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In this post we ask a question about a natural phenomenon called humans attempt to find decision by counting votes. The specific incident of such natural phenomenon that this question is about is the case of Brexit.

Note: the question is not about politics. The goal is to try to discuss such natural phenomenon from a statistical point of view based on observations.

The specific question is:

  • Question: What does the $51.9\%$ Brexit vote to leave mean? E.g. does it mean that the public really wants to leave EU? Does it simply mean that the public is unsure and needs more time to think? Or is it something else?

Assumption 1: there is no error in the voting process.

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closed as primarily opinion-based by COOLSerdash, Silverfish, Greenparker, amoeba says Reinstate Monica, gung - Reinstate Monica Jun 25 '16 at 14:05

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Democracy is not about statistical significance. The 51.9% result means that 51.9% of those who voted, voted "leave". This is not an opinion poll. Those who didn't vote, voted by (not) using their feet. Interpreting 51.9% as "public is unsure and needs more time to think" is simply lying with statistics. Brexit happened with probability 1. $\endgroup$ – Tim Jun 24 '16 at 20:07
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    $\begingroup$ This thread is destined to be non-statistical, opinionated, and possibly even polemical. It's just no fit for this site, regardless of how popular it might be. We have a chat room populated by people who would be happy to engage further in such conversations: check it out! $\endgroup$ – whuber Jun 24 '16 at 20:51
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    $\begingroup$ I believe the current discussion is statistically focused and is a good example of interpreting voting results as it applies to statistical testing. $\endgroup$ – Underminer Jun 24 '16 at 20:56
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    $\begingroup$ You bring up an important issue: measurement error of public opinion metrics such as polls. I'm afraid that main source of error is not from the sample size. $\endgroup$ – Aksakal Jun 24 '16 at 22:03
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    $\begingroup$ IMHO, this is a non-statistical question with a thin veneer of statistics added to disguise that fact. As I read it, the assumption "there is no error in the voting process" eliminates all statistical considerations and necessarily channels the discussion into what "voting ... means" in a democracy. That's a matter of political science and philosophy, not statistics. $\endgroup$ – whuber Jun 25 '16 at 14:10
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I agree with @Underminer that there is no sampling error, but not because sample is large, but because there was no sampling involved. Nobody was sampled to vote. There obviously was some negligible fraction of people who wanted to vote but, weren't able to (e.g. had car accident on this day), or who made invalid votes, but that's the only "sampling" in here.

The result is exact, there is no error involved since the whole population took part in vote (some took part by not taking part in it). Some people decided to vote, some didn't. Some decided to vote on leave, some didn't. Democracy is not about statistical significance, but about what really happened. Voting is not intended to learn about people opinion, but to make a decision. Actually, people sometimes do not vote according to what they think, but to manifest, or achieve something. For example, in election people may vote not to their preferred candidate, but to their second preferred one if they think he has greater chances of winning.

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  • $\begingroup$ Consider the case of a grey area where the voting population isn't very sure about what's good for them. For example, the case of having 2 candidates that are almost equally good. In such case, I think those who vote, will probably differ unsystematically as I think their votes might have a distribution close to a uniform one. My goal here is not to re-define democracy (a political topic) but rather to see what can we say about whether Brexit was a grey area? $\endgroup$ – caveman Jun 24 '16 at 20:54
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    $\begingroup$ @caveman no matter if they are sure, or not, what matters is how they voted since voting is about actual votes. For sure, some people did not have clear opinion, with some of them voting and some not, but this also doesn't matter since what counts is the actual votes of those who voted. $\endgroup$ – Tim Jun 24 '16 at 21:05
  • $\begingroup$ If I understand it correctly, your point is about how democracy interprets votes? I agree with you. However, I am not interpreting it in the way politicians do. I am trying to use the population to identify whether a decision is good, bad, or not very clear. This is a different usage of voting. $\endgroup$ – caveman Jun 24 '16 at 21:09
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    $\begingroup$ @caveman people change their minds all over the time, psychologists wrote thousands of papers about this... Yes, 51.9% doesn't mean that exactly 51.9% of Brits is 100% sure about leaving EU. People can even be unsure about comparing lengths of lines (en.wikipedia.org/wiki/Asch_conformity_experiments)... $\endgroup$ – Tim Jun 24 '16 at 21:37
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    $\begingroup$ @Aksakal I am not going to comment on who is eligible to vote and who isn't. I am also not going to comment on how difficult it might be to obtain the necessary credentials. That is politics and as such not on-topic here. From a statistical perspective, each eligible voter has a certain probability of not voting. This probability might be influenced by certain factors that may or may not be related to their preferences, but each eligible voter chooses (not) to exercise that right at his/her discretion. $\endgroup$ – user3697176 Jun 27 '16 at 15:03
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51.9% is the percentage of voters who want to leave. Since the sample size is so large (>33 million), there is virtually no random sampling error.

