/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 For the past decade, rubber powd... [FREE SOLUTION] | 91Ó°ÊÓ

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For the past decade, rubber powder has been used in asphalt cement to improve performance. The article "Experimental Study of Recycled Rubber-Filled High- Strength Concrete" (Magazine of Concrete Res., 2009: 549-556) includes a regression of \(y=\) axial strength (MPa) on \(x=c u b e\) strength (MPa) based on the following sample data: $$ \begin{array}{c|cccccccccc} x & 112.3 & 97.0 & 92.7 & 86.0 & 102.0 & 99.2 & 95.8 & 103.5 & 89.0 & 86.7 \\ \hline y & 75.0 & 71.0 & 57.7 & 48.7 & 74.3 & 73.3 & 68.0 & 59.3 & 57.8 & 48.5 \end{array} $$ a. Obtain the equation of the least squares line, and interpret its slope. b. Calculate and interpret the coefficient of determination. c. Calculate and interpret an estimate of the error standard deviation \(\sigma\) in the simple linear regression model.

Short Answer

Expert verified
The regression line equation is determined, with the slope showing the strength relationship. \(R^2\) shows model fit, and \(\sigma\) measures prediction accuracy.

Step by step solution

01

Calculate the Mean Values for x and y

First, we need to find the mean of the x-values (cube strength) and y-values (axial strength). Calculate \(\bar{x}\) and \(\bar{y}\) using:\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \] \[ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i \] where \(n\) is the number of data points, which is 10.
02

Use the Least Squares Formula to Find the Slope and Y-Intercept

The slope \(b\) of the least squares line is calculated using the formula: \[ b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \] Once the slope \(b\) is found, calculate the y-intercept \(a\) using: \[ a = \bar{y} - b\bar{x} \] Thus, the least squares regression line is \( y = a + bx \).
03

Interpret the Slope

The slope \( b \) represents the change in the axial strength (MPa) for every one MPa increase in cube strength. If \( b > 0 \), the relationship is directly proportional; \( b < 0 \), the relationship is inversely proportional.
04

Calculate the Coefficient of Determination \(R^2\)

The coefficient of determination \(R^2\) is calculated as:\[ R^2 = 1 - \frac{SS_{res}}{SS_{tot}} \] where \( SS_{res} \) is the sum of the squares of the residuals, and \( SS_{tot} \) is the total sum of squares. This value indicates the proportion of the variance in the dependent variable that is predictable from the independent variable.
05

Interpret the Coefficient of Determination \(R^2\)

The value of \( R^2 \) indicates how well the regression line fits the data. If \( R^2 \) is close to 1, a large portion of the variance in \(y\) is explained by \(x\). If \( R^2 \) is close to 0, the regression line does not explain much of the variance in \(y\).
06

Estimate the Error Standard Deviation \(\sigma\)

The error standard deviation (\(\sigma\)) is calculated using:\[ \sigma = \sqrt{\frac{1}{n-2}SS_{res}} \]This provides a measure of the typical distance that the observed values fall from the regression line, indicating the spread of the residuals.
07

Interpret the Error Standard Deviation \(\sigma\)

The error standard deviation \(\sigma\) gives an estimate of the typical size of the prediction errors (residuals) when using the regression line. A smaller \(\sigma\) indicates the data points are closer to the fitted line.

