Statistics 1 for Economics - Proportions and Hypothesis Testing

Introduction to Statistical Variables

• Population vs. Sample: Statistics involves making inferences about a larger population based on data collected from a sample.
• Quantitative Variables: Variables that measure a numerical quantity. Examples include:
  • How many hours of sleep do you get?
  • How often do you get up during the night?
  • How high is the blood pressure measured in numerical values?
  • Mean number of hours that adolescents sleep per $24$\ hours.
• Nominal Variables: Categorical variables represented by qualitative descriptions, often resulting in "yes/no" responses. Examples include:
  • Do you use painkillers or sleeping tablets?
  • Do you usually dream about school?
  • Did you see a scary movie before you went to bed?

Hypothesis Testing for Proportions (Normal Approximation)

• Assumptions and Conditions: For a large-sample test of a population proportion $p$ using the Normal distribution, the following must be met:
  • A random sample must be used.
  • $np_0 \geq 5$
  • $n(1 - p_0) \geq 5$
• Null and Alternative Hypotheses:
  • $H_0: p = p_0$ (The null hypothesis, representing no change or a specific status quo).
  • $H_1: p > p_0$, $H_1: p < p_0$, or $H_1: p \neq p_0$ (The research or alternative hypothesis).
• Test Statistic: The standardized distance between the sample proportion and the hypothesized proportion:
  • $Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$
  • Distribution: $Z \sim N(0, 1)$.
• Case Study: Question 1 (RTVL Market Share):
  • Scenario: An advertiser buys time if RTVL's market share is greater than $5\%$.
  • Data: $n = 500$, $\hat{p} = 6.2\% = 0.062$, $\alpha = 1\%$.
  • Check Conditions:
    • $np_0 = 500 \times 0.05 = 25 \geq 5$
    • $n(1 - p_0) = 500 \times 0.95 = 475 \geq 5$
  • Hypotheses: $H_0: p = 0.05$ vs. $H_1: p > 0.05$.
  • Rejection Region: At $\alpha = 1\%$, reject $H_0$ if $Z \geq z_{crit} = 2.33$.
  • Sample Realization:
    • $z_{obs} = \frac{0.062 - 0.05}{\sqrt{\frac{0.05 \times 0.95}{500}}} = 1.23$
  • Decision: Since $z_{obs} < z_{crit}$ ($1.23 < 2.33$), do not reject $H_0$.
  • Conclusion: At a significance level of $1\%$, there is insufficient evidence to infer that the market share is more than $5\%$.\n

The p-value and Power of the Test

• p-value Definition: The probability of obtaining evidence at least as strong as the observed sample, assuming the null hypothesis is true.
  • Calculation for Question 1: $p\text{-value} = P(Z \geq z_{obs}) = P(Z \geq 1.23) = 1 - 0.8907 = 0.1093$.
• Power of the Test: The probability of correctly rejecting the null hypothesis when it is false ($P(\text{reject } H_0 \mid H_0 \text{ is false})$).
• Calculating Power (Example Q3):
  • Target: Find power if true $p = 0.09$ for the RTVL test at $\alpha = 1\%$.
  • Step 1: Find the critical sample proportion ($\hat{p}_{crit}$) necessary to reject $H_0$:
    • $\hat{p}_{crit} = p_0 + z_{crit} \times \sqrt{\frac{p_0(1 - p_0)}{n}}$
    • $\hat{p}_{crit} = 0.05 + 2.33 \times \sqrt{\frac{0.05 \times 0.95}{500}} = 0.0727$
  • Step 2: Calculate the probability of observing $\hat{p} \geq 0.0727$ given the new population proportion $p = 0.09$:
    • $Z = \frac{0.0727 - 0.09}{\sqrt{\frac{0.09 \times (1 - 0.09)}{500}} } = -1.35$
    • $\text{Power} = P(Z \geq -1.35) = 0.9115$.

Distributions and Test Statistics

• Counts (X):
  • $X \sim Binomial(n, p)$.
  • Mean: $\mu = E(X) = np$.
  • Variance: $\sigma^2 = V(X) = np(1 - p)$.
  • Standardization: $Z = \frac{X - np}{\sqrt{np(1 - p)}} \sim N(0, 1)$.
• Proportions (\hat{p}):
  • Sample proportion $\hat{p} = \frac{X}{n}$.
  • Expected Value: $E(\hat{p}) = p$.
  • Variance: $V(\hat{p}) = \frac{p(1 - p)}{n}$.
  • Standardization: $Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1 - p)}{n}}} \sim N(0, 1)$.

