My Math Forum Question on statistical evidence

 Probability and Statistics Basic Probability and Statistics Math Forum

 October 10th, 2018, 07:02 PM #1 Newbie   Joined: Oct 2018 From: Canada Posts: 4 Thanks: 1 Question on statistical evidence Hi! This is my first post on this forum. If I'm doing anything incorrectly, please let me know! Also, I realize that this is probably not the typical type of problem that you guys encounter here, so if for whatever reason it isn't fitting for this forum, let me know about it. I have next to no knowledge regarding probabilities, but there's one of its implications which made me want to understand how they work for that particular scenario It has to do with measuring the ''strength'' of statistical evidence. Consider this. We have a process which has two possible outcomes. We believe that this process is purely random and has 50/50 percent chances of going to either outcome. We could think of it as a coin toss, except we don't yet know that the outcome is purely 50/50. Let's say we replicate this process 100 times. We get the first outcome 40 times and the second outcome 60 times. I want to find how strong of an evidence this is to show that the event is purely 50/50 in odds. It probably isn't such strong evidence, because I would think that getting a distribution as uneven, or even more uneven that 40:60 is quite unlikely. But how unlikely is it exactly? I would think that we would have to add the odds of getting the distributions 0:100, 1:99, 2:98 etc. and the odds of getting the distributions 100:0, 99:1, 98,2 etc. This should give us the odds of getting such an uneven distribution. If the odds are really low we would know that our statistical evidence is really strong. -I think that the odds of getting 0:100 is 1/2^100, because there is only one possibility to get 0:100 -I think that the odds of getting 1:99 is 100/2^100, because there are 100 instances where we could get our one ''distinct'' event. -I think that the odds of getting 2:98 is (99+98+97...+1)/2^100 I am quite unsure of the last one, however. I sort of worked it out to the best of my abilities on paper and this is what I obtained. Now I was expecting to see some sort of trend pop out, and I was hoping to work it out from there, but I can't really see anything interesting. This is all the progress I have made, and this is where I need help. Feel free to redirect me to some of the theory related to these things if you think that I could figure it out myself based on what I said already. Thank you so much! Last edited by skipjack; October 11th, 2018 at 09:39 PM.
 October 11th, 2018, 12:05 AM #2 Senior Member   Joined: Oct 2009 Posts: 630 Thanks: 193 Do you know about binomial coefficients? $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ Then the odds of getting $k : (100-k)$ is just $$\frac{1}{2^{100}} \binom{100}{k}$$
 October 11th, 2018, 02:28 PM #3 Newbie   Joined: Oct 2018 From: Canada Posts: 4 Thanks: 1 Oh, so that's how it works! I put the theory into practice and I believe I found the correct answer. I did the sum of these binomial coefficients from k=0 to k=40. I then multiplied by 2 to get the opposite repartitions included, and then I divided by 2^100 to get the percentage. I obtained 0.0568 which is 5.68% So I conclude that we had only 5.68% chances of getting such an uneven repartition of outcomes. Next I will try to learn where these binomial coefficients come from, and I'll be all set! Thanks a lot for your help! Thanks from studiot Last edited by skipjack; October 11th, 2018 at 09:40 PM.
 October 11th, 2018, 02:50 PM #4 Senior Member   Joined: Oct 2009 Posts: 630 Thanks: 193 Yep. So let's throw in some terminology. What you are doing is called a hypothesis test. It basically considers two options against eachother. One option is called the null-hypothesis, the other is called the alternative hypothesis. You must be careful what is what. The null hypothesis is the one you are trying to DISprove. In our case, the coin being fair is the null hypothesis: we are trying to prove the coin is NOT fair. The 5.68% you found is called the p-value and is indicative of the strength of your evidence against the null hypothesis. In social sciences, a p-value of 0.05 or less is usually taken to be enough evidence to say you disproved the null-hypothesis (but these things are very debatable). Something to keep in mind: you have a p-value of 0.0568 which is (according to tradition) NOT enough to disprove the null hypothesis. This does NOT mean you get to accept the null hypothesis as true. Indeed: not being able to disprove the null $\neq$ accepting the null. Compare it with a murder trial. A jury might not have found the selected evidence enough to convict the guy of murder. Thus the jury declares him not guilty. This is NOT the same as the person being declared innocent. Thanks from Benit13 and studiot
 October 11th, 2018, 03:01 PM #5 Newbie   Joined: Oct 2018 From: Canada Posts: 4 Thanks: 1 Right, that makes a lot of sense. Thank you very much
 October 12th, 2018, 01:30 AM #6 Senior Member   Joined: Jun 2015 From: England Posts: 891 Thanks: 269 Datlemondoe, +1 for catching on so quickly Keep the enquiries coming. Micromass +1 for the developing explanation especially the Agatha Christie bit at the end. Perhaps you could incorporate the Scottish verdice of Not Proven next time?
October 12th, 2018, 02:17 AM   #7
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Quote:
 Originally Posted by studiot Perhaps you could incorporate the Scottish verdice of Not Proven next time?
Very interesting! Thanks for showing me this.

I really only understood this stuff by watching some skepticism stuff by Randi. The idea is: either unicorns exist or they don't exist. But just because you can't prove they exist, doesn't mean they don't exist. And just because you can't prove they don't exist, doesn't mean they exist. Only when contemplating this example did the entire hypothesis testing make sense to me.

October 12th, 2018, 03:34 PM   #8
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Quote:
 Originally Posted by studiot Datlemondoe, +1 for catching on so quickly Keep the enquiries coming.
Hahaha cheers Studiot

I will be back on this forum for sure whenever I need you guys' help.

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