Two-Population Statistical Inference: Mean, Proportion, and Variance Comparison

Foundations of Two-Population Comparative Inference

• Core Objective of Comparative Inference:

  • Statistical inference frequently extends beyond evaluating a single population parameter to comparing parameters across two distinct populations.

  • Comparative studies aim to determine whether an observed disparity between two sample metrics reflects a genuine underlying difference in the population parameters or simply sampling fluctuation.

• Primary Parameters Subject to Comparison:

  • Population Means ($\mu_1$ vs. $\mu_2$): Evaluated to assess differences in central tendency, continuous measurements, or performance levels across two treatments, environments, or groups.

  • Population Variances ($\sigma_1^2$ vs. $\sigma_2^2$): Analyzed to compare dispersion, consistency, or variability between two processes or populations.

  • Population Proportions ($P_1$ vs. $P_2$ or $\pi_1$ vs. $\pi_2$): Evaluated for binary or categorical attributes to compare prevalence, preference, success rates, or defect rates.

• Dual Methodological Pillars of Comparative Inference:

  • Confidence Intervals: Provide an estimated plausible range of numerical values for the true difference between two population parameters (e.g., $\mu_1 - \mu_2$ or $P_1 - P_2$) at a prescribed confidence level ($1 - \alpha$).

  • Hypothesis Testing: Evaluates probabilistic evidence against a specific null claim regarding the parameters (typically that there is no difference, $H_0: \mu_1 = \mu_2$), facilitating formal decision-making under uncertainty.

• Structural Transition from One-Sample to Two-Sample Design:

  • In a single population setting, a sample of size $n$ is extracted from a population with mean $\mu$ and variance $\sigma^2$, yielding the sample mean distribution:     $\bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right)$

  • In a two-population comparative framework, independent random samples are extracted from two separate populations:

  • Population I has parameter mean $\mu_1$, variance $\sigma_1^2$, sample size $n_1$, and sample mean $\bar{X}_1$.

  • Population II has parameter mean $\mu_2$, variance $\sigma_2^2$, sample size $n_2$, and sample mean $\bar{X}_2$.

  • The analytical focus shifts from individual point estimators to the distribution of their difference ($\bar{X}_1 - \bar{X}_2$).

Confidence Intervals for Comparing Two Population Means

• Theoretical Sampling Distribution of the Difference Between Means:

  • By the reproductive property of independent normal variables (and central limit theorem principles for sufficiently large samples):     $\bar{X}_1 \sim N\left(\mu_1, \frac{\sigma_1^2}{n_1}\right)$     $\bar{X}_2 \sim N\left(\mu_2, \frac{\sigma_2^2}{n_2}\right)$

  • The difference between the two independent sample estimators has the expectation:     $E(\bar{X}_1 - \bar{X}_2) = \mu_1 - \mu_2$

  • The variance of the difference is the sum of the individual sampling variances:     $\text{Var}(\bar{X}_1 - \bar{X}_2) = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}$

  • Therefore, the sampling distribution of the difference is:     $\bar{X}_1 - \bar{X}_2 \sim N\left(\mu_1 - \mu_2, \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}\right)$

• Case 1: Common Known Population Variance ($\sigma_1^2 = \sigma_2^2 = \sigma^2$ Known):

  • Distributional Model: When both populations share a common, known variance $\sigma^2$, the standard error simplifies to:     $\sigma_{\bar{X}_1 - \bar{X}_2} = \sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$

  • Standardized Pivot:     $Z = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \sim N(0, 1)$

  • Two-Sided $95\%$ Confidence Interval Derivation:     $P\left(-1.96 \le Z \le 1.96\right) = 0.95$

  • Interval Bounds Formulation:     $\left[ (\bar{X}_1 - \bar{X}_2) - 1.96 \times \sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} ,\; (\bar{X}_1 - \bar{X}_2) + 1.96 \times \sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \right]$

  • Comprehensive Worked Example (Skill Development Programs):

  • Scenario: Two groups of employees in an organization are sent to two skill development programs. The average performance levels reached at completion are $32.5$ and $25.7$. The standard deviation of performance across similar programs is known to be $3.6$. Group sizes are $n_1 = 27$ and $n_2 = 32$.

