Sampling Distributions, Unbiased Estimators, and Empirical Proportions

Foundations of Sampling Distributions and Unbiased Estimators

• Core Principle of Sampling Distributions:

  • The mean of the sample proportions ($\mu_{\hat{p}}$) targets the true population proportion ($p$).
  • In a balanced binary experiment (such as tossing a fair coin $100$ times), the population proportion of heads is $p = 0.5$.
  • When multiple repeated samples of size $n = 100$ are taken, each individual sample yields an empirical proportion (e.g., the proportion of heads). The arithmetic average of all these sample proportions converges precisely on the true parameter: $\mu_{\hat{p}} = 0.5$.

• Sample Means as Unbiased Estimators:

  • The mean of the sample means ($\mu_{\bar{x}}$) targets and equals the true population mean ($\mu$):     $\mu_{\bar{x}} = \mu$
  • If repeated samples of size $n = 100$ of human body temperatures are measured, each sample generates an individual sample mean (denoted as $\bar{x}$). Averaging the sample means across all possible samples yields the accepted human population baseline average of $98.6\,\text{F}$.

• Standard Deviation Exception (Biased Estimator):

  • The property of an estimator targeting the true population parameter does not hold for sample standard deviations.
  • Calculating the standard deviation ($s$) for multiple samples of $100$ body temperatures and computing the arithmetic average of those standard deviations fails to equal the true population standard deviation ($\sigma$):     $\text{Average of } s \neq \sigma$
  • Sample standard deviation is inherently a biased estimator when computed in this manner, distinguishing it from sample means and sample proportions.

Political Polling and Practical Sampling Distributions

• Context and Cultural Relevance:

  • Societal discourse frequently identifies three controversial topics: gender, politics, and pineapple on pizza.
  • Midterm elections (held around November) demonstrate direct, large-scale applications of sampling distributions via opinion polling.

• Polling Methodology vs. Public Perception:

  • Polling data is often colloquially dismissed as "fake polls" by individuals who misunderstand sampling methodologies.
  • Pollsters do not casually ask arbitrary pedestrians for their opinions; they use structured, systematic sampling distribution protocols to quantify public opinion accurately.

• Hypothetical Polling Mechanics:

  • Consider an election between Candidate A and Candidate B:
  • Candidate A reports a projected vote/approval share of $59\%$ ($0.59$).
  • Candidate B reports a projected vote/approval share of $45\%$ ($0.45$).
  • Candidate A is reported as possessing a $14$-point lead over Candidate B ($59\% - 45\% = 14\%$).
  • The reported percentages do not sum to $100\%$ due to unpublicized third-party candidates (Candidate C) or five to six minor candidates who individually register less than $1\%$ of the vote.

• Regional Aggregation Strategy:

  • Pollsters gather multiple discrete samples of $n = 100$ individuals from varied geographic areas across the country:
  • Region 1 sample: $64\%$ approval for Candidate A.
  • Region 2 sample: $39\%$ approval for Candidate A.
  • Region 3 sample: $55\%$ approval for Candidate A.
  • By calculating the average across all regional sample distributions (e.g., averaging $40$ distinct samples of $100$ individuals), the resulting aggregate proportion reliably estimates the true population proportion ($p$).

• Non-Sampling Bias and Polling Errors:

  • Polling errors occur when sample distributions encounter systematic non-response or selection bias.
  • Frequent unsolicited outreach (such as individuals receiving text surveys three times a day) leads many respondents to flag outreach as spam, systematically altering the respondent pool.

Mathematical Demonstration: The Discrete Spinner Model

• Motivation and Dice Analogy:

  • Rolling a pair of standard six-sided dice produces $36$ possible sample space outcomes ($6 \times 6 = 36$). Rolling two dice repeatedly yields outcomes such as $2$, $7$, or $11$, with the overall long-run expected average outcome stabilizing at $7$ (observed in casino craps, where a losing roll causes all players at the table to lose simultaneously).
  • To examine every single sample in an entire population without dealing with large sample spaces, a miniature, finite universe is constructed.

• Model Setup:

  • Instrument: A three-sided spinner (equivalent to a three-sided die).
  • Population values: $\mathbf{S} = \{2, 5, 11\}$.
  • Population size: $N = 3$.
  • Sampling scheme: Samples of size $n = 2$ are drawn with replacement (spinning the spinner twice).
  • Number of all possible distinct sample orders: $3^2 = 9$ total samples.

• Population Mean Calculation:$\mu = \frac{2 + 5 + 11}{3} = \frac{18}{3} = 6$

• Complete Exhaustive Sample Space ($n = 2$ with Replacement):

  • Sample 1: $(2, 2)$ with sample mean $\bar{x}_1 = \frac{2 + 2}{2} = 2.0$
  • Sample 2: $(2, 5)$ with sample mean $\bar{x}_2 = \frac{2 + 5}{2} = 3.5$
  • Sample 3: $(2, 11)$ with sample mean $\bar{x}_3 = \frac{2 + 11}{2} = 6.5$
  • Sample 4: $(5, 2)$ with sample mean $\bar{x}_4 = \frac{5 + 2}{2} = 3.5$
  • Sample 5: $(5, 5)$ with sample mean $\bar{x}_5 = \frac{5 + 5}{2} = 5.0$
  • Sample 6: $(5, 11)$ with sample mean $\bar{x}_6 = \frac{5 + 11}{2} = 8.0$
  • Sample 7: $(11, 2)$ with sample mean $\bar{x}_7 = \frac{11 + 2}{2} = 6.5$
  • Sample 8: $(11, 5)$ with sample mean $\bar{x}_8 = \frac{11 + 5}{2} = 8.0$
  • Sample 9: $(11, 11)$ with sample mean $\bar{x}_9 = \frac{11 + 11}{2} = 11.0$

• Summation and Mean of the Sample Means ($\mu_{\bar{x}}$):

  • Sum of all $9$ sample means:     $\sum_{i=1}^{9} \bar{x}_i = 2.0 + 3.5 + 6.5 + 3.5 + 5.0 + 8.0 + 6.5 + 8.0 + 11.0 = 54.0$
  • Mean of the sample means:     $\mu_{\bar{x}} = \frac{54.0}{9} = 6.0$
  • Conclusion:     $\mu_{\bar{x}} = \mu = 6.0$
  • The average of all possible sample means equals the population mean exactly, not merely as an approximation.