Confidence and Prediction Intervals

• Confidence Interval (CI): Used to estimate the unknown population parameter $p$.
  • Formula: $p = \hat{p} \pm z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
  • Example (NOB Market Share): $n = 500$, $\hat{p} = 0.332$, Confidence Level = $90\%$.
  • Calculation: $\alpha = 10\% \implies \alpha/2 = 0.05 \implies z_{0.05} = 1.645$.
  • $\text{Margin of Error} = 1.645 \times \sqrt{\frac{0.332 \times 0.668}{500}} = 0.035$.
  • Interval: $0.332 \pm 0.035 \implies [0.297 < p < 0.367]$.
  • Width of the interval: $2 \times \text{margin} = 2 \times 0.035 = 0.070$ (or $7.0\%$).
• Prediction Interval: Used to determine the range in which a future sample proportion $\hat{p}$ is expected to fall based on a known population parameter $p$.
  • Formula: $\hat{p} = p \pm z_{\alpha/2} \times \sqrt{\frac{p(1 - p)}{n}}$
  • Example (Lock-down Support): Known $p = 0.62$, $n = 100$, $90\%$\ probability.
  • $\hat{p} = 0.62 \pm 1.645 \times \sqrt{\frac{0.62 \times 0.38}{100}} = 0.62 \pm 0.08$.
  • Range: $[0.54 < \hat{p} < 0.70]$.

Sample Size Determination

• Objective: Find the minimum sample size $n$ to achieve a specific margin of error ($B$) or bound.
• Derivation:
  • $z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \leq B$
  • $n \geq \frac{z_{\alpha/2}^2 \times \hat{p}(1 - \hat{p})}{B^2}$
• Case 1: Available Estimate: If a previous sample proportion $\hat{p}$ exists (e.g., $0.332$), use it in the formula.
  • Example: $n \geq \frac{1.645^2 \times 0.332 \times 0.668}{0.01^2} = 6001.3 \implies n \geq 6002$.
• Case 2: No Information (Conservative Approach): If $\hat{p}$ is unknown, use $\hat{p} = 0.50$ to maximize the required $n$ because the product $\hat{p}(1 - \hat{p})$ is largest at $0.5$.
  • $\hat{p} = 0.50 \implies \hat{p}(1 - \hat{p}) = 0.25$
  • Example: $n \geq \frac{1.645^2 \times 0.50 \times 0.50}{0.01^2} = 6765.06 \implies n \geq 6766$.

Exact Hypothesis Testing using the Binomial Distribution

• Conditions: Used when the normal approximation conditions fail ($np_0 < 5$ or $n(1 - p_0) < 5$).
• Procedure:
  1. Define $X \sim Binomial(n, p_0)$.
  2. Calculate the exact p-value using binomial probabilities.
  3. Compare p-value to $\alpha$.
• Example: Question 8 (Tourist Agency Loyalty):
  • $n = 15$, $p_0 = 0.20$, $\alpha = 5\%$.
  • $np_0 = 15 \times 0.20 = 3 < 5$ (Condition fails).
  • $H_0: p = 0.20$, $H_1: p > 0.20$. Observed $x_{obs} = 7$.
  • $p\text{-value} = P(X \geq 7) = 1 - P(X \leq 6) = 1 - 0.982 = 0.018$.
  • Decision: Reject $H_0$ ($0.018 \leq 0.05$).
• Example: Question 9 (Statistics Course Rating):
  • $n = 20$, $p_0 = 0.90$, $\alpha = 5\%$.
  • $n(1 - p_0) = 20 \times 0.10 = 2 < 5$ (Condition fails).
  • $H_0: p = 0.90$, $H_1: p < 0.90$. Observed $x_{obs} = 17$.
  • $p\text{-value} = P(X \leq 17) = 0.323$.
  • Decision: Do not reject $H_0$ ($0.323 > 0.05$).
• Example: Question 10a (Train Delays - Two-sided):
  • $n = 25$, $p_0 = 0.80$, $\alpha = 5\%$. Observed $x_{obs} = 16$.
  • $H_0: p = 0.80$, $H_1: p \neq 0.80$.
  • $p\text{-value} = 2 \times P(X \leq 16) = 2 \times 0.047 = 0.094$.
  • Decision: Do not reject $H_0$ ($0.094 > 0.05$).

Normal Approximation and Continuity Correction

• Continuity Correction (CC): Adjusting a discrete value by $0.5$ when using a continuous distribution to approximate a discrete one.
• Application Rules:
  • If using Counts ($X$) and conditions $np \geq 5$ and $n(1 - p) \geq 5$ are met: Use Normal approximation with CC.
    • Left tail: $P(X \leq x) \approx P(X_{cc} \leq x + 0.5)$.
    • Right tail: $P(X \geq x) \approx P(X_{cc} \geq x - 0.5)$.
  • If using Proportions ($\hat{p}$) and conditions are met: Use the normal test statistic $Z$ without CC.
  • If conditions are not met: Use the exact binomial test.
• Case Study: Question 10b (Train Delay CC Comparison):
  • With CC: Approximation of $2 \times P(X \leq 16)$ becomes $2 \times P(X_{cc} \leq 16.5)$.
    • $Z = \frac{16.5 - (25 \times 0.80)}{\sqrt{25 \times 0.80 \times 0.20}} = \frac{16.5 - 20}{2} = -1.75$
    • $p\text{-value} = 2 \times P(Z \leq -1.75) = 0.0802$ (Do not reject $H_0$; aligns with exact test).
  • Without CC: Approximation of $2 \times P(X \leq 16)$ using raw values.
    • $Z = \frac{16 - 20}{2} = -2.00$
    • $p\text{-value} = 2 \times P(Z \leq -2.00) = 0.0456$ (Incorrectly reject $H_0$).