  • Identified Parameters: $\bar{X}_1 = 32.5$, $\bar{X}_2 = 25.7$, $\sigma = 3.6$, $n_1 = 27$, $n_2 = 32$, confidence level = $95\%$ ($Z_{0.025} = 1.96$).

  • Point Estimate of Mean Difference:       $\bar{X}_1 - \bar{X}_2 = 32.5 - 25.7 = 6.8$

  • Margin of Error Calculation:       $E = 1.96 \times 3.6 \times \sqrt{\frac{1}{27} + \frac{1}{32}} = 1.96 \times 3.6 \times \sqrt{0.037037 + 0.031250} = 1.96 \times 3.6 \times \sqrt{0.068287} = 1.96 \times 3.6 \times 0.261318 = 1.844$

  • Lower Limit:       $32.5 - 25.7 - 1.84 = 4.96$

  • Upper Limit:       $32.5 - 25.7 + 1.84 = 8.64$

  • Result: The $95\%$ confidence interval for the true difference in mean performance levels ($\mu_1 - \mu_2$) is $[4.96, 8.64]$.

• Case 2: Common Unknown Population Variance ($\sigma_1^2 = \sigma_2^2 = \sigma^2$ Unknown):

  • Methodological Principle: When population variances are assumed equal but their true value $\sigma^2$ is unknown, sample variances $S_1^2$ and $S_2^2$ are combined into an unbiased estimator called the pooled sample variance ($S_p^2$).

  • Pooled Variance Formulation:     $S_p^2 = \frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}$

  • Pooled Standard Deviation:     $S_p = \sqrt{S_p^2}$

  • Degrees of Freedom: The pooled variance pools squared deviations from two separate samples, resulting in degrees of freedom given by:     $\text{df} = n_1 + n_2 - 2$

  • Student's $t$-Distribution Pivot:     $t = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \sim t_{n_1 + n_2 - 2}$

  • General $(1 - \alpha)\%$ Confidence Interval Formulation:     $\left[ (\bar{X}_1 - \bar{X}_2) - t_{n_1 + n_2 - 2,\, \alpha/2} \times S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} ,\; (\bar{X}_1 - \bar{X}_2) + t_{n_1 + n_2 - 2,\, \alpha/2} \times S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \right]$

  • Comprehensive Worked Example (Television Picture Tube Molding Process):

  • Scenario: An executive developed a low-cost molding process for color television picture tube glass stems. To evaluate whether tube lifetime (in hours) decreases under the developed process compared to the existing process, two samples were assessed under the assumption of equal variance:

    • Process 1 (Existing): $n_1 = 22$, $\bar{X}_1 = 3200$, $S_1^2 = 120$

    • Process 2 (Developed): $n_2 = 20$, $\bar{X}_2 = 2950$, $S_2^2 = 100$

  • Degrees of Freedom & Critical Value: For a $90\%$ confidence level ($\alpha = 0.10$), the degrees of freedom are:       $\text{df} = 22 + 20 - 2 = 40$       $t_{40,\, 0.05} = 1.684$

  • Pooled Variance Calculation:       $S_p^2 = \frac{(22 - 1) \times 120 + (20 - 1) \times 100}{22 + 20 - 2} = \frac{21 \times 120 + 19 \times 100}{40} = \frac{2520 + 1900}{40} = \frac{4420}{40} = 110.5$       $S_p = \sqrt{110.5} \approx 10.51$

  • Standard Error of Difference:       $S_{\bar{X}_1 - \bar{X}_2} = S_p \sqrt{\frac{1}{22} + \frac{1}{20}} = 10.51 \times \sqrt{0.045455 + 0.05} = 10.51 \times \sqrt{0.095455} = 10.51 \times 0.308957 \approx 3.247$

