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The following summary statistics were obtained from study that used regression analysis to investigate the relationship between pavement deflection and surface temperature of the pavement at various locations on a state highway. Here \(x=\) temperature \(\left({ }^{\circ} \mathrm{F}\right)\) and \(y=\) deflection adjustment factor \((y \geq 0)\) : $$ \begin{aligned} &n=15 \quad \sum x_{i}=1425 \quad \sum y_{i}=10.68 \\ &\sum x_{i}^{2}=139,037.25 \quad \sum x_{i} y_{i}=987.645 \\ &\sum y_{i}^{2}=7.8518 \end{aligned} $$ (Many more than 15 observations were made in the study; the reference is "Flexible Pavement Evaluation and Rehabilitation," Transportation Eng. J., 1977: 75-85.) a. Compute \(\hat{\beta}_{1}, \hat{\beta}_{0}\), and the equation of the estimated regression line. Graph the estimated line. b. What is the estimate of expected change in the deflection adjustment factor when temperature is increased by \(1^{\circ} \mathrm{F}\) ? c. Suppose temperature were measured in \({ }^{\circ} \mathrm{C}\) rather than in \({ }^{\circ} \mathrm{F}\). What would be the estimated regression line? Answer part (b) for an increase of \(1^{\circ} \mathrm{C}\). [Hint: \({ }^{\circ} \mathrm{F}=(9 / 5)^{\circ} \mathrm{C}+32\); now substitute for the "old \(x\) " in terms of the "new \(x\)."] d. If a \(200^{\circ} \mathrm{F}\) surface temperature were within the realm of possibility, would you use the estimated line of part (a) to predict deflection factor for this temperature? Why or why not?

Step by step solution

01

Calculate the Slope (β1)

To find the slope of the regression line, we use the formula: \[ \hat{\beta}_1 = \frac{n \sum x_iy_i - (\sum x_i)(\sum y_i)}{n \sum x_i^2 - (\sum x_i)^2} \]Substitute the given values: \[ \hat{\beta}_1 = \frac{15 \cdot 987.645 - (1425)(10.68)}{15 \cdot 139037.25 - (1425)^2} \]Calculate this to find:\[ \hat{\beta}_1 \approx 0.00723 \]

02

Calculate the Intercept (β0)

To find the intercept, use the formula:\[ \hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x} \]First, calculate the means \( \bar{x} \) and \( \bar{y} \):\[ \bar{x} = \frac{\sum x_i}{n} = \frac{1425}{15} = 95 \]\[ \bar{y} = \frac{\sum y_i}{n} = \frac{10.68}{15} \approx 0.712 \]Now calculate the intercept:\[ \hat{\beta}_0 = 0.712 - 0.00723 \times 95 \]\[ \hat{\beta}_0 \approx 0.023 \]

03

Form the Regression Line

With \( \hat{\beta}_0 = 0.023 \) and \( \hat{\beta}_1 = 0.00723 \), the estimated regression line equation is:\[ \hat{y} = 0.023 + 0.00723x \]

04

Graph the Regression Line

To graph the regression line \( \hat{y} = 0.023 + 0.00723x \), plot the line over a range of \( x \) values. This line represents the relationship between temperature in Fahrenheit and the deflection adjustment factor.

05

Calculate the Change in Deflection Factor for 1°F

The slope \( \hat{\beta}_1 = 0.00723 \) represents the expected change in the deflection adjustment factor for a 1°F increase in temperature: So, for an increase of 1°F, the deflection adjustment factor increases by approximately 0.00723.

06

Convert the Equation to Celsius

First, convert the temperature from °F to °C using the formula :\[ ^\circ F = \frac{9}{5} \times {^\circ C} + 32 \]Substitute this into the regression equation:\[ \hat{y} = 0.023 + 0.00723 \left( \frac{9}{5}x + 32 \right) \]Simplify to get:\[ \hat{y} = 0.023 + 0.013014x_C + 0.23136 \]Thus, when calculating the relationship with Celsius, the equation becomes: \[ \hat{y} = 0.25436 + 0.013014x_C \]

07

Calculate Change in Deflection Factor for 1°C

With temperature in Celsius, \( \hat{\beta}_1 = 0.013014 \) explains the change in the deflection adjustment factor for a 1°C increase:For an increase of 1°C, the deflection adjustment factor increases by approximately 0.013014.

08

Consider Using the Regression Line at 200°F

To decide whether you can use the estimated regression line for a temperature of 200°F, consider the range of data used to estimate the line. If 200°F lies outside the observed data range (based on mean \( \bar{x} = 95 \) and assumed max/min values), the linear relationship may not hold.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pavement Deflection

Pavement deflection refers to the bending or flexing of the pavement surface when subjected to loads such as vehicles. This can be an important measure for engineers to assess the health and durability of a road. When the pavement supports a load, it experiences internal stress. If a pavement deflects too much, it indicates that it may not withstand further usage, which could lead to cracks or deeper damage.
Understanding pavement deflection is crucial for road maintenance and planning. It helps in predicting the lifespan of the pavement and making decisions about necessary interventions. In this context, it is important to determine how factors like temperature affect deflection, which is why regression analysis is employed.

Surface Temperature

Surface temperature of the pavement can greatly influence its physical properties, including its deflection under loads. Pavement surface temperature can vary based on several factors like ambient temperature, sunlight intensity, and the material composition of the pavement.
For example, a higher surface temperature can make the bituminous surface layer softer, increasing deflection. More deflection at higher temperatures suggests that the temperature might be a critical factor in pavement behavior. By examining the relationship between surface temperature and deflection, we can improve the design and maintenance strategies to mitigate potential damage.

Linear Regression

Linear regression is a statistical method used to model the relationship between a dependent variable and one or more independent variables. In the context of the given exercise, it involves understanding how pavement deflection (dependent variable) changes with surface temperature (independent variable).
The technique involves finding a line of best fit through a set of data points. This line is represented by an equation, \[ \hat{y} = \hat{\beta}_0 + \hat{\beta}_1x \], where \( \hat{\beta}_0 \) is the intercept and \( \hat{\beta}_1 \) is the slope. In this scenario, linear regression helps determine how pavement deflection changes as temperature increases, allowing for predictions and analysis based on observed data.

Slope and Intercept Calculations

To calculate the regression line's slope \( \hat{\beta}_1 \), we analyze how much the dependent variable, deflection, changes for each unit change in the surface temperature. This is achieved using\[ \hat{\beta}_1 = \frac{n \sum x_iy_i - (\sum x_i)(\sum y_i)}{n \sum x_i^2 - (\sum x_i)^2} \]. This formula ensures a precise calculation of the slope, depicting the effect temperature has on deflection.
The intercept \( \hat{\beta}_0 \) is found using:\[ \hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x} \], which indicates the value of deflection when the temperature is zero.
These calculations result in the best-fit line, essential for making predictions about deflection at varying temperatures.

Temperature Conversion

In scientific studies like this one, temperature might be recorded in either Fahrenheit or Celsius. However, these two units can be converted using the formula.: \[ ^\circ F = \frac{9}{5} \times {^\circ C} + 32 \]
When working with regression analysis, you might need to convert temperature units to maintain consistency, particularly when given data in different units. In this exercise, converting temperature from Fahrenheit to Celsius significantly affects the regression equation and understanding how temperature influences deflection.
It's essential to comprehend temperature conversion as it allows better global understanding and comparability in different scientific studies, making results more versatile and widely applicable.