Average of averages?

niravrph

New member
Joined
Aug 9, 2011
Messages
4
Hi,

How do I calculate the average of some averages? I have the following data:

# of Meetings# of Attendees totalAvg. Attendees per Meeting
84101112.0
8486610.3
6886012.6

How do I calculate the average of these Averages (third column in table above)?

Thanks!
NIRAV
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
22,083
Hi,

How do I calculate the average of some averages? I have the following data:

# of Meetings
# of Attendees total
Avg. Attendees per Meeting
84
1011
12.0
84
866
10.3
68
860
12.6


How do I calculate the average of these Averages (third column in table above)?

Thanks!
NIRAV
Depends on what are you using that number for.

One answer would be [(12.0+10.3+12.6)/3 =] 11.7

Another would be [(1011+866+860)/(84+84+68) =] 11.6

Pretty close - but different
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, niravrph!

How do I calculate the average of some averages? . You don't!

I have the following data:

. . \(\displaystyle \begin{array}{ccc}\text{Meetings} & \text{Attendees} & \text{Average} \\
84 & 1011 & 12.0 \\ 84 & 866 & 10.3 \\ 68 & 860 & 12.6 \end{array}\)

Subhotosh's second answer is the correct one.


. . \(\displaystyle \begin{array}{cccc} & \text{Meetings} & \text{Attendees} & \text{Average} \\
& 84 & 1011 & 12.0 \\ & 84 & 866 & 10.3 \\ & 68 & 860 & 12.6 \\ \hline \text{Total} & 236 & 2737 \end{array}\)


The average attendance per meeting is: .\(\displaystyle \dfrac{2737}{236} \:=\:11.59745763 \:\approx\:11.6\)


A safe policy is: NEVER average averages.
.
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
22,083
I would not say never. It is used all the time in product quality control.

In material testing, where testing are conducted on batches (group of specimens) and lots (groups of batches) and runs (groups of lots) - the statisctic is grouped and and analyzed accordingly.

That is why I used "depends".
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588

Suppose we have two machines, A and B.

Machine A produced 90 items and 9 were defective: 10% defective.
Machine B produced 10 items and 5 were defective: 50% defective.

Would you say that on the average the two machines
. . produced: \(\displaystyle \dfrac{10\%+ 50\%}{2} \:=\:30\%\) defective?


No, out of 100 items only 14 were defective.
. . The average is 14% defective.


There are certain conditions in which averages can be averaged.
. . As I said, it is safer to avoid averaging averages.
.
 

niravrph

New member
Joined
Aug 9, 2011
Messages
4
Thank you, both! I really appreciate it.
 
Top