14.1 Applied Statistical Distributions & Sample Size Determination

14.1 Applied Statistical Distributions & Sample Size Determination

PTOE Exam Focus: Applied statistics is an essential foundation for Domain 5. Candidates must master the selection and application of probability distributions (Normal vs. Poisson vs. Negative Exponential), conduct Chi-Square ($\chi^2$) goodness-of-fit tests for vehicular arrivals, perform Student's $t$-tests for before-and-after operational studies, and calculate required sample sizes ($N = \left(\frac{z \cdot s}{E}\right)^2$) to satisfy specified confidence levels and error tolerances.

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1. Statistical Distributions in Traffic Engineering

Traffic stream phenomena fall into two fundamental mathematical categories: continuous variables (such as vehicle speed, travel time, and delay) and discrete events (such as vehicle arrivals, pedestrian counts, and crash occurrences). Selecting the appropriate theoretical distribution is critical for data analysis, capacity modeling, and simulation.

+-----------------------------------------------------------------------------------+
|                 TRAFFIC ENGINEERING PROBABILITY DISTRIBUTIONS                     |
|                                                                                   |
|  1. CONTINUOUS DISTRIBUTIONS:                                                     |
|     • Normal (Gaussian) Distribution: Free-flow spot speeds, travel times.         |
|     • Negative Exponential Distribution: Time headways between random arrivals.   |
|     • Shifted Exponential / Lognormal: Headways with minimum tracking headway.    |
|                                                                                   |
|  2. DISCRETE DISTRIBUTIONS:                                                       |
|     • Poisson Distribution: Vehicle arrivals in time t at low/moderate flow.      |
|     • Binomial Distribution: Bernoulli trials, turning movement splits.           |
|     • Negative Binomial Distribution: Overdispersed crash data (HSM modeling).    |
+-----------------------------------------------------------------------------------+

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2. The Normal (Gaussian) Distribution

The Normal Distribution is the symmetric, bell-shaped continuous distribution governing random variables influenced by the additive sum of numerous independent factors (by the Central Limit Theorem). In traffic engineering, free-flow spot speeds under uninterrupted conditions almost universally follow a Normal distribution.

Mathematical Formulation

The continuous probability density function (PDF) is given by:

$f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}}$

Where:

• $\mu$ = Population mean
• $\sigma$ = Population standard deviation
• $\sigma^2$ = Population variance

Standard Normal Variable ($Z$-Score)

Any normally distributed variable $X \sim \mathcal{N}(\mu, \sigma^2)$ is transformed to the standard normal distribution $Z \sim \mathcal{N}(0, 1)$ via:

$Z = \frac{x - \mu}{\sigma}$

Key Empirical Properties:

• $68.27%$ of observations lie within $\mu \pm 1.0\sigma$ (approx. $\pm 1\sigma$)
• $95.00%$ of observations lie within $\mu \pm 1.960\sigma$ (approx. $\pm 2\sigma$)
• $99.73%$ of observations lie within $\mu \pm 3.0\sigma$ (approx. $\pm 3\sigma$)
• 85th Percentile Speed ($V_{85}$): Corresponds to $Z = +1.036 \approx 1.04$: $V_{85} = \mu + 1.036 \cdot \sigma$
• 15th Percentile Speed ($V_{15}$): Corresponds to $Z = -1.036 \approx -1.04$: $V_{15} = \mu - 1.036 \cdot \sigma$

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3. The Poisson Distribution & Vehicle Arrivals

The Poisson Distribution is a discrete probability distribution modeling the number of events (vehicle arrivals, pedestrian actuations, collisions) occurring within a fixed interval of time or space. It applies when:

1. Arrivals are completely random and independent of one another.
2. The probability of an arrival in a tiny time increment $\Delta t$ is proportional to the average arrival rate $\lambda$.
3. The probability of two or more arrivals in $\Delta t$ approaches zero.

Probability Mass Function (PMF)

$P(x) = \frac{\lambda^x e^{-\lambda}}{x!} = \frac{(\mu)^x e^{-\mu}}{x!}$

Where:

• $x$ = Number of observed arrivals in time interval $t$ ($x = 0, 1, 2, 3, \dots$)
• $\lambda = q \cdot t$ = Expected average number of arrivals in time interval $t$
• $q$ = Mean arrival rate (vehicles per unit time, e.g., veh/s or veh/min)
• $e \approx 2.71828$ = Euler's constant

Unique Mathematical Property:

For a pure Poisson process, the variance equals the mean: $\sigma^2 = \mu = \lambda$

• If $\sigma^2 / \mu \approx 1.0$, data is Poisson distributed (random arrivals).
• If $\sigma^2 / \mu > 1.0$, data is overdispersed (clustered platoons, crash data modeled via Negative Binomial).
• If $\sigma^2 / \mu < 1.0$, data is underdispersed (uniform, congested flow).