  • Margin of Error:       $E = 1.684 \times 3.247 \approx 5.47$

  • Lower Limit:       $3200 - 2950 - 1.684 \times 10.51 \times \sqrt{\frac{1}{22} + \frac{1}{20}} = 250 - 5.47 = 244.53$

  • Upper Limit:       $3200 - 2950 + 1.684 \times 10.51 \times \sqrt{\frac{1}{22} + \frac{1}{20}} = 250 + 5.47 = 255.47$

  • Result: The $90\%$ confidence interval for the decrease in mean tube lifetime ($\mu_1 - \mu_2$) is $(244.53, 255.47)$.

• Case 3: Different Known Population Variances ($\sigma_1^2 \ne \sigma_2^2$, Both Known):

  • Methodological Principle: When population variances are unequal but known independently, no pooling is conducted. Standard errors use the exact known variances.

  • Standard Error:     $\sigma_{\bar{X}_1 - \bar{X}_2} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$

  • General $(1 - \alpha)\%$ Confidence Interval Formulation:     $\left[ (\bar{X}_1 - \bar{X}_2) - Z_{\alpha/2} \times \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} ,\; (\bar{X}_1 - \bar{X}_2) + Z_{\alpha/2} \times \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \right]$

  • Comprehensive Worked Example (Task Efficiency):

  • Scenario: In a large organization, $15$ employees completed a skills development program, while $10$ did not. Efficiency scores are recorded below with known population variances $\sigma_1^2 = 3.6$ and $\sigma_2^2 = 2.8$.

    • Group 1 (Trained, $n_1 = 15$): $8, 5, 7, 9, 5, 6, 7, 6, 8, 9, 6, 8, 5, 7, 9$

    • Group 2 (Untrained, $n_2 = 10$): $7, 6, 5, 8, 5, 4, 6, 8, 3, 5$

  • Sample Means Calculation:       $\bar{X}_1 = \frac{8 + 5 + 7 + 9 + 5 + 6 + 7 + 6 + 8 + 9 + 6 + 8 + 5 + 7 + 9}{15} = \frac{105}{15} = 7.0$       $\bar{X}_2 = \frac{7 + 6 + 5 + 8 + 5 + 4 + 6 + 8 + 3 + 5}{10} = \frac{57}{10} = 5.7$

  • Standard Error Calculation:       $\sigma_{\bar{X}_1 - \bar{X}_2} = \sqrt{\frac{3.6}{15} + \frac{2.8}{10}} = \sqrt{0.24 + 0.28} = \sqrt{0.52} \approx 0.7211$

  • Confidence Level Critical Value: For $90\%$ confidence ($\alpha = 0.10$), $Z_{0.05} = 1.64$.

  • Lower Limit:       $7.0 - 5.7 - 1.64 \times \sqrt{\frac{3.6}{15} + \frac{2.8}{10}} = 1.3 - 1.64 \times 0.7211 = 1.3 - 1.1826 = 0.12$

  • Upper Limit:       $7.0 - 5.7 + 1.64 \times \sqrt{\frac{3.6}{15} + \frac{2.8}{10}} = 1.3 + 1.1826 = 2.48$

  • Result: The $90\%$ confidence interval for the average efficiency improvement ($\mu_1 - \mu_2$) is $[0.12, 2.48]$.

• Case 4: Different Unknown Population Variances ($\sigma_1^2 \ne \sigma_2^2$, Both Unknown):

  • Methodological Principle: When variances cannot be assumed equal and remain unknown, each population variance is estimated by its respective sample variance ($S_1^2$ and $S_2^2$). Pooling is strictly invalid.

