and 8 are happy. (40% average) Non-Veg Shop 1 has 15 customers and 5 are happy. (33% average) Non-Veg Shop 2 has 25 customers and 10 are happy. (40% average)

Weighted average of Veg Shops is 43% (13 / 30), unweighted average is 45%. Weighted average of Non-Veg Shops is 37.5% (15 / 40), unweighted average is 36.5%.

I know that with weighted average distribution matters and I feel that it implies that there is correlation between the values aggregated. However, in this example, if I count weighted average I accept that there is correlation, but I discard the correlation between Veg and Non-Veg shops.

I can't really phrase the probabilities, but it feels that unweighted average is an answer to "my expected performance (based on customer feedback) among other shops of the same type", and weighted average is an answer to "the chance of a customer being happy, when going to any of the shops of the same type".

It is still not really clear for me, though. Does anybody have some clearer ideas on what the probabilities would mean?

Also, is it possible at all to assign meaningful values to both weighted and unweighted average on the same dataset?

Thanks, Norbert

• Please register &/or merge your accounts (you can find information on how to do this in the My Account section of our help center), then you will be able to edit & comment on your own question. – Silverfish Mar 24 '16 at 10:56

What you are calling the "weighted average" is the only proper way to calculate the percentage of satisfied customers in both shops. Taking average of percentages (what you call "unweighted" average) will give you useless results if your samples differ in size.

Imagine extreme case: you have two shops A and B, in shop A there was one customer and he was not satisfied and in shop B there was 100 customers and 90 of them were satisfied -- would you conclude that 45% of customers of both shops were satisfied? Obviously not!

Smaller sample is much less reliable, so it should not be included in the final estimate with the same weight as the larger one. Speaking more formally, estimate from the smaller sample has a greater error.

Two shops: A and B, in A there are two customers, one happy, in B, there were 100 and 75 were happy. So they have 50% and 75% happy customers, so the average is 62.5%. Using your example from the comment:

Standard error for the estimate from the first shop is $\frac{0.5(1-0.5)}{\sqrt 2} = 0.18$, while for the second one $\frac{0.75(1-0.75)}{\sqrt{100}} = 0.02$. So the possible deviation from the true proportion of satisfied customers in shop A is much larger. If the samples differ that much in their reliability, you shouldn't give them equal trust and weight them equally when taking their average.

You can easily convince yourself why using pooled mean is much better idea that using the raw one by conducting a simple simulation study. I simulate two shops, with $n=8$ and $n=100$ customers, both having the same proportion of satisfied customers. If you compare the results obtained using raw mean ("unweighted") and the pooled ("weighted") estimates, you'll see that when using the raw mean the variability of difference between the estimate and the true value is much greater. Saying this in simple English: using raw mean you are in greater risk of obtaining the wrong estimate, than in the case of pooled mean.

For much more advanced methods of pooling different probabilities you can check Combining probabilities/information from different sources thread, or this one Combining two estimates .

• What I'm saying is that there could be information in arithmetic mean as well. Two shops: A and B, in A there are two customers, one happy, in B, there were 100 and 75 were happy. So they have 50% and 75% happy customers, so the average is 62.5%. That's a baseline for a new shop when it enters the market, and can be used to compare itself against it. Am I above or below the average performance of other shops? – norbertk Mar 25 '16 at 7:02
• @norbertk you could, but this also could be very misleading. If there was only two customers, it might have happened by chance that they were happy -- 75 happy within 100 customers is much more likely to happen by chance. Standard error for the first estimate is 17%, while for the second one 1.9%... – Tim Mar 25 '16 at 8:27
• @norbertk I edited my answer to add more details. – Tim Mar 25 '16 at 9:47

So there is a student Joe. Student Joe has a deal with his mom that he will get a good allowance as long as he keeps a grade of a B or better. Joe is graded based on tests and quizzes. His scores are as follows:

Quizzes:

8:10 - 80%

9:10 – 90%

9:10 – 70%

10:10 – 100%

Tests:

50:100 – 50%

70:100 – 70%

85:100 – 85%

When Joe’s mom asks how hes doing in school he says, “Good, I have an 80.7% average!!”. His Mom says “Great, here is your allowance”. A month later Joe’s mom gets his report card and sees a 70.9%. Furious, Joe’s mom gets a meeting with Joe and the teacher. Joe explains he was averaging his test percents and Joe’s teacher shows him how the “weighted average” is calculated. Coincidentally Joe's teacher also lowers Joe’s grade to a D.

• I think this answer would be improved by showing where the 81.4% and 70.8% come from. – Silverfish Aug 11 '16 at 18:44
• I agree with @Silverfish. Also, I think your numbers are slightly off (241/340 = 70.9%) but I can't figure out how you got 81.4%: a straight average of the percent scores gets me 80.7%, averaging the tests only gets me 68.3%, and averaging the tests and quizzes separately, then weighing them 50:50 gets me 79.2%). – Matt Krause Aug 11 '16 at 19:14
• Otherwise though, I think the overall idea is a good one, since it relates to something almost everyone has experienced. Also, welcome to Cross Validated. Hope to see you around! – Matt Krause Aug 11 '16 at 19:15
• Thanks for the comments guys, you are absolutely right about the percentages. 80.7% for the averaging of percents and 70.9% is the proper way to round the weighted average. I've made the changes in the answer above. – Mat Mar 8 '17 at 0:27