/*! 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 7 The accompanying scatterplot is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The accompanying scatterplot is based on data provided by authors of the article "'Spurious Correlation in the USEPA Rating Curve Method for Estimating Pollutant Loads" (J. of Emvir. Engra, 2008: 610-618); here discharge is in \(\mathrm{ft}^{3} / \mathrm{s}\) as opposed to \(\mathrm{m}^{3} / \mathrm{s}\) used in the article. The point on the far right of the plot corresponds to the observation \((140,1529.35)\). The resulting standardized residual is 3.10. Minitab flags the observation with an \(R\) for large residual and an \(X\) for potentially influential observation. a line to the following data on \(x=\) prepreg thickness \((\mathrm{mm}\) ) and \(v=\) core crush \((\%)\) : $$ \begin{array}{c|cccccccc} x & .246 & .250 & .251 & .251 & .254 & .262 & .264 & .270 \\ \hline y & 16.0 & 11.0 & 15.0 & 10.5 & 13.5 & 7.5 & 6.1 & 1.7 \\ x & .272 & .277 & .281 & .289 & .290 & .292 & .293 & \\ \hline y & 3.6 & 0.7 & 0.9 & 1.0 & 0.7 & 3.0 & 3.1 & \end{array} $$ a. Fit the simple linear regression model. What proportion of the observed variation in core crush can be attributed to the model relationship? b. Construct a scatterplot. Does the plot suggest that a linear probabilistic relationship is appropriate? c. Obtain the residuals and standardized residuals, and then construct residual plots. What do these plots suggest? What type of function should provide a better fit to the data than does a straight line?

Short Answer

Expert verified
a. Calculate \( R^2 \) from the regression model. b. Check scatterplot for linearity. c. Analyze residual plots for non-random patterns suggesting better model fit.

Step by step solution

01

Understanding the Simple Linear Regression Model

The simple linear regression model relates two variables: a predictor (independent variable) and a response (dependent variable). The equation is given by \( y = \beta_0 + \beta_1 x + \epsilon \) where \( \beta_0 \) is the intercept, \( \beta_1 \) is the slope, and \( \epsilon \) is the error term.
02

Determining the Regression Equation

To fit the simple linear regression model, we need to calculate the slope (\( \beta_1 \)) and the intercept (\( \beta_0 \)) using the least squares method. This involves using formulas for \( \beta_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \) and \( \beta_0 = \bar{y} - \beta_1 \bar{x} \). After calculations based on the provided data, the estimated equation would look like \( y = \beta_0 + \beta_1 x \).
03

Proportion of Observed Variation

The proportion of variation explained by the model is given by the coefficient of determination \( R^2 \). It is calculated as \( R^2 = 1 - \frac{SS_{residual}}{SS_{total}} \), where \( SS_{residual} \) is the sum of squares of residuals and \( SS_{total} \) is the total sum of squares. This value will tell us how much of the variation in core crush is due to the linear relationship with thickness.
04

Constructing and Analyzing the Scatterplot

Plot the data points of prepreg thickness (x) and core crush (y) to visually examine their relationship. A scatterplot should show whether the points roughly align in a linear pattern. If they do, a linear regression model may be appropriate. If not, consider alternative models.
05

Calculating Residuals and Standardized Residuals

The residual for each observation is calculated as \( e_i = y_i - \hat{y}_i \), where \( \hat{y}_i \) is the predicted value from the model. Standardized residuals are calculated using \( e_i / \hat{\sigma} \), where \( \hat{\sigma} \) is the standard deviation of the residuals. This helps identify outliers.
06

Constructing and Analyzing Residual Plots

Create residual plots plotting residuals and standardized residuals against the predicted response values. Look for patterns; if residuals display a non-random pattern, it suggests the linear model may not be appropriate and a different function, like a polynomial, might be better.

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Key Concepts

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

Scatterplot
A scatterplot is an essential tool in visualizing the relationship between two quantitative variables. Here, we plot data points with one variable on the x-axis (prepreg thickness) and the other on the y-axis (core crush percentage). This plot helps us determine if there's a linear relationship.
A well-organized scatterplot will show data points that, ideally, form a straight line if the relationship between the variables is linear.

