3.2 Parameter Estimation, Linear Regression, Correlation, and Queuing Models

3.2 Parameter Estimation, Linear Regression, Correlation, and Queuing Models

The AMSTHC TOS area "Engineering Data Analysis" has five competencies:

1. Describe data and data sets.
2. Identify important discrete and continuous distributions.
3. Estimate parameters.
4. Interpret a linear regression equation.
5. Apply stochastic processes and queuing models.

Section 3.1 covers descriptive statistics, the main distributions and confidence intervals. This section completes the area.

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Estimating Parameters

A parameter describes a population, such as the true mean strength $\mu$ of a concrete supply. A statistic computed from a sample estimates it.

An interval estimate adds a margin of error, $\bar{x} \pm t_{\alpha/2,\,n-1}\, s/\sqrt{n}$. The sample size needed to estimate a mean within $\pm E$ at confidence $1 - \alpha$ is:

$n = \left(\frac{z_{\alpha/2}\,\sigma}{E}\right)^2$

Example. Estimate mean daily traffic within $\pm 200$ vehicles at 95% confidence when $\sigma \approx 1{,}000$. Then $n = (1.96 \times 1{,}000/200)^2 = 96.04$, so count 97 days. Always round up.

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Simple Linear Regression

To fit $\hat{y} = a + bx$ by least squares, minimize $\sum (y_i - \hat{y}_i)^2$:

$b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - \left(\sum x\right)^2}, \qquad a = \bar{y} - b\bar{x}$

Because $a = \bar{y} - b\bar{x}$, the line always passes through the point $(\bar{x}, \bar{y})$.

Correlation coefficient:

$r = \frac{n\sum xy - \sum x \sum y}{\sqrt{\left[n\sum x^2 - (\sum x)^2\right]\left[n\sum y^2 - (\sum y)^2\right]}}, \qquad -1 \le r \le 1$

Coefficient of determination: $R^2 = r^2$ for simple linear regression. It is the fraction of the variation in $y$ explained by $x$. The standard error of estimate is $s_e = \sqrt{\sum (y_i - \hat{y}_i)^2/(n - 2)}$.

Worked example: strength versus curing age

Five concrete cylinder pairs give:

Sums: $\sum x = 73$, $\sum y = 107$, $\sum xy = 1{,}814$, $\sum x^2 = 1{,}479$, $\sum y^2 = 2{,}457$, with $n = 5$, $\bar{x} = 14.6$ and $\bar{y} = 21.4$.

$b = \frac{5(1{,}814) - 73(107)}{5(1{,}479) - 73^2} = \frac{9{,}070 - 7{,}811}{7{,}395 - 5{,}329} = \frac{1{,}259}{2{,}066} = 0.609\text{ MPa/day}$

$a = \bar{y} - b\bar{x} = 21.4 - 0.609(14.6) = 12.50\text{ MPa}$

$r = \frac{1{,}259}{\sqrt{2{,}066\,[5(2{,}457) - 107^2]}} = \frac{1{,}259}{\sqrt{2{,}066 \times 836}} = \frac{1{,}259}{1{,}314.2} = 0.958, \qquad R^2 = 0.918$

The line $\hat{y} = 12.50 + 0.609x$ explains about 92% of the variation in strength. As a check, the predicted strength at $\bar{x} = 14.6$ days is $12.50 + 0.609(14.6) = 21.4\text{ MPa} = \bar{y}$.

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Stochastic Processes and Arrivals

A stochastic process is a sequence of random events in time. Two models dominate engineering applications:

• Poisson arrivals. Vehicles reaching a toll plaza, or trucks reaching a batch plant, arrive independently at an average rate $\lambda$. The number arriving in time $t$ is Poisson: $P(n) = \dfrac{(\lambda t)^n e^{-\lambda t}}{n!}$.
• Exponential headways. The time between Poisson arrivals is exponential: $P(h \ge t) = e^{-\lambda t}$. This is the probability that a gap is long enough for a pedestrian or a merging driver.

Example. At $\lambda = 360$ vehicles/h $= 0.1$ veh/s, the probability of a gap of at least $8\text{ s}$ is $e^{-0.8} = 0.449$.

A Markov chain models systems that move between states with fixed transition probabilities. For example, a pavement section's condition rating moves from good to fair to poor from year to year. Multiplying a state vector by the transition matrix gives the next year's distribution.

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Queuing Models (M/M/1)

An M/M/1 queue has Poisson arrivals at rate $\lambda$, exponential service at rate $\mu$, and one server. With utilization $\rho = \lambda/\mu < 1$:

Little's law, $L = \lambda W$ and $L_q = \lambda W_q$, holds for almost any queue in steady state.

Worked example: a toll booth

Cars arrive at $\lambda = 240$/h, and one booth serves $\mu = 300$/h.

• $\rho = 0.80$.
• $L = 0.8/0.2 = 4$ cars.
• $L_q = 0.64/0.2 = 3.2$ cars.
• $W = 1/(300 - 240) = 1/60\text{ h} = 60\text{ s}$.
• $W_q = 240/(300 \times 60) = 0.0133\text{ h} = 48\text{ s}$.

Check with Little's law: $L = \lambda W = 240(1/60) = 4$.

If arrivals rise to $290$/h, then $\rho = 0.967$ and $W = 1/10\text{ h} = 6$ minutes. Delay grows explosively as $\rho \to 1$, which is why facilities are designed well below capacity.