Results 1 to 3 of 3

Math Help - weighted/adjusted average - is this a sound method?

  1. #1
    Newbie
    Joined
    Jan 2012
    Posts
    2

    weighted/adjusted average - is this a sound method?

    I have been presented with the following challenge in a software program that I am changing. Among other things, the software reads in records that contain four scores, let’s label these scores ‘A’, ‘B’, ‘C’, and ‘D’, (each score can range from 000 to 100). The program calculates the average of these four scores and if the average is greater or equal to 076, it will pass or fail the record. The milestone of 076 had been chosen because based on various other factors, it lead to the fairest outcomes in terms of whether or not these records 'truly' deserved to pass or fail, all other things being considered.
    Choosing a milestone that is less than 076 caused too many records to be passed (when they really deserve to fail); and choosing a milestone that is greater than 076, caused too many records to be inappropriately failed.

    So for example, if a record contained these values: Score 'A' => 065, Score 'B' => 079, Score 'C' => 090 and Score 'D' => 070, well the average of these scores would be: [(065 + 079 + 090 + 070 ) / 4] = 076. Since 076 is greater or equal to the milestone of: 076, then the record would pass.

    Recent changes have caused my customer to re-assess the program's scoring criteria. They stated that due to their nature, scores 'A' and 'B' should be given 'double' the weight in determining the combined average that is compared to the '076' milestone to determine a pass or fail. In English what they are saying, is that when scores ‘A’ and ‘B’ contain high values or low values, that is a greater indicator that the record deserves to pass or fail, then when scores ‘B’ and ‘D’ contain high or low values. So in light of that, the customer simply requested that the formula give scores ‘A’ or ‘B’, double the weight. Without researching the matter, I ‘intuitively’ interpreted the customer’s request as exemplified in the following formula (using the previous example). Note how I am achieving the ‘double weight’ affect by double counting scores ‘A’ and ‘B’. But to equalize things I am using ‘6’ (instead of ‘4’) in the denominator:

    [(065 + 065 + 079 + 079 + 090 + 070 ) / 6] = 074.6. Since 074.6 is less than 076, the record would fail under these new rules.

    When I ran tests using this formula, the results were favorable and it appeared that the new formula was achieving the desired results.
    But I still felt the above formula looked a little ‘dorky’ and unprofessional. I then discovered that I could arrive at the same value by
    expressing this way. Note that the ‘double weight’ affect is achieved by adjusting scores ‘A’ and ‘B’ by 67%, and scores ‘C’ and ‘D’ by ‘33%’.
    However, I am using ‘2’ in the denominator. Here is the more professional version: [ (.67 * (065 + 079 )) + (.33 * (090 + 070)) ] / 2 = 074.6.

    My question is: Is my interpretation of ‘double the weight’ correct, and consistent with sound statistical or mathematical theory ? After all, it is leading to good results. If not, what would be a better formula to use? And if my interpretation is correct, then is there a better way to express the formula?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28

    Re: weighted/adjusted average - is this a sound method?

    Quote Originally Posted by dcs48 View Post
    My question is: Is my interpretation of ‘double the weight’ correct, and consistent with sound statistical or mathematical theory ? After all, it is leading to good results. If not, what would be a better formula to use? And if my interpretation is correct, then is there a better way to express the formula?
    I think what you have done is sound, you have thought about the problem and tested your theory.

    I would use \frac{2A+2B+C+D}{6} as \frac{0.67(A+B)+0.33(C+D)}{2} contains a slight error as 0.67 is not exactly double 0.33
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2012
    Posts
    2

    Re: weighted/adjusted average - is this a sound method?

    Thank-you Pickslides for your quick and comforting response and your advice.

    dcs48
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Weighted average?
    Posted in the Business Math Forum
    Replies: 0
    Last Post: October 24th 2011, 04:05 AM
  2. Weighted Average
    Posted in the Statistics Forum
    Replies: 2
    Last Post: April 19th 2011, 07:25 PM
  3. Calculate Weighted Average
    Posted in the Statistics Forum
    Replies: 2
    Last Post: August 3rd 2010, 11:44 PM
  4. Weighted Average / Pop Mean
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: February 16th 2010, 07:03 PM
  5. weighted average
    Posted in the Business Math Forum
    Replies: 1
    Last Post: January 20th 2010, 01:02 PM

Search Tags


/mathhelpforum @mathhelpforum