To create a scatterplot, simply mark each observation as a point on the graph, where the x-coordinate is the value of the predictor variable, and the y-coordinate is the value of the response variable. Once plotted:
  • Look for a pattern or trend.
  • Check if the points suggest an increasing or decreasing relationship.
  • If the points are scattered widely with no visible pattern, a linear model might not fit well.
  • Moreover, any significant deviations, such as outliers, should be noted, as they can affect the model significantly.
Scatterplots are the first step in deciding the appropriateness of using simple linear regression by providing a visual cue to detect a possible linear relationship.
Residuals
Residuals are the differences between observed values and the values predicted by the regression model. Calculating residuals is crucial for assessing the accuracy of a simple linear regression model.
The residual for each data point is formulated as the observed value minus the predicted response (i.e., \( e_i = y_i - \hat{y}_i \)). Each residual provides a measure of how far the actual value is from what the model predicts.
  • Residuals help identify whether a linear model is appropriate.
  • A small residual means the model predicts closely; a large residual indicates a greater prediction error.
  • Standardized residuals, computed as \( \frac{e_i}{\hat{\sigma}} \), allow for comparison across different units or scales.
A residual plot, where residuals are plotted against predicted values, can uncover patterns indicating a poor model fit. If the residuals display a non-random pattern, it suggests issues with the linear fit, indicating that another model type, like polynomial regression, may be more appropriate.
Coefficient of Determination (R²)
The coefficient of determination, denoted as \(R^2\), quantifies how well the independent variable explains the variance in the dependent variable in simple linear regression. It ranges from 0 to 1, where a value close to 1 implies a strong relationship.
Mathematically, \(R^2\) is expressed as \(R^2 = 1 - \frac{SS_{residual}}{SS_{total}}\), where:
  • \(SS_{residual}\) is the sum of squares of the residuals, reflecting the unexplained variation.
  • \(SS_{total}\) is the total sum of squares, indicating the overall variability in the response data.
An \(R^2\) of 0.8, for example, would suggest that 80% of the response variable's variance is accounted for by the linear relationship with the predictor variable.
A higher \(R^2\) value generally indicates a better fit, although it doesn’t assure causality. Checking the \(R^2\) value alongside residual plots gives a comprehensive overview of the model's appropriateness and accuracy.
Least Squares Method
The Least Squares Method is a mathematical technique used to find the best-fitting line in simple linear regression by minimizing the sum of the squares of the residuals. It helps in estimating the slope and intercept of the regression line.
The goal is to make the difference between the observed values and predicted values as small as possible. The method relies on minimizing the sum of squared deviations, \(\sum (y_i - \hat{y}_i)^2\), ensuring a minimal average squared distance between data points and the regression line.
  • The slope \( \beta_1 \) is computed using \( \beta_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \).
  • The intercept \( \beta_0 \) follows from \( \beta_0 = \bar{y} - \beta_1 \bar{x} \).
Once calculated, \(\beta_0\) and \(\beta_1\) form the linear regression equation \( y = \beta_0 + \beta_1 x \). This line represents the average predicted response for given values of the predictor variable, guiding inferences and predictions in applied settings.

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Most popular questions from this chapter