Statistical significance testing would try to determine if the difference in remain and leave could be explained by random sampling error alone, and the difference would certainly be significant (see @caveman's answer).

The problem with this approach is that statistical significance makes a very strong assumption that the sample is representative of the entire population (all of Britain), not just those who vote.

The non-response rate (those that do not vote) is enormously important in determining if more than half of all of Britain wants to 'leave', and is difficult to measure. Non-response bias is created when subgroups who are less likely to vote have systematically different views. Based on exit-polls, for example, millennials were less likely to vote, but more likely to vote to remain, which biases the results when trying to represent the population of all of Britain.

For this reason, statistical significance testing in its traditional sense is largely inappropriate.


Assumptions: We need to define some terms for any of this to make sense and avoid political discussion of what voting is trying to accomplish. Here are my definitions:

Population: Every person living in Britain

Sampling Frame: Every voting eligible person capable of voting

Sampling Methodology: Voluntary response, the act of voting is participating in the survey

Sample: The individuals who actually vote

In this setup, the sample proportion could be used (for better or worse) to estimate the percentage of all people who lean towards remain (or leave).

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You ask

What does the 51.9% Brexit vote to leave mean?

It means 51.9% of the voters voted to leave.

E.g. does it mean that the public really wants to leave EU? Does it simply mean that the public is unsure and needs more time to think? Or is it something else?

The votes comprised $17\,421\,887$ "leave" votes and $16\,146\,297$ "remain" votes, indicating $12\,931\,353$ eligible voters did not vote and approximately $18$ million inhabitants are not eligible voters. Since neither the collection of actual voters nor the collection of eligible voters is "the public" and neither is a representative (random, unbiased, pick a relevant adjective) sample of "the public", the 51.9% Brexit vote is noninforming to your second and subsequent questions.

It might have been possible to construct a questionnaire responsive to your questions. This does not seem to have been what happened in the referendum as implemented.

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    $\begingroup$ Could you please discuss the meaning of the votes in relation to the voters (i.e. not the entire population), beyond the surface conclusion that it means "51.9% voted leave"? I wonder what is the extent of information we can extract from this. $\endgroup$ – caveman Jun 25 '16 at 13:53
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    $\begingroup$ Caveman, this comment, more than any other, demonstrates your question is non-statistical. Because 51.9% (together with the total counts) constitute all the data in evidence about the voters, and there is no uncertainty (unless you want to challenge the accuracy of the counting, which is a separate issue), your rejection of this answer implies you are looking for non-statistical conclusions. $\endgroup$ – whuber Jun 25 '16 at 14:13
  • $\begingroup$ What if we model Brexit as a binary classification problem, and consider voters as estimates of classifiers that are a member of an ensemble. In this model, the goal is not to identify what the majority of citizens want, but rather the goal is to identify the optimal classifier from the space of classifiers. We can then use some measures to test the goodness of such human-voter-based classifier ensemble. E.g. we may use Perplexity or something else that is suitable for this binary classification task where ground truth is unknown (e.g. we clearly don't know if leave is better than remain). $\endgroup$ – caveman Jun 25 '16 at 18:56
  • $\begingroup$ @caveman : Given that the ground truth is (correctly) unknown, what metric would you use to "identify the optimal classifier from the space of classifiers"? Any such metric encodes the biases of the analyst that picks the metric, except for the metric "reproduces the result of the vote", for which metric you already know the answer: 51.9%/48.1%. $\endgroup$ – Eric Towers Jun 26 '16 at 21:21
  • $\begingroup$ @EricTowers I've taken this to politics.stackexchange.com where I talked about different methods -- politics.stackexchange.com/questions/11433/… $\endgroup$ – caveman Jun 26 '16 at 22:10
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TL;DR

I simulated an unsure population below (under details) for $R=1000$ times, and then measured the probability of observing a leave vote of $\ge 51.9\%$ under such unsure simulated population. This gave me the simulated probability that an unsure population can reach a leave vote that is $51.9\%$ or greater.

This simulated probability of leave under the unsure population is $0$.

Maybe redundant, but I also did the same but with remain to measure the probability that such unsure population to get a $\le 48.1\%$ vote remain.

This simulated probability of remain under the unsure population is also $0$.

Therefore I conclude that the Brexit vote is not a noisy side effect of an unsure or confused population. There seems to be a systematic reason that is deriving them to leave EU.

