10.5: Elections in Ruritania

Even though it is an absolute monarchy, national elections are held in Ruritania to elect members of the Ruritanian parliament, the Národní Shromáždění (National Assembly). There are many parties represented in the parliament, but the party that consistently receives a majority of the seats and votes is the monarchist Pohyb pro Ruritánii (PR; Movement for Ruritania). The main opposition party is the Demokratické Hnutí (DH; Democratic Movement) party, but votes are also usually received by the Socialistická (SP; Socialist Party), Křstanská Demokratická (KD, Christian Democratic), and Republikánská (RS, Republican) parties.

It is fortuitous that Ruritania does not use computerized ballots. They use ballot papers for the parliamentary election that consist of the party names, symbols, and abbreviations... and a box for the voter to place their inked fingerprint next to the party. After voting, the ballot is placed in a ballot box to await counting.

At the end of the evening, the ballot papers at each precinct are securely transported to the division headquarters, where they are counted by electoral officials. Each ballot is checked by that official to ensure that it was lawfully cast and that the "will of the voter" can be discerned.

When the division is finished counting the ballots, the totals are then telephoned to the Independent Electoral Commission (Nezávislá Volební Komise, NVK) in the capital. With much pomp and circumstance, and not a little fanfare, the division totals are added and reported to the people.

After the last election, the exiles in Denmark claimed that the ballot boxes were stuffed. That is, the ballot boxes had votes for the PR party in them before voting began. Because guarantees of the secret ballot are built into the Ruritanian Constitution, the ballot boxes are opaque. In other words, direct evidence of ballot box stuffing does not exist, only claims by those who live in exile in another country (Denmark). However, if ballot box stuffing existed in this election to any great extent, it would leave evidence. Why/how?

Question:

What do stuffing ballots have that naturally cast ballots do not?

Easy: The stuffing ballots are all for the ruling PR party and they are all completed (filled in) correctly. The naturally cast ballots will consist of votes for all parties and will include ballots not filled-in correctly.

And so, in the presence of systematic and significant ballot box stuffing, there will be a relationship between the invalidation rate and the level of support for the ruling party.

That is the theory. The exiles are paying for this analysis. We like the money, so we need to be confident — and clear — in our conclusions. The NVK is providing the official counts in the rur2013parl data file, so we need to ensure that the statistical analysis is clean. That is, it is up to us to do the analysis correctly, neither concluding too much nor too little.

And, as always, being clear in our reasoning.

Ordinary Least Squares

The first analysis we will do is ordinary least squares. The dependent variable is the invalidation rate; the independent variable is the support for the Movement for Ruritania (PR) party. Why them? Since they were in charge before the election, they were in position to stuff the ballot boxes.

Here is the code to load the data, create the variables, fit the model, and determine if a relationship between the invalidation rate and PR support rate can be detected.

votes = read.csv("http://rur.kvasaheim.com/data/rur2013parl.csv")
attach(votes)

Valid  = Total-Invalid
pPR    = PR/Valid
pInv   = Invalid/Total

modOLS = lm(pInv~pPR)

summary(modOLS)

These results indicate that we did not detect a relationship... at the \(\alpha=0.05\) level \((\texst{p-value}=0.0668)\). Thus, ordinary least squares did not detect unfairness in the vote.

It is important to emphasize here that the correct terminology is that we did not detect unfairness. We cannot say there was no fairness. We can only say we didn't detect it.

Remember to check the assumptions. This point cannot be over-emphasized. If the assumptions are not met, then the model is not correct. Well, not perfectly correct. See Chapter 6: Dood! Check the Requirements for a discussion of this point.

Weighted Least Squares

Note that ordinary least squares is not be the best option here. The invalidation rate has greater inherent variability in smaller divisions than in larger. We know this because of the distribution of the invalidation rate. Invalidation counts follow something akin to a Binomial distribution. Its two parameters are sample size (number of votes cast) and success probability (invalidation rate). The variance of a Binomial random variable is \(n \pi (1-\pi)\).

Dividing the invalidation count by the number of votes cast gives the invalidation rate. The distribution of the invalidation rate can be approximated with a Normal distribution (see the Central Limit Theorem). The expected value of the observed invalidation rate is \(\pi\), the inherent invalidation rate. The variance is \(\pi(1-\pi)/n \propto 1/n\). Because the data are heteroskedastic in nature, and because the structure of the heteroskedasticity is known, weighted least squares will be more appropriate here.

Here is the code. Compare it to the ordinary least squares code from above.

Valid = Total-Invalid
pPR   = PR/Valid
pInv  = Invalid/Total

modWLS = lm(pInv~pPR, weights=Total)

summary(modWLS)

This produces the following (abbreviated) output:

Coefficients:
            Estimate  Std. Error  t value  Pr(>|t|)
(Intercept)   2.0726      0.7109    2.915    0.0225 *
pPR          -2.8808      1.0124   -2.845    0.0249 *

Note that this method did detect a relationship between the invalidation rate and the PR party support rate (\(\text{p-value}< 0.05 = \alpha\)). Furthermore, that differential invalidation helped the ruling party. The negative coefficient indicates that those kraj with higher PR support also tended to count more of the votes (reject fewer). Thus, we can conclude that the data are consistent with the exile claim of ballot box stuffing. Figure \(\PageIndex{1}\) illustrates this.

Figure \(\PageIndex{1}\): An invalidation plot for the 2013 Ruritanian parliamentary election. The lines of best fit are provided. The OLS fit is in brown and the WLS is in red.

Question:

How much is the differential invalidation?

Well, from the regression output, we know that when the support for PR increases by 10 percentage points (from 65 to 75%, for instance), the invalidation drops by an average of 27pp. That seems rather substantial to me.

Caution

Nothing in statistics ever constitutes proof. Nothing. Ever. Period. End of thought. Žádné další!

Statistics only provides evidence in favor of — or against — the null hypothesis. In this case, the p-value is 0.0249. If the null hypothesis is correct, then we would observed results this extreme or more so 2.49% of the time. This is not incredibly rare, especially when you realize you are claiming the government cheated. Cheating is a more serious claim than just that someone was mistaken.

It is always better to report the results, interpret the results, be explicit that there is no proof and that the null hypothesis has a non-zero probability of being reality.

Do not live your statistical life ruled by \(\alpha=0.05\). Realize — and accept — that the p-value is a measure of how well the data support the null hypothesis, the hypothesis of no relationship/difference/effect/evidence.