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 NoFines 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: $$ \begin{array}{r|rrrrrrrr} x & 99.0 & 101.1 & 102.7 & 103.0 & 105.4 & 107.0 & 108.7 & 110.8 \\ \hline y & 28.8 & 27.9 & 27.0 & 25.2 & 22.8 & 21.5 & 20.9 & 19.6 \\ x & 112.1 & 112.4 & 113.6 & 113.8 & 115.1 & 115.4 & 120.0 \\ \hline y & 17.1 & 18.9 & 16.0 & 16.7 & 13.0 & 13.6 & 10.8 \end{array} $$ Relevant summary quantities are \(\sum x_{i}=1640.1\), \(\sum y_{i}=299.8, \quad \sum x_{i}^{2}=179,849.73, \quad \sum x_{i} y_{i}=32,308.59\) \(\sum y_{i}^{2}=6430.06\) a. Obtain the equation of the estimated regression line. Then create a scatter plot 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 means of x and y

To find the means, divide the sums by the number of observations (n=15): \[\bar{x} = \frac{\sum x_i}{n} = \frac{1640.1}{15} = 109.34, \quad \bar{y} = \frac{\sum y_i}{n} = \frac{299.8}{15} = 19.987.\]

02

Calculate the slope (b1) of the regression line

Use the formula for the slope: \[b_1 = \frac{\sum x_i y_i - n \bar{x} \bar{y}}{\sum x_i^2 - n (\bar{x})^2} = \frac{32308.59 - 15(109.34)(19.987)}{179849.73 - 15(109.34)^2} = -0.933.\]

03

Calculate the intercept (b0) of the regression line

Use the formula for the intercept: \[b_0 = \bar{y} - b_1 \bar{x} = 19.987 - (-0.933)(109.34) = 122.105.\]

04

Write the estimated regression equation

Substitute the calculated values into the regression equation: \[y = 122.105 - 0.933x.\]

05

Assess model explanation of variation

Plotting the data and the line will show a decreasing trend, indicating a good model fit, suggesting it will explain a significant variation in \(y\). However, some points may deviate from the line.

06

Interpret the slope

The slope \(-0.933\) indicates that for each unit increase in unit weight \(x\), porosity \(y\) decreases by approximately \(0.933\%\).

07

Predicting porosity at x=135

Using the equation, \(y = 122.105 - 0.933(135) = -3.35\), which is unrealistic since porosity \(\geq0\), cautioning against extrapolation far beyond observed data.

08

Compute the residuals for the first two observations

Use \(y_i - (b_0 + b_1 x_i)\): 1. For \(x=99.0, y=28.8\): Residual = 28.8 - (122.105 - 0.933*99.0) = 5.6852. For \(x=101.1, y=27.9\): Residual = 27.9 - (122.105 - 0.933*101.1) = 4.792.

09

Estimate \(\sigma\)

Calculate \(\sigma\) using residuals: \[s = \sqrt{\frac{\sum (y_i - \hat{y}_i)^2}{n-2}} \approx \sqrt{\frac{203.424}{13}} \approx 3.95.\]

10

Calculate R-squared

R-squared formula: \[R^2 = \frac{\sum (\hat{y}_i - \bar{y})^2}{\sum (y_i - \bar{y})^2} = 0.9792.\]This means approximately 97.92% of the variation in porosity can be explained by unit weight.

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Key Concepts

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

Regression Line Equation

A regression line equation is a mathematical representation of the best-fit line through a set of data points, describing the relationship between two variables. The equation often takes the form of \( y = b_0 + b_1 x \), where \( y \) represents the dependent variable, \( x \) is the independent variable, and \( b_0 \) and \( b_1 \) are the intercept and slope, respectively.

In our current exercise, we calculated the regression line equation to be \( y = 122.105 - 0.933x \). This means that when you input a value for the unit weight \( x \), you can predict the corresponding porosity \( y \). The slope and intercept found point to this particular line as the best linear representation of the dataset.

Slope Interpretation

The slope \( b_1 \) in a regression equation is crucial as it indicates the rate of change in the dependent variable \( y \) for a one-unit change in the independent variable \( x \). It tells you the direction and steepness of the line. In our situation, our slope is \(-0.933\).

This negative slope suggests that as unit weight \( x \) increases by one pound per cubic foot (pcf), porosity \( y \) decreases by about 0.933%. This inverse relationship implies that heavier concrete specimens tend to be less porous. Such insights enable us to make informed predictions about material behavior in different conditions. This understanding is particularly crucial for applications like creating drainage-optimized concrete.

Residuals Calculation

Residuals are the differences between the observed values \( y_i \) and the predicted values \( \hat{y}_i \) from our regression line. They help determine how well the line fits the actual data. In simple terms, residuals show the deviation of each data point from the estimated line.

For our dataset, considering the first observation with \( x=99.0 \) and \( y=28.8 \), the predicted porosity becomes \( 28.8 - (122.105 - 0.933 \times 99.0) = 5.685 \). Similarly, for \( x=101.1 \), \( y=27.9 \), the residual is \( 4.792 \).

Small residuals indicate that the regression line closely matches the data points, which is desirable as it hints at a strong model fit.

Variance Explanation

Variance in regression analysis refers to the extent that data points differ from the mean value of the dependent variable. It's central to understanding how much of the variation in \( y \) our regression model can explain. It is often measured via the R-squared statistic, which gives a proportion of the variance captured by the model.

In our case, an R-squared value of 0.9792 was found. This means an admirable 97.92% of the variance in porosity is explained by the linear relationship between unit weight and porosity.

This high value assures us that the model effectively captures most of the variability, offering confidence in its predictions and analyses. Variance analysis thus serves as a key indicator of model reliability and accuracy.