  • Standard Error:     $S_{\bar{X}_1 - \bar{X}_2} = \sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}$

  • Degrees of Freedom Formulation: Utilizes standard two-sample degrees of freedom approximation:     $\text{df} = n_1 + n_2 - 2$

  • Test Statistic / Pivot:     $t = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}} \sim t_{n_1 + n_2 - 2}$

  • General $(1 - \alpha)\%$ Confidence Interval Formulation:     $\left[ (\bar{X}_1 - \bar{X}_2) - t_{n_1 + n_2 - 2,\, \alpha/2} \times \sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}} ,\; (\bar{X}_1 - \bar{X}_2) + t_{n_1 + n_2 - 2,\, \alpha/2} \times \sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}} \right]$

  • Comprehensive Worked Example (Puppy Food Variety Weights):

  • Scenario: To compare the quality of two food varieties provided for puppies, two samples were fed the respective diets over one month. End-of-month puppy weights (in grams) yielded:

    • Food I: $n_1 = 18$, $\bar{X}_1 = 1250\,\text{g}$, $S_1^2 = 240\,\text{g}$

    • Food II: $n_2 = 15$, $\bar{X}_2 = 950\,\text{g}$, $S_2^2 = 210\,\text{g}$

  • Degrees of Freedom & Critical Value: For a $95\%$ confidence interval:       $\text{df} = 18 + 15 - 2 = 31$       $t_{31,\, 0.025} = 2.08$

  • Standard Error Calculation:       $\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}} = \sqrt{\frac{240}{18} + \frac{210}{15}} = \sqrt{13.333 + 14.0} = \sqrt{27.333} \approx 5.228$

  • Margin of Error:       $E = 2.08 \times 5.228 \approx 10.87$

  • Lower Limit:       $1250 - 950 - 2.08 \times \sqrt{\frac{240}{18} + \frac{210}{15}} = 300 - 10.87 = 289.1$

  • Upper Limit:       $1250 - 950 + 2.08 \times \sqrt{\frac{240}{18} + \frac{210}{15}} = 300 + 10.87 = 310.9$

  • Result: The $95\%$ confidence interval for the weight difference between the two diet groups is $[289.1, 310.9]$.

Confidence Intervals for Comparing Two Population Proportions

• Distributional Foundations of Proportions:

  • Let $P_1$ and $P_2$ denote true population proportions. Independent random samples of size $n_1$ and $n_2$ produce sample proportions $\pi_1$ (or $p_1$) and $\pi_2$ (or $p_2$).

  • By the Normal approximation to the binomial distribution (valid when $n_i P_i \ge 5$ and $n_i (1 - P_i) \ge 5$):     $\pi_1 \sim N\left(P_1, \frac{P_1(1 - P_1)}{n_1}\right)$     $\pi_2 \sim N\left(P_2, \frac{P_2(1 - P_2)}{n_2}\right)$

  • The distribution of the difference in sample proportions satisfies:     $\pi_1 - \pi_2 \sim N\left(P_1 - P_2, \frac{P_1(1 - P_1)}{n_1} + \frac{P_2(1 - P_2)}{n_2}\right)$

  • The standardized $Z$-pivot is:     $Z = \frac{(\pi_1 - \pi_2) - (P_1 - P_2)}{\sqrt{\frac{P_1(1 - P_1)}{n_1} + \frac{P_2(1 - P_2)}{n_2}}} \sim N(0, 1)$

• Confidence Interval Formulation for $P_1 - P_2$:

  • Because population parameters $P_1$ and $P_2$ are unknown, they are substituted by sample proportions $\pi_1$ and $\pi_2$ inside the standard error radical.

  • General $(1 - \alpha)\%$ Confidence Interval:     $\left[ (\pi_1 - \pi_2) - Z_{\alpha/2} \times \sqrt{\frac{\pi_1(1 - \pi_1)}{n_1} + \frac{\pi_2(1 - \pi_2)}{n_2}} ,\; (\pi_1 - \pi_2) + Z_{\alpha/2} \times \sqrt{\frac{\pi_1(1 - \pi_1)}{n_1} + \frac{\pi_2(1 - \pi_2)}{n_2}} \right]$

• Comprehensive Worked Example 1 (Displeasure with Government Policies):

  • Scenario: A survey in Verginia measured displeasure toward government policies by selecting independent random samples from two provinces:

  • Southern province: $n_1 = 450$, opposition proportion $\pi_1 = 0.68$

  • Central province: $n_2 = 650$, opposition proportion $\pi_2 = 0.55$

  • Confidence Level: $95\%$ confidence implies $Z_{0.025} = 1.96$.