Verification of Sample Proportions

• Definition of Success:

  • Let the event of interest be spinning an odd number.
  • In the population $\mathbf{S} = \{2, 5, 11\}$, the odd values are $5$ and $11$.

• True Population Proportion ($p$):$p = \frac{2}{3} = 0.\bar{6}$

• Sample Proportions ($\hat{p}$) Across All $9$ Samples ($n = 2$):

  • Sample 1: $(2, 2)$ contains $0$ odd numbers $\rightarrow \hat{p}_1 = \frac{0}{2} = 0.0$
  • Sample 2: $(2, 5)$ contains $1$ odd number $\rightarrow \hat{p}_2 = \frac{1}{2} = 0.5$
  • Sample 3: $(2, 11)$ contains $1$ odd number $\rightarrow \hat{p}_3 = \frac{1}{2} = 0.5$
  • Sample 4: $(5, 2)$ contains $1$ odd number $\rightarrow \hat{p}_4 = \frac{1}{2} = 0.5$
  • Sample 5: $(5, 5)$ contains $2$ odd numbers $\rightarrow \hat{p}_5 = \frac{2}{2} = 1.0$
  • Sample 6: $(5, 11)$ contains $2$ odd numbers $\rightarrow \hat{p}_6 = \frac{2}{2} = 1.0$
  • Sample 7: $(11, 2)$ contains $1$ odd number $\rightarrow \hat{p}_7 = \frac{1}{2} = 0.5$
  • Sample 8: $(11, 5)$ contains $2$ odd numbers $\rightarrow \hat{p}_8 = \frac{2}{2} = 1.0$
  • Sample 9: $(11, 11)$ contains $2$ odd numbers $\rightarrow \hat{p}_9 = \frac{2}{2} = 1.0$

• Summation and Mean of the Sample Proportions ($\mu_{\hat{p}}$):

  • Sum of all $9$ sample proportions:     $\sum_{i=1}^{9} \hat{p}_i = 0.0 + 0.5 + 0.5 + 0.5 + 1.0 + 1.0 + 0.5 + 1.0 + 1.0 = 6.0$
  • Mean of the sample proportions:     $\mu_{\hat{p}} = \frac{6.0}{9} = \frac{2}{3} = 0.\bar{6}$
  • Conclusion:     $\mu_{\hat{p}} = p = \frac{2}{3} = 0.\bar{6}$
  • Even with a minimal population ($N = 3$) and small sample size ($n = 2$), the average of all possible sample proportions matches the population proportion with zero margin of error.

Classroom Dialogue and Discussion

• Inquiry on Expected Average of Coin Toss Proportions:

  • Prompt: In a class exercise where every student completes $100$ coin tosses and submits their results, what should the overall average of these sample averages be?
  • Response: One half ($0.5$).
  • Verification: Displaying class data confirms that the mean of the sample proportions equals $0.5$.

• Inquiry on Three Controversial Topics:

  • Prompt: What are three major controversial topics?
  • Response: Gender, politics, and pineapple on pizza.

• Inquiry on Possible Outcomes of Spinning Twice ($n = 2$):

  • Prompt: For a spinner with outcomes $2, 5, 11$, what is one possible outcome when spinning twice?
  • Response: Two and a five ($2, 5$).
  • Prompt: What is another possible outcome?
  • Response: Five and eleven ($5, 11$).
  • Prompt: What else can happen?
  • Response: Five and two ($5, 2$).
  • Prompt: Where else might it land?
  • Response: Eleven and two ($11, 2$).
  • Prompt: Could the spinner land on eleven and eleven ($11, 11$) or two and two ($2, 2$)?
  • Response: Yes, both are valid outcomes (though rolling two twos would yield minimal movement on the game board).

• Inquiry on the Average of Sample Means:

  • Prompt: Summing all nine sample means from the spinner table and dividing by nine, what is the resulting average?
  • Response: Six ($6$).
  • Prompt: What is the true average of the population ($\{2, 5, 11\}$)?
  • Response: Six ($6$).

• Inquiry on Sample Proportions of Odd Numbers:

  • Prompt: How many odd numbers are in the sample $(2, 2)$?
  • Response: Zero ($0$).
  • Prompt: What is the proportion of odd numbers in that sample?
  • Response: Zero ($0.0$).
  • Prompt: How many odd numbers are in the sample $(2, 5)$?
  • Response: One ($1$).
  • Prompt: What is the proportion of odd numbers in that sample?
  • Response: One half ($0.5$).
  • Prompt: How many odd numbers exist in the total population of three numbers ($2, 5, 11$)?
  • Response: Two ($2$), yielding two-thirds or $0.6$ repeating ($0.\bar{6}$).
  • Prompt: What is the calculated average when all nine sample proportions are summed and divided by nine?
  • Response: Two-thirds or $0.6$ repeating ($0.\bar{6}$).