The article "'The Influence of Honing Process Parameters on Surface Quality, Productivity, Cutting Angle, and Coefficient of Friction" (Industrial Lubrication and Tribology, 2012: 77-83) included the following data on \(x_{1}=\) cutting speed \((\mathrm{m} / \mathrm{s}), x_{2}=\) specific pressure of pre-honing process \(\left(\mathrm{N} / \mathrm{mm}^{2}\right), x_{3}=\) specific pressure of finishing honing process, and \(y=\) productivity in the honing process ( \(\mathrm{mm}^{3} / \mathrm{s}\) for a particular tool; productivity is the volume of the material cut in a second.a. The article proposed a multivariate power model \(Y=\alpha x_{1}^{\beta_{1}} x x_{2}^{\beta_{2}} x_{3}^{\beta_{i}} \epsilon\). The implied linear regression model involves regressing \(\ln (y)\) against the three predictors \(\ln \left(x_{1}\right), \ln \left(x_{2}\right)\), and \(\ln \left(x_{3}\right)\). Partial Minitab output from fitting this latter model is as follows (the corresponding estimated power regression function appeared in the cited article). Carry out the model utility test at significance level \(.05\). b. The large \(P\)-value corresponding to the \(t\) ratio for \(\ln \left(x_{2}\right)\) suggests that this predictor can be eliminated from the model. Doing so and refitting yields the following Minitab output. c. Fit the simple linear regression model implied by your conclusion in (b) to the transformed data, and carry out a test of model utility. d. The standardized residuals from the fit referred to in (c) are .03,.33. \(1.69, .33,-.49, .96, .57, .33,-, 25\), \(-1.28, .29,-2.26\). Plot these against \(\ln \left(x_{1}\right)\). What does the pattern suggest? e. Fitting a quadratic regression model to relate \(\ln (y)\) to \(\ln \left(x_{1}\right)\) gave the following Minitab output. Carry out a test of model utility at significance level \(.05\) (the pattern in residual plots is satisfactory). Then use the fact that \(s_{\ln \left(\tilde{Y}^{\prime}\right)}=.0178\left[Y^{\prime}=\ln (Y)\right]\) when \(x_{1}=1\) to obtain a \(95 \%\) prediction interval for productivity.

Continuous recording of heart rate can be used to obtain information about the level of exercise intensity or physical strain during sports participation, work, or other daily activities. The article "The Relationship Between Heart Rate and Oxygen Uptake During Non-Steady State Exercise" (Ergonomics, 2000: 1578-1592) reported on a study to investigate using heart rate response \((x\), as a percentage of the maximum rate) to predict oxygen uptake ( \(y\), as a percentage of maximum uptake) during exercise. The accompanying data was read from a graph in the article. $$ \begin{array}{l|llllllll} \mathrm{HR} & 43.5 & 44.0 & 44.0 & 44.5 & 44.0 & 45.0 & 48.0 & 49.0 \\ \hline \mathrm{VO}_{2} & 22.0 & 21.0 & 22.0 & 21.5 & 25.5 & 24.5 & 30.0 & 28.0 \\\ \mathrm{HR} & 49.5 & 51.0 & 54.5 & 57.5 & 57.7 & 61.0 & 63.0 & 72.0 \\ \hline \mathrm{VO}_{2} & 32.0 & 29.0 & 38.5 & 30.5 & 57.0 & 40.0 & 58.0 & 72.0 \end{array} $$ Use a statistical software package to perform a simple linear regression analysis, paying particular attention to the presence of any unusual or influential observations.

The article "Bank Full Discharge of Rivers" (Water 91Ó°ÊÓ \(\left.J_{.}, 1978: 1141-1154\right)\) reports data on discharge amount \(\left(q\right.\), in \(\left.\mathrm{m}^{3} / \mathrm{sec}\right)\), flow area \(\left(a\right.\), in \(\left.\mathrm{m}^{2}\right)\), and slope of the water surface \((b\), in \(\mathrm{m} / \mathrm{m})\) obtained at a number of floodplain stations. A subset of the data follows. Let \(y=\ln (q), x_{1}=\ln (a)\), and \(x_{2}=\ln (b)\). Consider fitting the model \(Y=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\epsilon\). a. The resulting \(h_{i i}\) 's are \(.138, .302, .266, .604, .464\), \(.360, .215, .153, .214\), and \(.284\). Does any observation appear to be influential? b. The estimated coefficients are \(\hat{\beta}_{0}=1.5652, \hat{\beta}_{1}=\) \(.9450\), and \(\hat{\beta}_{2}=.1815\), and the corresponding estimated standard deviations are \(s_{\hat{\beta}_{1}}=.7328, s_{\hat{\beta}_{t}}=\) \(.1528\), and \(s_{\dot{H}_{2}}=.1752\). The second standardized residual is \(e_{2}^{*}=2.19\). When the second observation is omitted from the data set, the resulting estimated coefficients are \(\hat{\beta}_{0}=1.8982, \hat{\beta}_{1}=1.025\), and \(\hat{\beta}_{2}=.3085\). Do any of these changes indicate that the second observation is influential? c. Deletion of the fourth observation (why?) yields \(\hat{\beta}_{0}=1.4592, \hat{\beta}_{1}=.9850\), and \(\hat{\beta}_{2}=.1515\). Is this observation influential?

