Comprehensive Study Notes on Population Proportions: Confidence Intervals and Hypothesis Testing

The Statistical Setting:
- Take a Simple Random Sample (SRS) of size $n$ from a large population containing an unknown proportion of "successes," denoted by $p$ .
Large Sample Confidence Interval for Population Proportion:
- An approximate level $C$ confidence interval for the unknown proportion $p$ is calculated as: $\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
- Structure of the Interval:
  - The formula follows the standard format: $\text{Estimate} \pm \text{Margin of Error}$ .
  - The Margin of Error consists of: $(\text{Critical Value}) \times (\text{Standard Error})$ .
- The Critical Value ( $z^*$ ): This is the level $C$ critical value, representing the area $C$ under the density curve between $-z^*$ and $z^*$ in the $z$ distribution with the appropriate degrees of freedom.
- Usage Requirement: This method for a confidence interval should only be used when the hard count of the number of successes and the number of failures in the sample data are both at least $15$ .

Test Statistic: To test the null hypothesis that $p = p_0$ , the sample $z$ statistic ( $z_{\hat{p}}$ ) is used in the standard normal distribution: $z_{\hat{p}} = \frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$
Null and Alternative Hypotheses:
- Null Hypothesis ( $H_0$ ): Always takes the form $H_0: p = p_0$ . It establishes a set of expectations/benchmarks against which sample data is weighed.
- Alternative Hypothesis ( $H_a$ ): Defines what we are looking for (the suspicion).
  - One-sided (larger): $H_a: p > p_0$ (The true proportion of successes is greater than $p_0$ ; Right-tailed).
  - One-sided (smaller): $H_a: p < p_0$ (The true proportion of successes is less than $p_0$ ; Left-tailed).
  - Two-sided (different): $H_a: p \neq p_0$ (The true proportion of successes is different from $p_0$ ; Two-tailed).
Conditions for Significance Tests:
1. Randomness: Sample data must come from an SRS (or at least a random and representative sample).
2. Sample Size/Large Enough: Both the hypothesized number of successes and failures must be at least $10$ .
  - $n p_0 \geq 10$
  - $n(1 - p_0) \geq 10$

Problem Context: Frequent tournaments attempt to neutralize "home field advantage." If no such advantage existed, home teams would win approximately $50\%$ of all games played.
Data Analysis (2013 MLB Season):
- Total Games ( $n$ ): $2431$
- Home Team Wins ( $x$ ): $1308$
- Sample Proportion ( $\hat{p}$ ): $53.81\%$
Research Question: Does the deviation from $50\%$ represent natural sampling variability, or is it evidence of a true home field advantage in professional baseball?

National Benchmark: According to the National Institute on Alcohol Abuse and Alcoholism, $41\%$ of college students nationwide engage in binge drinking behavior (five or more drinks on one occasion within the past two weeks).
Study Details:
- P (Population & Parameter):
  - Population: All students enrolled at the specific college.
  - Parameter ( $p$ ): The true proportion of students at this college who engage in binge drinking.
- H (Hypotheses):
  - $H_0: p = 0.41$
  - $H_a: p < 0.41$ (The true proportion is less than the national average).
- A (Assumptions and Conditions):
  1. Random Samples: $462$ students selected randomly from an enrollment list.
  2. Large Enough:
    - $n p_0 = 462 \times 0.41 \approx 189.42 \geq 10$
    - $n(1 - p_0) = 462 \times 0.59 \approx 272.58 \geq 10$
- N (Name the Test): One-proportion $z$ -test.
- T (Test Statistic): $z = -2.59$ . This value follows the standard normal distribution.
- O (Obtain P-value): $P\text{-value} = 0.005$ .
- M (Make Decision):
  - Significance Level ( $\alpha$ ): $0.05$ .
  - Comparison: $0.005 < 0.05$ .
  - Decision: Reject the null hypothesis in favor of the alternative. The evidence is statistically significant.
- S (Summary): There is statistically significant evidence at the $\alpha = 0.05$ level that the proportion of students at this college who engage in binge drinking is less than $0.41$ .

Context: In 1965, approximately $44\%$ of the U.S. adult population had never smoked.
Study Details (2010 Survey):
- P (Population & Parameter):
  - Population: U.S. adult population in 2010.
  - Parameter ( $p$ ): The true proportion of adults who have never smoked.
  - Sample Data: $n = 1472$ , Successes ( $x$ ) = $677$ , $\hat{p} \approx 0.4565$ .
- H (Hypotheses):
  - $H_0: p = 0.44$ (Benchmark establish by 1965 results).
  - $H_a: p \neq 0.44$ (Testing for a change/difference since 1965).
- A (Assumptions and Conditions):
  1. Random Samples: Unclear if the survey was an SRS, though it is presumably random.
  2. Large Enough:
    - $n p_0 = 1472 \times 0.44 \approx 647.68 \geq 10$
    - $n(1 - p_0) = 1472 \times 0.56 \approx 824.32 \geq 10$
- N (Name the Test): One-sample, one-proportion $z$ -test.
- T (Test Statistic): $z = 1.54$ .
- O (Obtain P-value): $P\text{-value} = 0.1237$ .
- M (Make Decision):
  - Comparison: $0.1237 > 0.05$ .
  - Decision: Fail to reject the null hypothesis. The $P\text{-value}$ is not statistically significant.
- S (Summary): We do not find statistically significant evidence that the proportion of U.S. adults who have never smoked has changed since 1965.

Historical Context: In the 1980s, congenital abnormalities were believed to affect approximately $5\%$ of children.
Current Study:
- P (Population & Parameter):
  - Population: Nations' children.
  - Parameter ( $p$ ): The true proportion of children with a congenital abnormality.
  - Sample Data: $n = 384$ , Successes ( $x$ ) = $46$ , $\hat{p} \approx 0.1197$ ( $11.97\%$ \text{)}.
- H (Hypotheses):
  - $H_0: p = 0.05$
  - $H_a: p > 0.05$ (One-sided, testing for an increase since the 1980s).
- A (Assumptions and Conditions):
  1. Random Samples: Unclear if data comes from a random sample.
  2. Large Enough:
    - $n p_0 = 384 \times 0.05 = 19.2 \geq 10$
    - $n(1 - p_0) = 384 \times 0.95 = 364.8 \geq 10$
- N (Name the Test): One-proportion $z$ -test.
- T (Test Statistic): $z = 6.28$ .
- O (Obtain P-value): $P\text{-value} = 1.7 \times 10^{-10}$ .
- M (Make Decision):
  - Comparison: $1.7 \times 10^{-10} < 0.05$ .
  - Decision: Reject the null hypothesis in favor of the alternative. Statistically significant evidence against the null.
- S (Summary): We have statistically significant evidence that the incidence rate of congenital abnormalities has increased since the 1980s.