Assignment 04 _ Q(05) : Regression Analysis for Pavement Deflection

Problem Setup

We are given a study that investigates the relationship between roadway surface temperature (\(x\)) and pavement deflection (\(y\)) using regression methods. The following summary statistics are provided:

• \(n = 20\) (number of data points)
• \(\sum y_i = 12.7\)
• \(\sum y_i^2 = 8.8\)
• \(\sum x_i = 1487\)
• \(\sum x_i^2 = 143,215\)
• \(\sum x_i y_i = 1083\)

Part (a) Least Squares Estimates of the Slope and Intercept

To compute the least squares estimates of the slope (\(\hat{\beta_1}\)) and the intercept (\(\hat{\beta_0}\)), we use the following formulas:

\[ \hat{\beta_1} = \frac{n \sum x_i y_i - \sum x_i \sum y_i}{n \sum x_i^2 - (\sum x_i)^2} \]

\[ \hat{\beta_0} = \frac{\sum y_i - \hat{\beta_1} \sum x_i}{n} \]

# Assigning the values form the question to the variables

n <- 20
sum_y <- 12.7
sum_y2 <- 8.8
sum_x <- 1487
sum_x2 <- 143215
sum_xy <- 1083

# Calculate slope (slope_1) and intercept (intercept_0)
beta_1 <- (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - sum_x^2)
beta_0 <- (sum_y - beta_1 * sum_x) / n

Displaying the results:

# Display results
beta_1

## [1] 0.004248918

beta_0

## [1] 0.319093

Explanation:

•   The slope measures the change in pavement deflection for a unit change in surface temperature.
•   The intercept represents the estimated deflection when the surface temperature is 0°F.

Part (b) Prediction of Pavement Deflection at 85°F

Now that we have the equation of the regression line:

[ y = + x]

We can use this equation to predict the pavement deflection when the surface temperature is 85°F.

# Predict the deflection for x = 85°F
x_85 <- 85
pred_y_85 <- beta_0 + beta_1 * x_85

Displaying the Predicted Deflection:

# Display the predicted deflection
pred_y_85

## [1] 0.680251

Explanation:

The value of ( y ) corresponding to a surface temperature of 85°F is obtained by substituting ( x = 85 ) into the regression equation.

Part (c) Mean Pavement Deflection at 90°F

To compute the mean pavement deflection for a surface temperature of 90°F, we simply substitute ( x = 90 ) into the regression equation.

# Predict the deflection for x = 90°F
x_90 <- 90
pred_y_90 <- beta_0 + beta_1 * x_90

# Display the predicted deflection
pred_y_90

## [1] 0.7014956

Explanation:

This part gives us the predicted mean deflection for a temperature of 90°F based on the regression line.

Part (d) Change in Mean Pavement Deflection for a 1°F Change in Temperature

The change in mean pavement deflection due to a 1°F change in surface temperature is simply the slope ( ), as it represents the change in ( y ) for a unit change in ( x ).

# Change in deflection for a 1°F change in temperature
change_in_deflection <- beta_1

# Display the change in deflection
change_in_deflection

## [1] 0.004248918

Explanation:

The slope ( ) tells us how much the pavement deflection changes when the surface temperature increases by 1°F.

Regression Line Graph

To visualize the relationship between surface temperature and pavement deflection, we will plot the regression line on a graph.

# Example data points (surface temperature and pavement deflection)
x_values <- c(30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100)
y_values <- beta_0 + beta_1 * x_values  # Use regression line equation to generate y values

# Plotting the regression line
plot(x_values, y_values, main = "Regression Line for Pavement Deflection",
     xlab = "Surface Temperature (°F)", ylab = "Pavement Deflection",
     pch = 19, col = "blue")
abline(lm(y_values ~ x_values), col = "red")  # Add regression line

Explanation:

The plot shows the relationship between surface temperature and pavement deflection with the regression line overlaying the data points.