  • Point Estimate of Difference:     $\pi_1 - \pi_2 = 0.68 - 0.55 = 0.13$

  • Standard Error Calculation:     $\text{SE} = \sqrt{\frac{0.68(1 - 0.68)}{450} + \frac{0.55(1 - 0.55)}{650}} = \sqrt{\frac{0.68 \times 0.32}{450} + \frac{0.55 \times 0.45}{650}} = \sqrt{\frac{0.2176}{450} + \frac{0.2475}{650}}$     $\text{SE} = \sqrt{0.00048356 + 0.00038077} = \sqrt{0.00086433} \approx 0.0294$

  • Margin of Error:     $E = 1.96 \times 0.0294 \approx 0.0576$

  • Lower Limit:     $0.13 - 1.96 \times 0.0294 = 0.13 - 0.058 = 0.072$

  • Upper Limit:     $0.13 + 1.96 \times 0.0294 = 0.13 + 0.058 = 0.188$

  • Result: The $95\%$ confidence interval for the difference between the two population proportions is $[0.072, 0.188]$.

• Comprehensive Worked Example 2 (Late Attendance Decline via Fingerprint Machine):

  • Scenario: Ashwee Private Limited introduced fingerprint machines to reduce late attendance. Employees unwilling to accept the system were permitted to leave with compensation.

  • Initial state (Sample 1): $n_1 = 256$, late attendance proportion $\pi_1 = 0.46$

  • Post-implementation state (Sample 2): $39$ employees left, leaving $n_2 = 256 - 39 = 217$ employees, among whom late attendance dropped to $\pi_2 = 0.27$

  • Confidence Level: $90\%$ confidence implies $Z_{0.05} = 1.64$.

  • Point Estimate of Decline:     $\pi_1 - \pi_2 = 0.46 - 0.27 = 0.19$

  • Standard Error Calculation:     $\text{SE} = \sqrt{\frac{0.46(1 - 0.46)}{256} + \frac{0.27(1 - 0.27)}{217}} = \sqrt{\frac{0.46 \times 0.54}{256} + \frac{0.27 \times 0.73}{217}} = \sqrt{\frac{0.2484}{256} + \frac{0.1971}{217}}$     $\text{SE} = \sqrt{0.0009703 + 0.0009083} = \sqrt{0.0018786} \approx 0.04334$

  • Margin of Error:     $E = 1.64 \times 0.04334 \approx 0.071$

  • Lower Limit:     $0.19 - 1.64 \times 0.04334 = 0.19 - 0.071 = 0.119$

  • Upper Limit:     $0.19 + 1.64 \times 0.04334 = 0.19 + 0.071 = 0.261$

  • Result: The $90\%$ confidence interval for the percentage decline in late attendance is $[0.119, 0.261]$ (or $11.9\%$ to $26.1\%$).

Hypothesis Testing Framework for Comparing Two Population Means

• General Hypothesis Structures:

  • Two-Tailed Test (Testing for Any Significant Disparity):     $H_0: \mu_1 = \mu_2 \quad \iff \quad H_0: \mu_1 - \mu_2 = 0$     $H_1: \mu_1 \ne \mu_2 \quad \iff \quad H_1: \mu_1 - \mu_2 \ne 0$

  • One-Tailed Test (Right-Tailed / Greater Than):     $H_0: \mu_1 \le \mu_2 \quad \iff \quad H_0: \mu_1 - \mu_2 \le 0$     $H_1: \mu_1 > \mu_2 \quad \iff \quad H_1: \mu_1 - \mu_2 > 0$

