— **Closed**. Many satisfying answers. Thanks! —
Let’s say people on the street are asked to rate three different movies on a scale of 1-10, but only if they saw it. We get the following, *already averaged* data:
* Movie A: 1000 people gave A 8.1 on avg.
* Movie B: 10’000 people gave B a 6.9 on avg.
* Movie C: 20’000 people gave C a 7.4 on avg.
Simply according to the averaged ratings the movie to watch would be movie A. But taking into account that way less people watched A, its rating does not seem as accurate, i.e. trustworthy as the ratings of the other two movies: C seems to be best suited for the average viewer, while A seems to be the choice if you like that type of movie (or it is a hidden gem).
So back to my question: Is there a way to calculate one value (preferably in the original rating range) for each movie that allows accurate (approximate) comparison of these movies – i.e. that takes the count of participants into account?
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*What I’ve tried:*
* Dividing and multiplying these values:
|Movie|Participants: *ps*|Avg. rating: *rg*|*rg/ps*1000*|*ps/rg*|*ps*rg/1000*|
|:-|:-|:-|:-|:-|:-|
|A|1000|8.1|8.10|123.5|8.1|
|B|10000|6.9|0.69|1449.3|69.0|
|C|20000|7.4|0.37|2702.7|148.0|
||||useless, I think|better, but probably not fair|maybe even better, but still unfair|
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* Getting the total of all participants and averaging the ratings accordingly
In total 31000 people have participated and have given a 7.3 to all three movie on average [sum of (ps*rg for A,B,C) divided by 31000]. I feel like this could be helpful, but then again I have not gained much with this new average. I simply cannot grasp how I could find such a number…
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Any help/explanation appreciated!
In: Mathematics
Should you buy a product with 100% positive ratings from 2 people, 80% positive ratings from 10 people, or 60% from 100 people.
What you can do is add one positive and one negative review. So it would be 2/4 on the first. 9/12 for the second. And 61/102 with the last. That would be 50%, 82% and 60%. So the second product would be your goto.
Dk if it works with so big numbers, as it would merely affect the result.
Source: 3Blue1Brown
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