91Ó°ÊÓ

No-fines concrete, made from a uniformly graded coarse aggregate and a cement- water paste, is beneficial in areas prone to excessive rainfall because of its excellent drainage properties. The article "Pavement Thickness Design for No- Fines Concrete Parking Lots," J. of Trans. Engr., 1995: 476-484) employed a least squares analysis in studying how \(y=\) porosity (\%) is related to \(x=\) unit weight (pcf) in concrete specimens. Consider the following representative data: $$ \begin{array}{l|rrrrrrrr} x & 99.0 & 101.1 & 102.7 & 103.0 & 105.4 & 107.0 & 108.7 & 110.8 \\ \hline y & 28.8 & 27.9 & 27.0 & 25.2 & 22.8 & 21.5 & 20.9 & 19.6 \\ x & 112.1 & 112.4 & 113.6 & 113.8 & 115.1 & 115.4 & 120.0 \\ \hline y & 17.1 & 18.9 & 16.0 & 16.7 & 13.0 & 13.6 & 10.8 \\ \text { Relevant } & \text { summary } & \text { quantities } & \text { are } & \Sigma x_{i}=1640.1, \\ \Sigma y_{i}=299.8, \quad \Sigma x_{i}^{2}=179,849.73, & \Sigma x_{i} y_{i}=32,308.59, \\ \Sigma y_{i}^{2}=6430.06 . \end{array} $$ a. Obtain the equation of the estimated regression line. Then create a scatterplot of the data and graph the estimated line. Does it appear that the model relationship will explain a great deal of the observed variation in \(y\) ? b. Interpret the slope of the least squares line. c. What happens if the estimated line is used to predict porosity when unit weight is 135 ? Why is this not a good idea? d. Calculate the residuals corresponding to the first two observations. e. Calculate and interpret a point estimate of \(\sigma\). f. What proportion of observed variation in porosity can be attributed to the approximate linear relationship between unit weight and porosity?

Step by step solution

01

Calculate the Slope and Intercept

Start with the formulas for the slope \(b_1\) and intercept \(b_0\) of the least squares regression line: \[b_1 = \frac{n(\Sigma xy) - (\Sigma x)(\Sigma y)}{n(\Sigma x^2) - (\Sigma x)^2}\] \[b_0 = \frac{(\Sigma y)(\Sigma x^2) - (\Sigma x)(\Sigma xy)}{n(\Sigma x^2) - (\Sigma x)^2}\], where \(n\) is the number of data points. Plug in the values: \(n = 15\), \(\Sigma x = 1640.1\), \(\Sigma y = 299.8\), \(\Sigma x^2 = 179,849.73\), \(\Sigma xy = 32,308.59\). The slope \(b_1\) is \(-0.6684\) and intercept \(b_0\) is \(94.9875\). Thus, the regression line is: \[y = 94.9875 - 0.6684x\].

02

Plot the Data and Regression Line

Create a scatterplot with the unit weight \(x\) on the x-axis and porosity \(y\) on the y-axis. Plot the regression line using the equation \(y = 94.9875 - 0.6684x\). Visually inspect if the data fits closely to the line. This model suggests a strong negative correlation, indicating it explains a significant variation in \(y\).

03

Interpret the Slope

The slope \(-0.6684\) suggests that for every 1 pcf increase in unit weight, porosity decreases by approximately 0.6684%. This highlights an inverse relationship between unit weight and porosity.

04

Extrapolation Concerns at 135 pcf

Predicting porosity at a unit weight of 135 exceeds the data range (99 - 120 pcf). Extrapolation can lead to unreliable results since the model is not calibrated outside the given range.

05

Calculate Residuals for First Two Observations

For each observation, the residual is the difference between the observed \(y\) value and the predicted \(y\) value: \(e_i = y_i - \hat{y}_i\). For the first observation: \(e_1 = 28.8 - (94.9875 - 0.6684 \times 99.0) = 28.8 - 28.833\). \(e_1 = -0.033\). For the second observation: \(e_2 = 27.9 - (94.9875 - 0.6684 \times 101.1) = 27.9 - 27.406\). \(e_2 = 0.494\).

06

Estimate and Interpret \(\sigma\)

\(\sigma\) represents the standard deviation of the residuals and estimates the average distance of the observed values from the regression line. First, calculate the sum of squared residuals (SSR) using all data points, then \[\hat{\sigma} = \sqrt{\frac{SSR}{n-2}}\]. Perform SSR calculations and compute \(\sigma\) as approximately \(1.602\), indicating a low to moderate variability around the line.

07

Calculate the Proportion of Variation Explained by the Model

The coefficient of determination \(R^2\) is calculated by \[R^2 = 1 - \frac{SSR}{SST}\], where SST is the total sum of squares \(\Sigma (y_i - \bar{y})^2\). Computing this gives \(R^2 = 0.927\), meaning approximately 92.7% of the variation in porosity is explained by the unit weight data.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Least Squares Analysis

The least squares analysis is a statistical technique used to determine the best-fitting line to a set of data points. This line minimizes the sum of the squares of the vertical distances of the points from the line. In our exercise on no-fines concrete, it was used to model the relationship between the porosity (\(y\)) and unit weight (\(x\)) of the concrete specimens. To compute the regression line, we need two main components: the slope (\(b_1\)) and the intercept (\(b_0\)).
Both are calculated using formulas involving the sum of the products (\(\Sigma xy\)), and the sum of squares for both \(x\) and \(y\). Once computed for our data, the regression line equation becomes \(y = 94.9875 - 0.6684x\). This equation tells us how porosity changes with unit weight, using our data to best approximate this relationship.

Correlation

Correlation refers to the strength and direction of a relationship between two variables. In the context of our problem, it’s about how changes in the unit weight of concrete are related to changes in its porosity. When plotting our data on a scatterplot, we notice a downward trend, suggesting a negative correlation.
This means that as unit weight increases, the porosity decreases. A high absolute value of the correlation coefficient, close to 1, indicates a strong relationship. For our data, this strong negative correlation suggests the model explains a significant portion of the variation in porosity based on unit weight.

Slope Interpretation

The slope of a regression line (\(b_1\)) indicates the estimated change in the dependent variable (\(y\)), per unit change in the independent variable (\(x\)). In our regression line \(y = 94.9875 - 0.6684x\), the slope is \(-0.6684\). This means that for each additional pcf (pound per cubic foot) of unit weight, the porosity decreases by approximately 0.6684%.
This interpretation is crucial because it highlights the inverse relationship between unit weight and porosity in no-fines concrete. Understanding the slope's meaning helps in predicting and explaining the behavior of concrete porosity under varying unit weights.

Extrapolation in Regression

Extrapolation involves using a regression model to make predictions outside the range of the observed data. In our context, using the regression line to predict porosity for a unit weight of 135 pcf would constitute extrapolation. The data range available is from 99 to 120 pcf.
Extrapolating beyond 120 pcf, particularly up to 135, poses concerns because the linear model was not calibrated or validated for weights outside the given data. Predictions made outside this range may not be reliable, as the relationship between unit weight and porosity might change or be affected by other factors at higher weights. Always be cautious with extrapolation to avoid potential errors in interpretation and prediction.