The article "Analysis of the Modeling Methodologies for Predicting the Strength of Air-Jet Spun Yarns" (Textile Res. \(J ., 1997: 39-44\) ) reported on a study carried out to relate yarn tenacity ( \(y\), in \(\mathrm{g} /\) tex \()\) to yarn count \(\left(x_{1}\right.\), in tex), percentage polyester \(\left(x_{2}\right)\), first nozzle pressure \(\left(x_{3}\right.\), in \(\left.\mathrm{kg} / \mathrm{cm}^{2}\right)\), and second nozzle pressure \(\left(x_{4}\right.\), in \(\mathrm{kg} /\) \(\left.\mathrm{cm}^{2}\right)\). The estimate of the constant term in the corresponding multiple regression equation was \(6.121\). The estimated coefficients for the four predictors were \(-.082\), \(.113, .256\), and \(-.219\), respectively, and the coefficient of multiple determination was 946 . a. Assuming that the sample size was \(n=25\), state and test the appropriate hypotheses to decide whether the fitted model specifies a useful linear relationship between the dependent variable and at least one of the four model predictors. b. Again using \(n=25\), calculate the value of adjusted \(R^{2}\). c. Calculate a \(99 \%\) confidence interval for true mean yarn tenacity when yarn count is \(16.5\), yarn contains \(50 \%\) polyester, first nozzle pressure is 3 , and second nozzle pressure is 5 if the estimated standard deviation of predicted tenacity under these circumstances is . 350 .

The article "Validation of the Rockport Fitness Walking Test in College Males and Females" (Research Quarterly for Exercise and Sport, 1994: 152-158) recommended the following estimated regression equation for relating \(y=\mathrm{VO}_{2} \max (\mathrm{L} / \mathrm{min}\), a measure of cardiorespiratory fitness) to the predictors \(x_{1}=\) gender \((\) female \(=0\), male \(=1), x_{2}=\) weight \((\mathrm{lb})\), \(x_{3}=1\)-mile walk time \((\mathrm{min})\), and \(x_{4}=\) heart rate at the end of the walk (beats/min): $$ \begin{aligned} y=& 3.5959+.6566 x_{1}+.0096 x_{2} \\ &-.0996 x_{3}-.0080 x_{4} \end{aligned} $$ a. How would you interpret the estimated coefficient \(\hat{\beta}_{3}=-.0996 ?\) b. How would you interpret the estimated coefficient \(\hat{\beta}_{1}=.6566 ?\) The article "Validation of the Rockport Fitness Walking Test in College Males and Females" (Research Quarterly for Exercise and Sport, 1994: 152-158) recommended the following estimated regression equation for relating \(y=\mathrm{VO}_{2} \max (\mathrm{L} / \mathrm{min}\), a measure of cardiorespiratory fitness) to the predictors \(x_{1}=\) gender \((\) female \(=0\), male \(=1), x_{2}=\) weight \((\mathrm{lb})\), \(x_{3}=1\)-mile walk time \((\mathrm{min})\), and \(x_{4}=\) heart rate at the end of the walk (beats/min): $$ \begin{aligned} y=& 3.5959+.6566 x_{1}+.0096 x_{2} \\ &-.0996 x_{3}-.0080 x_{4} \end{aligned} $$ a. How would you interpret the estimated coefficient \(\hat{\beta}_{3}=-.0996 ?\) b. How would you interpret the estimated coefficient \(\hat{\beta}_{1}=.6566 ?\)

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