  • One-Tailed Test (Left-Tailed / Less Than):     $H_0: \mu_1 \ge \mu_2 \quad \iff \quad H_0: \mu_1 - \mu_2 \ge 0$     $H_1: \mu_1 < \mu_2 \quad \iff \quad H_1: \mu_1 - \mu_2 < 0$

• Case 1: Common Known Variance ($\sigma^2$ Known):

  • Null Distribution & Test Statistic: Under $H_0: \mu_1 = \mu_2$, the test statistic simplifies to:     $Z = \frac{\bar{X}_1 - \bar{X}_2}{\sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \sim N(0, 1)$

  • Comprehensive Worked Example (Durability of Road Crossing Paints):

  • Scenario: To evaluate the durability of two road-marking paints on passenger crossings, twelve $100\,\text{cm}^2$ plots were painted with each paint on a heavily traveled road. After one week of use, average durability readings were $10.4$ and $9.6$. Historical experience establishes a common variance of $\sigma^2 = 2.4$. Test for equality of durability at $\alpha = 0.10$.

  • Hypotheses:       $H_0: \mu_1 = \mu_2$       $H_1: \mu_1 \ne \mu_2$

  • Given Values: $\bar{X}_1 = 10.4$, $\bar{X}_2 = 9.6$, $\sigma^2 = 2.4$, $n_1 = 12$, $n_2 = 12$, $\alpha = 0.10$.

  • Test Statistic Calculation:       $Z = \frac{10.4 - 9.6}{\sqrt{2.4 \left(\frac{1}{12} + \frac{1}{12}\right)}} = \frac{0.8}{\sqrt{2.4 \times \frac{2}{12}}} = \frac{0.8}{\sqrt{2.4 \times 0.16667}} = \frac{0.8}{\sqrt{0.4}} = \frac{0.8}{0.632456} \approx 1.26$

  • Decision Rule & Critical Value: For a two-tailed test at $\alpha = 0.10$, the benchmark critical value for fail-to-reject bounds is verified against the rejection region (tabulated standard threshold $Z_{0.05} = 1.645$, with an upper bound test threshold of $1.96$). Since calculated $Z = 1.26$ falls well below the critical threshold ($1.26 < 1.645 < 1.96$), it does not fall into the critical region.

  • Conclusion: Do not reject $H_0$. There is insufficient statistical evidence to establish that the durability of the two road-marking paints differs significantly.

• Case 2: Common Unknown Variance ($\sigma^2$ Unknown, Equal Variance Assumed):

  • Test Statistic Formulation:     $t = \frac{\bar{X}_1 - \bar{X}_2}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \sim t_{n_1 + n_2 - 2}$     where:     $S_p = \sqrt{\frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}}$

  • Comprehensive Worked Example (Mobile Phone Recharge Card Sales):

  • Scenario: To evaluate demand for Rs. 100 recharge cards across North and Eastern provinces during November 2011, sales per shop were recorded:

    • Province I ($n_1 = 10$): $12, 15, 8, 21, 18, 10, 5, 17, 12, 14$

    • Province II ($n_2 = 8$): $18, 12, 10, 16, 20, 12, 10, 6$

  • Hypotheses:       $H_0: \mu_1 = \mu_2$       $H_1: \mu_1 \ne \mu_2$

  • Sample Statistics:       $\bar{X}_1 = 13.2, \quad S_1^2 = 4.83, \quad n_1 = 10$       $\bar{X}_2 = 13.0, \quad S_2^2 = 4.66, \quad n_2 = 8$

  • Degrees of Freedom:       $\text{df} = 10 + 8 - 2 = 16$

  • Pooled Variance Computation:       $S_p^2 = \frac{(10 - 1)(4.83) + (8 - 1)(4.66)}{16} = \frac{9(4.83) + 7(4.66)}{16} = \frac{43.47 + 32.62}{16} = \frac{76.09}{16} \approx 4.756$       $S_p = \sqrt{4.756} \approx 2.1808$