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Key Concepts

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

Least Squares Method
The least squares method is a fundamental approach in linear regression analysis. Its purpose is to find the best-fitting line through a set of data points by minimizing the sum of the squared differences between the observed values and the values predicted by the line. These differences are known as residuals.
Here's how you calculate it:
  • Calculate the means: First, we find the average values of the x and y data sets, called \(\bar{x}\) and \(\bar{y}\).
  • Determine the slope (\(b\)): This tells us how much \(y\) (axial strength) is expected to change with a one-unit change in \(x\) (cube strength). It's given by the formula: \[b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}\]
  • Calculate the y-intercept (\(a\)): Once we have the slope, the y-intercept can be found using: \[a = \bar{y} - b\bar{x}\]
  • Form the equation: The equation of the least squares line is then \(y = a + bx\).
The resulting line is the one that best fits the data, where the sum of the squared residuals is at its minimum.
This is very useful for making predictions based on trends in the data.
Coefficient of Determination
The coefficient of determination, represented as \(R^2\), is a key metric that evaluates the goodness of fit of your regression line. It tells us what portion of the variance in the dependent variable can be predicted from the independent variable.
This is how it is calculated:
  • Calculate \(SS_{res}\): This involves measuring the sum of squares of the residuals.
  • Calculate \(SS_{tot}\): This is the total sum of squares, or the total variance in the data.
  • Compute \(R^2\): The formula is \[R^2 = 1 - \frac{SS_{res}}{SS_{tot}}\].
Interpreting \(R^2\):
  • Ranges from 0 to 1: An \(R^2\) closer to 1 implies that a significant extent of the variation in the output (\(y\)) is explained by the input (\(x\)).
  • If \(R^2 = 0.8\), for instance, 80% of the variance in \(y\) can be accounted for by the line fit to \(x\). The remaining 20% is due to other factors or randomness.

Higher \(R^2\) values generally indicate a better fit, meaning the line closely follows the real data points.
Error Standard Deviation
The error standard deviation, denoted as \(\sigma\), is crucial in understanding the variability or spread of residuals. Residuals are essentially the differences between observed values and those predicted by the regression line.
  • Calculation: It is determined through the formula: \[\sigma = \sqrt{\frac{1}{n-2}SS_{res}}\]where \(SS_{res}\) is the sum of the squared residuals, and \(n\) is the number of observations.
  • Interpreting \(\sigma\): A small \(\sigma\) implies that the observed data points hover closely around the line of best fit, which means reliable predictions.
Understanding \(\sigma\) helps us comprehend how much prediction error exists on average when using the regression model.
By minimizing \(\sigma\), we aim for a tighter fit between the model and actual data, enhancing the predictability and reliability of regression analyses.

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Most popular questions from this chapter

The flow rate \(y\left(\mathrm{~m}^{3} / \mathrm{min}\right)\) in a device used for air-quality measurement depends on the pressure drop \(x\) (in. of water) across the device's filter. Suppose that for \(x\) values between 5 and 20 , the two variables are related according to the simple linear regression model with true regression line \(y=-.12+.095 x\) a. What is the expected change in flow rate associated with a 1-in. increase in pressure drop? Explain. b. What change in flow rate can be expected when pressure drop decreases by 5 in.? c. What is the expected flow rate for a pressure drop of 10 in.? A drop of 15 in.? d. Suppose \(\sigma=.025\) and consider a pressure drop of 10 in. What is the probability that the observed value of flow rate will exceed .835? That observed flow rate will exceed \(.840\) ? e. What is the probability that an observation on flow rate when pressure drop is 10 in. will exceed an observation on flow rate made when pressure drop is 11 in.?

Suppose that in a certain chemical process the reaction time \(y(\mathrm{hr})\) is related to the temperature \(\left({ }^{\circ} \mathrm{F}\right)\) in the chamber in which the reaction takes place according to the simple linear regression model with equation \(y=5.00-.01 x\) and \(\sigma=.075\) a. What is the expected change in reaction time for a \(1^{\circ} \mathrm{F}\) increase in temperature? For a \(10^{\circ} \mathrm{F}\) increase in temperature? b. What is the expected reaction time when temperature is \(200^{\circ} \mathrm{F}\) ? When temperature is \(250^{\circ} \mathrm{F}\) ? c. Suppose five observations are made independently on reaction time, each one for a temperature of \(250^{\circ} \mathrm{F}\). What is the probability that all five times are between \(2.4\) and \(2.6 \mathrm{hr}\) ? d. What is the probability that two independently observed reaction times for temperatures \(1^{\circ}\) apart are such that the time at the higher temperature exceeds the time at the lower temperature?