I uploaded the simulator code here: https://github.com/Al-Caveman/Brexit

Details

Given Assumption 1, the possible answers (or hypothesis) are:

  • $H_0$: The public is unsure.
  • $H_1$: The public confidently wants to leave.

Note: that it is impossible that the public confidently wants to remain because we have ruled out voting errors.

To answer this question (i.e. whether $H_0$ or $H_1$), I try to measure:

  • The probability that an unsure population can achieve $\ge 51.9\%$ leave vote.
  • Or, probability that an unsure population can achieve $\le 1-51.9\%$ remain vote.

If this probability is low enough, we can conclude that the public confidently wants to leave (i.e. $H_1$). However, if this probability is large enough, we can conclude that the public is unsure about deciding Brexit (i.e. $H_0$).

In order to measure this probability, we need to know the distribution of an unsure British population in such a binary voting system as Brexit. Therefore, my first step is to this is to simulate this distribution by following the assumption below:

  • Assumption 2: a population that is composed of unsure individuals will have a random chance vote. I.e. every possible answer has an equal chance of being chosen.

In my view this assumption is fair/reasonable.

Additionally, we model the leave and remain campaigns as two distinct processes as follows:

  • Process $P_{\text{leave}}$ with the output $O_{\text{leave}} = [l_1, l_2, \ldots, l_n]$.
  • Process $P_{\text{remain}}$ with the output $O_{\text{remain}} = [r_1, r_2, \ldots, r_n]$.

where:

  • $n$ is the total population of UK (includes non-voters).
  • For any $i \in \{1,2,\ldots,n\}$, $l_i,r_i \in \{0, 1\}$. An output value of $0$ signifies that a voter has voted no for the subject process, and $1$ significances that a voter has voted yes for the same process.

subject to the following constraint:

  • For any $i \in \{1,2,\ldots,n\}$, $l_i$ and $r_i$ cannot simultaneously be $1$ at the same time. I.e. $l_i=1$ necessarily implies that $r_i = 0$, and $r_i=1$ necessarily implies that $l_i=0$. This is due to the fact that a voter $i$ among the population $\{1,2,\ldots,n\}$ cannot vote to both leave and remain at the same time.

For example, if $O_{\text{leave}} = [1,0,0]$, it means that out of a population of $3$, one has voted yes to leave and two have voted no to leave.

Likewise, if $O_{\text{remain}} = [0,1,0]$, it means that out of a population of $3$, one has voted yes to remain and two have voted no to remain.

Note that in both of the examples above, there is one member of the population that has not voted for any of the processes (or campaigns). Specifically, the third voter (i.e. $O_{\text{leave}}[3] = O_{\text{remain}}[3] = 0$).

What we know from here is that out of $33,568,184$ ballot papers, $51.9\%$ have voted to leave EU (i.e. $100-51.9=48.1\%$ voted to remain). This means:

  • $n = 33,568,184$.
  • $33,568,184 \times 0.519 = 17,421,887.496$ have voted yes to the leave campaign. I.e. $$ \sum_{i=1}^{33,568,184}O_{\text{leave}}[i] = 17,421,887.496 \approx 17,421,887 $$
  • $33,568,184 \times (1-0.519) = 16,146,296.504$ have voted yes to the remain campaign. I.e. $$ \sum_{i=1}^{33,568,184}O_{\text{remain}}[i] = 16,146,296.504 \approx 16,146,297 $$

Therefore, we define the output arrays as follows:

  • For all $i \in \{1,2,\ldots, 17421887\}$, $O_{\text{leave}}[i] = 1$.
  • For all $i \in \{17421887+1,17421887+2,\ldots, 33568184\}$, $O_{\text{leave}}[i] = 0$.
  • For all $i \in \{1,2,\ldots, 17421887\}$, $O_{\text{remain}}[i] = 0$.
  • For all $i \in \{17421887+1,17421887+2,\ldots, 33568184\}$, $O_{\text{remain}}[i] = 1$.
  • By Assumption 2, for all $i \in \{1,2,\ldots, 33568184\}$, $O_{\text{unsure},m}[i] = C$, where $C$ is a uniformly distributed random variable that takes values in $\{0,1\}$ (e.g. a fair coin toss), and $m$ is a number that identifies a particular random instantiation of $O_{\text{unsure},m}$. In other words, the probability that two distinct random instantiations of $O_{\text{unsure},m}$ equal each other, i.e. $O_{\text{unsure},1} = O_{\text{unsure},2}$, is $0.5^{33,568,184}$.

