Data Here

It has been some time since my last Stats course and I have what I am sure is a relatively simple problem. My goal is to understand the fundamental difference here.

Take Column A-D. Each row is a new day. Each cell in Columns A and B are independent day to day. Each cell in Column C is equal to A/B. The goal is to provide a Month to Date recording in Column D.

I saw 2 ways of going about this:

Option 1:

for any given cell of $D$, take $D_i = \frac{\sum_{n=A_o}^{A_i}n}{\sum_{m=B_o}^ {B_i}m}$

Option 2: (ignoring the 2nd row with all 0's)

for any given cell of $D$, take $D_i = \frac{\sum_{n=C_o}^{C_i}n}{C_i}$

These give different values however. Option 1 returns a final $D$ value of 13.62% vs Option 2 13.54%.

What fundamental aspect am I missing between this two options. Additionally, which would ideal for the Month to Date value? I'm leaning towards taking the average (Option 2), but would appreciate input.

  • $\begingroup$ What exactly is the data representing? What do you mean by "month to date"? $\endgroup$
    – Joseph Eck
    Apr 19 '18 at 19:36
  • $\begingroup$ First: Thanks for the help with formatting. Column B is # of unique items removed. Column A is the # of unique items replenished. Column C is the % of these unique items that were replenished each day. And Column D is desired to show the Month-to-Date "average" of Column C for each given day, which resets at the end of the month. Column D is the most important since any day the % in Column C can shift, but the overall "average" is the target (say below 14%). $\endgroup$
    – Sterks
    Apr 19 '18 at 19:43
  • $\begingroup$ It seems to me like I want to take the Option 2 approach since that is by definition Averaging, but my intuition initially led me towards Option 1 of Summing A, Summing B and dividing. These do not equal, and I can see it on my hand drawn notes. But what is Option 1 doing then if not averaging??? $\endgroup$
    – Sterks
    Apr 19 '18 at 19:46
  • $\begingroup$ In case anyone comes across this question, this link is the equivalent of what I was asking and has a suitable answer. $\endgroup$
    – Sterks
    Apr 20 '18 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.