Need help with Calculating Averages

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In summary, user Mfb is trying to calculate the weighted average of hours per class for each subject at school, but is having difficulty getting the answer to be identical. When he calculates the weighted average directly, the answer is different from when he calculates it by averaging the individual averages. However, when he calculates the weighted average by taking into account the "busy weeks", the answer is identical.
  • #1
Astro
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1
Hello,

To anyone who knows a lot about calculating averages:

I NEED HELP WITH:

I need some help with some statistical calculations regarding how much time is spent on classes, subjects, and homework at school. I am trying to calculate the mean average from some raw data and what puzzles me is that when I use two different ways to calculate the answer I get two different results.

I think both of my methods are correct so I'm guessing that the reason for the discrepancy is because I used slightly different ways to derive my answer. However, I can't shake the feeling that the answers should be identical and I don't really understand why they are not.

I WOULD LIKE TO KNOW / NEED HELP WITH:

Which method is correct? Are both methods correct? Is one more accurate then the other, and if so, why? And if one method is wrong, then why?

Please see attached PDF file for all my calculations and the specific problem I'm having.

Thank you in advance for anyone who helps. I'm stumped atm. x3

 

Attachments

  • Calculating Averages (Help).pdf
    1.6 MB · Views: 273
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  • #2
This is a typical problem where a weighted average is needed if you average between averages.

Let's consider a simplified example. We collect two datasets, raw values:
set A: 4, 8, 7, 8, 5, 4
set B: 76, 80

All values together have an average of 24.
Set A has an average of 6.
Set B has an average of 78.
The average of the averages is (6+78)/2=42, much higher than the average of all values because the two large values had the same "influence" as the 6 small values in this average.
 
  • #3
Hello and thank you for your answer.

I thought I understood your answer but when I applied it, this happened (see Page 3). Why is this happening?
 

Attachments

  • 2015 April 16.pdf
    288 KB · Views: 196
  • #4
Check the average for engineering.
 
  • #5
One problem is that the busy weeks have more classes than the other weeks. Since your numbers are on a per-class basis, the high and low averages shouldn't count equally. Why don't you just total the hours and the classes and calculate the average hours per class directly?
 
  • #6
mfb said:
Check the average for engineering.

Oh! x3 I didn't see that; thanks for pointing it out. :3

However, even when I make the correction, the two final calculations still aren't identical and I'm still not sure why.

In the PDF in my previous post, the correct average hours/class for Engineering is=[(1.5x2)+(3x1)]/(2+1)=2 h/class . When I put the corrected value of 2 into the "weighted average of averages sets #1-#5" (see Page 3 of the PDF) , I now get 1.4. 1.4 is still ≠ 1.4117... derived from the other calculation. :-{ Can someone please explain why the answers are not identical? Shouldn't they be identical?
 
Last edited:
  • #7
FactChecker said:
One problem is that the busy weeks have more classes than the other weeks. Since your numbers are on a per-class basis, the high and low averages shouldn't count equally. Why don't you just total the hours and the classes and calculate the average hours per class directly?

Hello,

In his initial post in this thread, user "Mfb" stated that "a weighted average is needed if you average between averages". I wanted to test that so I decided to first try to solve for the weighted average by calculating the individual averages of hours per class for each subject and then averaging the averages.

When I calculated the weighted average directly as you suggest, the answer I got was different. I'm just trying to understand why the two approaches aren't producing the same answer. As far as I can tell, they should--but clearly I'm still not understanding something.

Note: With regard to the "low averages", I'm ignoring those figures for now. For the sake of simplicity, sets #1 to #5 that I indicated in my PDF of April 16th only only use figures for the busiest possible week. Right now, I'm just trying to test the concept of "a weighted average is needed if you average between averages" .

I still need help though. Can someone please explain why the answers which were calculated slightly differently are not identical? (ie. Why is one answer 1.4 and the other is 1.4117... ?? Shouldn't they be identical? What's wrong with my logic? What am I missing?)
 
  • #8
I think I solved it!

User Mfb, I like your explanation. :) However, the concept is so simple it's deceptive. x3

See the PDF for my solution (along with the explanation).

Thank you everyone for your help! :3
 

Attachments

  • 2015 APRIl 18 ver3.pdf
    60.1 KB · Views: 179

1. How do I calculate the average of a set of numbers?

To calculate the average, add all of the numbers in the set together and then divide by the total number of numbers in the set. For example, if you have the numbers 3, 5, 7, and 9, you would add them together (3+5+7+9 = 24) and then divide by 4 (since there are 4 numbers in the set). The average in this case would be 24/4 = 6.

2. Can I include negative numbers when calculating an average?

Yes, you can include negative numbers when calculating an average. The process is the same as when working with positive numbers. Just make sure to include the negative sign when adding them together and dividing.

3. What is the difference between mean, median, and mode?

The mean is the average of a set of numbers, calculated by adding all the numbers together and then dividing by the total number of numbers. The median is the middle number in a set of numbers when they are arranged in order. The mode is the number that appears most frequently in a set of numbers.

4. How do I handle outliers when calculating an average?

Outliers, which are extreme values that are significantly different from the rest of the data, can skew the average. One way to handle outliers is to remove them from the set of numbers before calculating the average. Another option is to use a different measure of central tendency, such as the median, instead of the mean.

5. Can I calculate an average without knowing all of the numbers in the set?

No, in order to calculate an average, you need to know all of the numbers in the set. If you are missing some numbers, you can estimate the average by using the numbers you do have and assuming that the missing numbers are similar in value to the ones you have. However, this will not give you an accurate average.

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