Duality with Negative Exponential Headways

If vehicle arrivals in time follow a Poisson distribution with mean rate $q$ (veh/s), the time headways ($h$) between successive vehicles are continuous and follow the Negative Exponential Distribution:

$P(h \ge t) = e^{-q \cdot t}$ $P(h < t) = 1 - e^{-q \cdot t}$

Where $P(h \ge t)$ represents the probability of finding a headway greater than or equal to $t$ seconds (critical for gap acceptance analysis at stop signs and unsignalized crossings).

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4. Chi-Square ($\chi^2$) Goodness-of-Fit Testing

The Chi-Square Goodness-of-Fit Test verifies whether empirical traffic field data (such as 1-minute arrival counts or binned spot speeds) statistically conforms to a hypothesized theoretical distribution (such as Poisson or Normal).

+-----------------------------------------------------------------------------------+
|                     CHI-SQUARE GOODNESS-OF-FIT WORKFLOW                           |
|                                                                                   |
|  [ Step 1: Bin Field Observations into k Frequency Classes (O_i) ]                |
|                                  |                                                |
|                                  v                                                |
|  [ Step 2: Compute Theoretical Expected Frequencies E_i = N * P_i ]               |
|                                  |                                                |
|                                  v                                                |
|  [ Step 3: Pool Adjacent Bins if Expected Frequency E_i < 5 ]                     |
|                                  |                                                |
|                                  v                                                |
|  [ Step 4: Calculate Test Statistic: Chi^2 = Sum [ (O_i - E_i)^2 / E_i ] ]        |
|                                  |                                                |
|                                  v                                                |
|  [ Step 5: Determine Degrees of Freedom: df = k - 1 - p ]                         |
|                                  |                                                |
|                                  v                                                |
|  [ Step 6: Compare Chi^2_calc with Chi^2_critical (alpha = 0.05) ]                |
|       • If Chi^2_calc <= Chi^2_critical: Fail to Reject H0 (Distribution fits)     |
|       • If Chi^2_calc > Chi^2_critical: Reject H0 (Distribution does not fit)     |
+-----------------------------------------------------------------------------------+

Mathematical Formulation

$\chi^2_{\text{calc}} = \sum_{i=1}^{k} \frac{(O_i - E_i)^2}{E_i}$

Where:

• $O_i$ = Observed frequency in category/bin $i$
• $E_i = N \cdot P_i$ = Expected frequency in category/bin $i$ under theoretical distribution
• $k$ = Final number of categories/bins (after pooling small cells where $E_i < 5$)

Degrees of Freedom ($df$)

$df = k - 1 - p$

Where:

• $k$ = Number of bins used in the summation
• $1$ = Constraint that $\sum E_i = \sum O_i = N$
• $p$ = Number of distribution parameters estimated directly from sample data (for Poisson, $p = 1$ because $\lambda = \bar{x}$; for Normal, $p = 2$ because $\mu = \bar{x}$ and $\sigma = s$)

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5. Student's $t$-Test for Before-and-After Studies

To evaluate whether a traffic engineering countermeasure (e.g., dynamic speed feedback signs, road diets, or signal retiming) has produced a statistically significant change in operating speeds, engineers perform a Two-Sample Independent Student's $t$-Test.

Hypotheses Formulation:

• Null Hypothesis ($H_0$): $\mu_1 = \mu_2$ (no change in mean operating speed)
• Alternative Hypothesis ($H_a$): $\mu_1 \ne \mu_2$ (two-tailed) or $\mu_1 > \mu_2$ (one-tailed speed reduction)

Pooled Variance ($s_p^2$) Formulation:

Assuming equal population variances (homoscedasticity):

$s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}$ $s_p = \sqrt{s_p^2}$

Test Statistic ($t_{\text{calc}}$):

$t_{\text{calc}} = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

With degrees of freedom: $df = n_1 + n_2 - 2$.

Decision Rule (Two-Tailed at $\alpha = 0.05$):

• If $|t_{\text{calc}}| > t_{\text{crit}}$, reject $H_0$ and conclude the speed change is statistically significant at the $95%$ confidence level.
• If $|t_{\text{calc}}| \le t_{\text{crit}}$, fail to reject $H_0$ (difference could be random noise).