"Mode-mixity" refers to how much of crack propagation is attributable to the three conventional fracture modes of opening, sliding, and tearing. For plane problems, only the first two modes are present, and the mode-mixity angle is a measure of the extent to which propagation is due to sliding as opposed to opening. The article "Increasing Allowable Flight Loads by Improved Structural Modeling" (AIAA J., 2006: 376-381) gives the following data on \(x=\) mode- mixity angle (degrees) and \(y=\) fracture toughness \((\mathrm{N} / \mathrm{m})\) for sandwich panels use in aircraft construction. $$ \begin{array}{l|llllllll} x & 16.52 & 17.53 & 18.05 & 18.50 & 22.39 & 23.89 & 25.50 & 24.89 \\ \hline y & 609.4 & 443.1 & 577.9 & 628.7 & 565.7 & 711.0 & 863.4 & 956.2 \\ x & 23.48 & 24.98 & 25.55 & 25.90 & 22.65 & 23.69 & 24.15 & 24.54 \\ \hline y & 679.5 & 707.5 & 767.1 & 817.8 & 702.3 & 903.7 & 964.9 & 1047.3 \end{array} $$ a. Obtain the equation of the estimated regression line, and discuss the extent to which the simple linear regression model is a reasonable way to relate fracture toughness to mode-mixity angle. b. Does the data suggest that the average change in fracture toughness associated with a one-degree increase in mode-mixity angle exceeds \(50 \mathrm{~N} / \mathrm{m}\) ? Carry out an appropriate test of hypotheses. c. For purposes of precisely estimating the slope of the population regression line, would it have been preferable to make observations at the angles \(16,16,18,18,20,20\), \(20,20,22,22,22,22,24,24,26\), and 26 (again a sample size of 16)? Explain your reasoning. d. Calculate an estimate of true average fracture toughness and also a prediction of fracture toughness both for an angle of 18 degrees and for an angle of 22 degrees, do so in a manner that conveys information about reliability and precision, and then interpret and compare the estimates and predictions.

Verify that if each \(x_{i}\) is multiplied by a positive constant \(c\) and each \(y_{i}\) is multiplied by another positive constant \(d\), the \(t\) statistic for testing \(H_{0}: \beta_{1}=0\) versus \(H_{\mathrm{a}}: \beta_{1} \neq 0\) is unchanged in value (the value of \(\hat{\beta}_{1}\) will change, which shows that the magnitude of \(\hat{\beta}_{1}\) is not by itself indicative of model utility).

Toughness and fibrousness of asparagus are major determinants of quality. This was the focus of a study reported in "Post-Harvest Glyphosphate Application Reduces Toughening, Fiber Content, and Lignification of Stored Asparagus Spears" (J. of the Amer. Soc. of Hort. Science, 1988: 569–572). The article reported the accompanying data (read from a graph) on \(x=\) shear force \((\mathrm{kg})\) and \(y=\) percent fiber dry weight. $$ \begin{array}{l|ccccccccc} x & 46 & 48 & 55 & 57 & 60 & 72 & 81 & 85 & 94 \\ \hline y & 2.18 & 2.10 & 2.13 & 2.28 & 2.34 & 2.53 & 2.28 & 2.62 & 2.63 \\ x & 109 & 121 & 132 & 137 & 148 & 149 & 184 & 185 & 187 \\ \hline y & 2.50 & 2.66 & 2.79 & 2.80 & 3.01 & 2.98 & 3.34 & 3.49 & 3.26 \end{array} $$ a. Calculate the value of the sample correlation coefficient. Based on this value, how would you describe the nature of the relationship between the two variables? b. If a first specimen has a larger value of shear force than does a second specimen, what tends to be true of percent dry fiber weight for the two specimens? c. If shear force is expressed in pounds, what happens to the value of \(r\) ? Why? d. If the simple linear regression model were fit to this data, what proportion of observed variation in percent fiber dry weight could be explained by the model relationship? e. Carry out a test at significance level \(.01\) to decide whether there is a positive linear association between the two variables.

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