  • Standard Error Calculation:       $\text{SE} = S_p \sqrt{\frac{1}{10} + \frac{1}{8}} = 2.1808 \times \sqrt{0.10 + 0.125} = 2.1808 \times \sqrt{0.225} = 2.1808 \times 0.47434 = 1.0345$

  • Calculated Test Statistic:       $t = \frac{13.2 - 13.0}{1.0345} = \frac{0.2}{1.0345} \approx 0.062$

  • Decision Rule: Two-tailed critical value at $\alpha = 0.05$ with $16$ degrees of freedom is $t_{16,\, 0.025} = \pm 2.12$.

  • Decision: The calculated test statistic $t = 0.062$ falls within the non-rejection interval $[-2.12, 2.12]$.

  • Conclusion: Fail to reject $H_0$ at the $5\%$ significance level ($95\%$ confidence). The sample data provide no evidence that average mobile card sales per shop differ between the two provinces.

• Case 3: Different Known Variances ($\sigma_1^2 \ne \sigma_2^2$, Both Known):

  • Test Statistic Formulation:     $Z = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \sim N(0, 1)$

  • Under $H_0: \mu_1 - \mu_2 = 0$, this becomes:       $Z = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}$

  • Comprehensive Worked Example (Daily Production at Manufacturing Plants):

  • Scenario: A manufacturing firm operates two plants producing identical products. Management investigates whether Plant I has significantly higher daily production than Plant II. Operating data from the past month are summarized:

    • Plant I: $n_1 = 20$ days operated, $\bar{X}_1 = 62$ units, known $\sigma_1^2 = 24$

    • Plant II: $n_2 = 22$ days operated, $\bar{X}_2 = 60$ units, known $\sigma_2^2 = 36$

  • Hypotheses (One-Tailed Test):       $H_0: \mu_1 \le \mu_2$       $H_1: \mu_1 > \mu_2$

  • Test Statistic Calculation:       $Z = \frac{62 - 60}{\sqrt{\frac{24}{20} + \frac{36}{22}}} = \frac{2}{\sqrt{1.20 + 1.6364}} = \frac{2}{\sqrt{2.8364}} = \frac{2}{1.68415} \approx 1.19$

  • Decision Rule: For a right-tailed test at $\alpha = 0.10$, the critical standard normal value is $Z_{0.10} = 1.28$.

  • Decision: The calculated $Z = 1.19$ is less than the critical value $1.28$ and does not fall in the critical region ($Z < 1.28$).

  • Conclusion: Fail to reject $H_0$ at the $10\%$ level. Sample data provide no statistically significant evidence that Plant I produces more items daily than Plant II.

• Case 4: Different Unknown Variances ($\sigma_1^2 \ne \sigma_2^2$, Both Unknown):

  • Test Statistic Formulation:     $t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}} \sim t_{n_1 + n_2 - 2}$

  • Large Sample Equivalence: When total degrees of freedom exceed $30$ ($n_1 + n_2 - 2 > 30$), the Student's $t$-distribution converges toward the standard normal distribution $N(0, 1)$, allowing the use of normal critical thresholds.

  • Comprehensive Worked Example (Sri Lankan Share Market Equity Prices):

  • Scenario: An investor compares the closing share prices of two companies over the previous month to test whether Company II has a higher mean close price at $\alpha = 0.05$:

    • Company I: $n_1 = 20$ trading days, $\bar{X}_1 = 37.50$, $S_1^2 = 12.5$

    • Company II: $n_2 = 22$ trading days, $\bar{X}_2 = 42.50$, $S_2^2 = 15.0$

  • Hypotheses (Left-Tailed Test for $\mu_1 - \mu_2$):       $H_0: \mu_1 \ge \mu_2 \quad (\text{or } \mu_1 - \mu_2 \ge 0)$       $H_1: \mu_1 < \mu_2 \quad (\text{Company II closing price is higher})$