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6. Sample Size Determination for Traffic Studies

To ensure field data achieves a designated level of statistical precision without wasting resources on unnecessary data collection, sample size $N$ is derived from the margin of error of the mean.

Fundamental Sample Size Equation

For a normally distributed variable with known or estimated standard deviation $s$, confidence level $1 - \alpha$, and maximum allowable error (tolerance) $E$:

$E = z_{\alpha/2} \cdot \text{SE} = z_{\alpha/2} \cdot \frac{s}{\sqrt{N}}$

Solving for required sample size $N$:

$N = \left( \frac{z \cdot s}{E} \right)^2 = \frac{z^2 \cdot s^2}{E^2}$

Where:

• $N$ = Minimum required sample size (always rounded up to the nearest integer)
• $z$ = Critical value corresponding to the desired confidence level:
  • $90%$ Confidence ($\alpha = 0.10$): $z = 1.645$
  • $95%$ Confidence ($\alpha = 0.05$): $z = 1.960$
  • $99%$ Confidence ($\alpha = 0.01$): $z = 2.576$
• $s$ = Estimated standard deviation of the population (typically $4.0\text{ to }6.0\text{ mph}$ for spot speeds if unknown; default $s = 5.0\text{ mph}$)
• $E$ = Allowable margin of error (tolerance), e.g., $\pm 1.0\text{ mph}$ or $\pm 0.5\text{ mph}$

Finite Population Correction (FPC)

If the sample size $N$ represents more than $5%$ of the total finite population $N_{\text{pop}}$ ($N / N_{\text{pop}} > 0.05$, common in small vehicle fleets or parking turnover counts), the required sample size is adjusted downward:

$N_{\text{adj}} = \frac{N}{1 + \frac{N - 1}{N_{\text{pop}}}}$

Statistical Testing and Distribution Reference for Traffic Engineering

7. Worked Calculation Examples

Example 1: Spot Speed Sample Size Determination

Problem: A traffic engineer is planning a spot speed study on an urban arterial. Historical data indicates a standard deviation of $s = 5.2\text{ mph}$. The study must achieve a $95%$ confidence level ($z = 1.96$) with an allowable error of $E = \pm 1.0\text{ mph}$. What is the minimum required sample size?

Step-by-Step Solution:

1. Identify Given Parameters:

   • $s = 5.2\text{ mph}$
   • $z = 1.96$ ($95%$ confidence)
   • $E = 1.0\text{ mph}$

2. Apply Sample Size Formulation: $N = \left( \frac{z \cdot s}{E} \right)^2 = \left( \frac{1.96 \times 5.2}{1.0} \right)^2 = (10.192)^2 = 103.88$

3. Rounding Rule: Always round up to the nearest integer: $N = 104\text{ vehicles}$.

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Example 2: Student's $t$-Test for Speed Feedback Sign Evaluation

Problem: A dynamic speed feedback sign is installed in a school zone. Spot speed studies conducted before and after installation yield the following data:

• Before: $n_1 = 64$, $\bar{x}_1 = 34.2\text{ mph}$, $s_1 = 4.8\text{ mph}$
• After: $n_2 = 64$, $\bar{x}_2 = 31.6\text{ mph}$, $s_2 = 4.2\text{ mph}$

Test at $\alpha = 0.05$ ($t_{\text{crit}} = 1.980$ for $df = 126$) whether the sign caused a statistically significant reduction in mean operating speed.

Step-by-Step Solution:

1. Compute Pooled Standard Deviation ($s_p$): $s_p^2 = \frac{(64 - 1)(4.8)^2 + (64 - 1)(4.2)^2}{64 + 64 - 2} = \frac{63(23.04) + 63(17.64)}{126} = \frac{1451.52 + 1111.32}{126} = \frac{2562.84}{126} = 20.34$ $s_p = \sqrt{20.34} = 4.51\text{ mph}$

2. Compute Test Statistic ($t_{\text{calc}}$): $t_{\text{calc}} = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} = \frac{34.2 - 31.6}{4.51 \sqrt{\frac{1}{64} + \frac{1}{64}}} = \frac{2.60}{4.51 \sqrt{0.03125}} = \frac{2.60}{4.51 \times 0.17678} = \frac{2.60}{0.7973} = 3.26$

3. Statistical Decision: Since $|t_{\text{calc}}| = 3.26 > t_{\text{crit}} = 1.980$, reject $H_0$ ($p < 0.002$). The speed feedback sign produced a statistically significant reduction in mean operating speed.