Finally, we define the $p_{\text{leave}}$ value of the leave process as follows: $$ p_{\text{leave}} = \frac{1}{R}\sum_{m=1}^R \begin{cases} 1 & \text{if } \Big(\sum_{i=1}^{33,568,184} O_{\text{leave}}[i]\Big) \le \Big(\sum_{i=1}^{33,568,184} O_{\text{unsure},m}[i]\Big)\\ 0 & \text{else} \end{cases} $$ where $R$ is total number of simulation rounds by which at each time a random instance of $O_{\text{unsure},m}$ is defined.

Likewise, we define the $p_{\text{remain}}$ value of the remain process as follows: $$ p_{\text{remain}} = \frac{1}{R}\sum_{m=1}^R \begin{cases} 1 & \text{if } \Big(\sum_{i=1}^{33,568,184} O_{\text{remain}}[i]\Big) \ge \Big(\sum_{i=1}^{33,568,184} O_{\text{unsure},m}[i]\Big)\\ 0 & \text{else} \end{cases} $$

To answer that, I simulated the above in C using $R=1,000$ and the output is:

total leave votes: 17421887
total remain votes: 16146297
simulating p values............ ok
p value for leave: 0.000000
p value for remain: 0.000000

In other words:

  • $p_{\text{leave}} = 0$.
  • $p_{\text{remain}} = 0$.
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    $\begingroup$ Perhaps more important in this case is the non-response rate (i.e. individuals who do no vote). The margin of error (or measure of statistical significance) only takes into account random sample error. Non-response bias is NOT included in this, and it is much more impactful than random sampling error with a poll with such a large sample size. $\endgroup$ – Underminer Jun 24 '16 at 19:46
  • $\begingroup$ Here it says that UK has $46,499,537$ eligible voters. Meaning $46,499,537 - (17421887+16146297) = 12,931,353$ didn't vote. Any idea how to interpret such unvoting population? Source: en.wikipedia.org/wiki/… $\endgroup$ – caveman Jun 24 '16 at 20:09
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    $\begingroup$ There is no statistically satisfactory way to deal with non-random missing data. $\endgroup$ – Underminer Jun 24 '16 at 20:12
  • $\begingroup$ Those who haven't voted, could be composed of individuals that don't care about politics (e.g. no more trust). Alternatively, it such unvoters could be those who were not sure. Or, it could be a mixture of the two. What would happen if we assume that "all unvoters are unsure"? Would this be an upper bound for testing whether the current situation was one where the public was feeling that Brexit was a grey area? $\endgroup$ – caveman Jun 24 '16 at 21:03
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    $\begingroup$ There is a confusion here about the nature & scope of statistics. You are attempting to create a process model of voting, & how that can inform the mechanisms & validity of governance & public decision making. This is a worthwhile task in Political Science. It is simply not statistics (although statistics is involved). $\endgroup$ – gung - Reinstate Monica Jun 24 '16 at 21:36
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You could ask a slightly different question: Assuming that 50% of a very large population voted "Yes", and you asked a random sample of size S, what is the probability that 51.9% of your sample responded "Yes", depending on the sample size?

The expected value of number of "Yes" votes is 0.5 S. The variance is 0.25 S. The standard definition is 0.5 $S^{1/2}$. A deviation of the actual from the expected number of "Yes" votes more than 6.1 standard deviations has a chance of one in a billion.

We have this when 0.019 S (difference between 50% and 51.9%) is 6.1 * 0.5 * $S^{1/2}$, or S = $(6.1 * 0.5 / 0.019)^2$ or S ≈ 25,800.

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This is another solution using an analytical method instead of a simulation.

Previously, I have simulated an unsure population to be one that its vote is random chance guessing. So out of $n$ many voters, an unsure population would tend to vote leave or remain for $0.5$ of the time.

In order for an unsure population to get exactly $51.9\%$ vote on leave, there needs to be $17,421,887$ 1s in $O_{\text{leave}}$. The probability for this is $0.5^{33,568,184}$. Similarly, the probability of getting $17,421,887 + 1$ votes is also $0.5^{33,568,184}$. This goes on.

This is the probability of getting $\ge 17,421,887$ votes: $$\begin{split} \sum_{i=17,421,887}^{33,568,184} 0.5^{33,568,184} &= (33,568,184-17,421,887) \times 0.5^{33,568,184}\\ &= 8.39663381928984×10^-10105024\\ &\approx 0\\ \end{split} $$

($8.39663381928984×10^-10105024$ calculated by Wolframalpha)

And this is the probability of having $\ge 51.9\%$ of an unsure population vote leave.

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