  • Standard Error Calculation:       $\text{SE} = \sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}} = \sqrt{\frac{12.5}{20} + \frac{15.0}{22}} = \sqrt{0.625 + 0.68182} = \sqrt{1.30682} \approx 1.14316$

  • Test Statistic Calculation:       $t = \frac{37.50 - 42.50}{1.14316} = \frac{-5.0}{1.14316} \approx -4.37$

  • Degrees of Freedom & Critical Value:       $\text{df} = 20 + 22 - 2 = 40$

    • Because $\text{df} = 40 > 30$, the standard normal distribution provides the critical threshold. For a one-tailed test at $\alpha = 0.05$, the left-tail critical value is $-Z_{0.05} = -1.64$.

  • Decision: The calculated test statistic $t = -4.37$ is substantially less than the critical threshold $-1.64$ ($-4.37 < -1.64$), lying deep within the rejection region.

  • Conclusion: Reject $H_0$ at the $5\%$ significance level. There is strong statistical evidence ($95\%$ confidence) that the closing share price of Company II is significantly higher than that of Company I.

Systematic Summary of Estimators, Standard Errors, and Distributions

• Comparison Summary Across Variance Conditions for Two Population Means:

  • Common Known Variance ($\sigma_1 = \sigma_2 = \sigma$):

  • Standard Error Formulation:       $\sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$

  • Test Statistic / Pivot:       $Z = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

  • Sampling Distribution:       $N(0, 1)$

  • Confidence Interval:       $(\bar{X}_1 - \bar{X}_2) \pm Z_{\alpha/2} \times \sigma \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$

  • Common Unknown Variance ($\sigma_1 = \sigma_2 = \sigma$ Unknown):

  • Standard Error Formulation:       $S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$       where:       $S_p^2 = \frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}$

  • Test Statistic / Pivot:       $t = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

  • Sampling Distribution:       $t_{n_1 + n_2 - 2}$

  • Confidence Interval:       $(\bar{X}_1 - \bar{X}_2) \pm t_{n_1 + n_2 - 2,\, \alpha/2} \times S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$

  • Different Known Variances ($\sigma_1^2 \ne \sigma_2^2$ Known):

  • Standard Error Formulation:       $\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$

  • Test Statistic / Pivot:       $Z = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}$

  • Sampling Distribution:       $N(0, 1)$

  • Confidence Interval:       $(\bar{X}_1 - \bar{X}_2) \pm Z_{\alpha/2} \times \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$

  • Different Unknown Variances ($\sigma_1^2 \ne \sigma_2^2$ Unknown):

  • Standard Error Formulation:       $\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}$

  • Test Statistic / Pivot:       $t = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}}$

  • Sampling Distribution:       $t_{n_1 + n_2 - 2}$ (converges to $N(0, 1)$ when $n_1 + n_2 - 2 > 30$)

  • Confidence Interval:       $(\bar{X}_1 - \bar{X}_2) \pm t_{n_1 + n_2 - 2,\, \alpha/2} \times \sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}$

• Inference Summary for Difference Between Two Population Proportions ($P_1 - P_2$):

  • Standard Error Formulation:     $\sqrt{\frac{\pi_1(1 - \pi_1)}{n_1} + \frac{\pi_2(1 - \pi_2)}{n_2}}$

  • Test Statistic / Pivot:     $Z = \frac{(\pi_1 - \pi_2) - (P_1 - P_2)}{\sqrt{\frac{\pi_1(1 - \pi_1)}{n_1} + \frac{\pi_2(1 - \pi_2)}{n_2}}}$

  • Sampling Distribution:     $N(0, 1)$

  • Confidence Interval:     $(\pi_1 - \pi_2) \pm Z_{\alpha/2} \times \sqrt{\frac{\pi_1(1 - \pi_1)}{n_1} + \frac{\pi_2(1 - \pi_2)}{n_2}}$