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I work for a driving company who likes to hide what I consider important feedback.

Right now I have an average rating of 4.61 on a 1-5 star rating scale. Once I figured out they were not going to tell me the data,I have been keeping track of the up and downs. I have at least 11 ratings but I do not know the total number of ratings.I could have 20,30 or 50 ratings.

Started data collection with:
4.81
4.68
4.7
4.71
4.72
4.73
4.62
4.6
4.58
4.59
4.61

Is there a way to figure out what my total rating number is with this data provided so I can predict future data?

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  • $\begingroup$ Just to make sure I'm tracking your question right, I think it's something like this: after some mystery number of ratings (maybe 5, maybe 10, etc.), you checked your aggregate rating and it was 4.81. Then, after an unknown number of additional ratings came in, you checked your aggregate rating and it was 4.68. You did this several times (11 in total) and logged the results. Did I get that right? $\endgroup$
    – Aaron Montgomery
    Dec 9, 2019 at 17:17
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    $\begingroup$ Yes to unknown number of ratings before the 4,81.When I realized it was 4.81 I started tracking after every change.So the 11 following the 4.81 are in order and the only ratings I have had since the 4.81. $\endgroup$
    – Jennifer
    Dec 9, 2019 at 17:23
  • $\begingroup$ Got it. Last question: is there an upper bound on how many ratings you might have? (For instance, you have given 100 rides total?) $\endgroup$
    – Aaron Montgomery
    Dec 9, 2019 at 17:35
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    $\begingroup$ Sorry for delay had to look it up.It is on a rolling 100,newest score replaces oldest. I have only done 248 deliveries with this company but others say on average you are only rated on 1 out of every 10 deliveries but that could vary wildly for each driver. $\endgroup$
    – Jennifer
    Dec 9, 2019 at 17:39
  • $\begingroup$ Looking at the size of the changes, you've had at least 14 ratings: not that useful, but at least it's a start. $\endgroup$
    – user3482749
    Dec 9, 2019 at 17:52

1 Answer 1

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Let's start with the inference that the big drops ($4.81 \to 4.68$, $4.73 \to 4.62$) are caused by rankings of $1$. (This might be wrong, but it's a reasonable first guess to get us started.)

Call rating $i$ by the name $x_i$, and call the sum of the first $n$ ratings by the name $s_n := \sum_{i=1}^n x_i$. After $n$ total ratings, you had an average of $4.81$; that is, $\frac{s_n}{n} \approx 4.81$. You then received what may have been a rating of $1$, and it dragged your answer down; this would suggest that perhaps $\frac{s_n + 1}{n+1} \approx 4.68$. We now have two equations with two unknowns that we can solve. By substituting $4.81 n$ for $s_n$ in the second equation, we get $$\frac{4.81 n + 1}{n + 1} \approx 4.68 \implies n \approx 28.$$ But, this solution doesn't seem quite right; we need a whole number $s_{28}$ such that $s_{28} / 28 \approx 4.81$, and there isn't one. Using $s_{28} = 134$ would give an average of $4.79$, and using $s_{28} = 135$ would give an average of $4.82$. If we try the nearby number $n = 27$, we note that having $s_{27} = 130$ would give the average $130 / 27 \approx 4.81$ we wanted, and if you then received a one-star rating, the average $131/28$ would be $4.68$. This suggests we might be on to something.

Now, let's check whether this is consistent with the rest of the table, and we'll see if we can infer the individual ratings after number 27. To spoil a surprise, I think you missed one rating early on in the tracking. Individual ratings I inferred are in blue, and the rating I think you missed is in red.

$$\begin{array}{r | c| c | l} n & x_n & s_n & \text{avg} = s_n / n\\ \hline 27 & & 130 & 130 / 27 \approx 4.81 \\ 28 & \color{blue}{1} & 131 & 131 / 28 \approx 4.68 \\ \color{red}{29} & \color{blue}{5} & \color{red}{136} & \color{red}{4.69} \\ 30 & \color{blue}{5} & 141 & 4.70 \\ 31 & \color{blue}{5} & 146 & 4.71 \\ 32 & \color{blue}{5} & 151 & 4.72 \\ 33 & \color{blue}{5} & 156 & 4.73 \\ 34 & \color{blue}{\mathbb 1^*} & 157 & 4.62 \\ 35 & \color{blue}{4} & 161 & 4.60 \\ 36 & \color{blue}{4} & 165 & 4.58 \\ 37 & \color{blue}{5} & 170 & 4.59 \\ 38 & \color{blue}{5} & 175 & 4.61 \end{array}$$

So this all fits perfectly, if I'm right that you missed a single rating early on. I think you may have around $\fbox{38}$ ratings so far.

Of course, I'm drawing some inferences. Maybe you missed even more ratings, for instance. But these data tracked well enough that I feel good about this answer.

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  • $\begingroup$ Other possible sources of error: maybe those large drops were explainable by ratings of $2$ or $3$, not $1$. But like I said, the rest of the data (minus that one missing row) fit too neatly for me to worry much about this. $\endgroup$
    – Aaron Montgomery
    Dec 9, 2019 at 18:27
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    $\begingroup$ Also, to make this a bit more fun, I'll try to predict your next rating; since you usually get 5's, I bet your next rating change will take you to $180 / 39 \approx 4.62$. If you get a $4$ instead, your average will drop to $4.59$, and if you get hit with another $1$, your average will drop to $4.51$. $\endgroup$
    – Aaron Montgomery
    Dec 9, 2019 at 18:30
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    $\begingroup$ Wow. Thank you! Transferring all of this to a spreadsheet as soon as I am done typing here. I love data just cannot do math. Well this may prove your theory I just went up to a 4.62 with 1 rating as I was typing the first part. So that would have been a 5 star. EDITED to add: I am too new to upvote but it said it recorded it.Thank you again. $\endgroup$
    – Jennifer
    Dec 9, 2019 at 18:33
  • $\begingroup$ I fixed a typo in the table (changed a 5 to a 1) and marked it with an asterisk. $\endgroup$
    – Aaron Montgomery
    Dec 9, 2019 at 19:19
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    $\begingroup$ This answer is so good! $\endgroup$
    – Harshal Gajjar
    Dec 9, 2019 at 23:56

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