Statistically Impossible

According to Grok:

The current population of Cherokee County, NC, as of the latest projections for 2024, is estimated to be 30,459.

I asked Grok how many people were registered to vote in Cherokee County but it declined and sent me to the North Carolina State Board of Elections site.

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After downloading the voting information from the state for Cherokee County, I put it into an Excel Spreading and sorted by voter status. There are 21,762 “active” voters in the county.

I asked Grok “What was the average percentage population voting in presidential elections from 1900 thought 2000”?

Calculating the average voter turnout for U.S. Presidential Elections from 1900 through 2000 involves looking at the percentage of the voting-age population that voted in each election:

• 1900: 65.2%
• 1904: 65.4%
• 1908: 65.8%
• 1912: 58.8%
• 1916: 61.6%
• 1920: 49.2% (This low turnout is often attributed to the first election where women could vote, which might have affected the percentage due to organizational unpreparedness.)
• 1924: 48.9%
• 1928: 56.9%
• 1932: 56.9%
• 1936: 61.0%
• 1940: 62.5%
• 1944: 55.9% (Lower due to World War II)
• 1948: 51.1%
• 1952: 63.3%
• 1956: 60.6%
• 1960: 63.8%
• 1964: 61.8%
• 1968: 60.9%
• 1972: 55.2%
• 1976: 53.6%
• 1980: 52.6%
• 1984: 53.1%
• 1988: 50.1%
• 1992: 55.9%
• 1996: 49.0%
• 2000: 51.2%

To find the average turnout over this period:

1. Sum all the percentages: 65.2 + 65.4 + 65.8 + 58.8 + 61.6 + 49.2 + 48.9 + 56.9 + 56.9 + 61.0 + 62.5 + 55.9 + 51.1 + 63.3 + 60.6 + 63.8 + 61.8 + 60.9 + 55.2 + 53.6 + 52.6 + 53.1 + 50.1 + 55.9 + 49.0 + 51.2 = 1,585.3
2. Divide by the number of elections (26):
   1,585.3 / 26 ≈ 60.97%

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From Grok:

If you’re interested in the percentage of registered voters who turned out to vote in those elections, those figures would be higher since they only account for people who could legally vote by being registered. However, exact figures for registered voter turnout for each year might not be as readily available or standardized historically due to changes in voter registration laws and practices over time.

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From Grok:

For a rough projection: If we consider that about 70% of eligible voters are registered and take into account the historical turnout’s lower bound within that group, you might see around 40-50% of the total eligible population voting. This number implies a turnout of 57-71% of registered voters, assuming the 70% registration rate holds.

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Based on 70% of registered voters actually voting, that means statistically 15,233 people should be voting in Cherokee County. When I went to vote yesterday in early voting for Trump, over 4000 people had already voted. In 2-1/2 days of early voting, over 25% of the active voters have already voted here based on historical data.

Since I had already received information that 70% of the population is eligible to vote and 70% of eligible voters actually vote, I applied this to the population and number of votes cast in the 2020 presidential election. I then asked for a p value based on 21 previous election ending in 2020 with a standard deviation 0.0358 compared to the 86.11% value of eligible voters voting in 2020.

From Grok:

If you have 20 data sets of data with a standard deviation of .0358 for an average of 0.7, what is the p value that a different set will be an average of .8611?

Answer
To find the p-value for an average of 0.8611 given a dataset with an average of 0.7 and a standard deviation of 0.0358, follow these steps:

Calculate the p-value:

For a two-tailed test (assuming we’re interested in both tails for significance), you look for the probability associated with z≥∣4.5∣.

Using the standard normal distribution table or a calculator:

The cumulative probability for z=4.5 is extremely close to 1, practically ≈1.

The probability of z>4.5 or z<−4.5 (for a two-tailed test) is:
P(Z>4.5)≈0P(Z<−4.5)≈0

Since we’re considering both tails:p≈2×P(Z>4.5)≈2×0≈0

Interpretation:

A p-value of approximately 0 indicates that it is extremely unlikely (almost impossible, for practical purposes) to see an average of 0.8611 if the true population mean is 0.7 with a standard deviation of 0.0358. This suggests that such an occurrence would be considered highly significant or anomalous.

Remember, in practical terms, a p-value close to zero (like this) often indicates that either:

• The sample is from a different population with a different mean than expected.
• There’s an error in the data or assumptions (like the standard deviation or normality of distribution).

Given the context provided, if you’re working with real data, this extreme result might suggest reviewing your data collection method, assumptions about the population, or considering if there’s an external factor affecting this particular result.

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In order words, the number of people who voted in the 2020 presidential election is not statistically possible.

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