/*! 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 24 The invasive diatom species Didy... [FREE SOLUTION] | 91影视

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The invasive diatom species Didymosphenia geminata has the potential to inflict substantial ecological and economic damage in rivers. The article "Substrate Characteristics Affect Colonization by the BloomForming Diatom Didymosphenia geminata (Aquatic Ecology, 2010: 33-40) described an investigation of colonization behavior. One aspect of particular interest was whether \(y=\) colony density was related to \(x=\) rock surface area. The article contained a scatterplot and summary of a regression analysis. Here is representative data: $$ \begin{array}{l|rrrrrrrr} x & 50 & 71 & 55 & 50 & 33 & 58 & 79 & 26 \\ \hline y & 152 & 1929 & 48 & 22 & 2 & 5 & 35 & 7 \\ x & 69 & 44 & 37 & 70 & 20 & 45 & 49 & \\ \hline y & 269 & 38 & 171 & 13 & 43 & 185 & 25 & \end{array} $$ a. Fit the simple linear regression model to this data, predict colony density when surface area \(=70\) and when surface area \(=71\), and calculate the corresponding residuals. How do they compare? b. Calculate and interpret the coefficient of determination. c. The second observation has a very extreme \(y\) value (in the full data set consisting of 72 observations, there were two of these). This observation may have had a substantial impact on the fit of the model and subsequent conclusions. Eliminate it and recalculate the equation of the estimated regression line. Does it appear to differ substantially from the equation before the deletion? What is the impact on \(r^{2}\) and \(s\) ?

Short Answer

Expert verified
Fit a linear regression model, calculate and compare residuals, interpret \( R^2 \), and assess the impact of an extreme observation on the model fit.

Step by step solution

01

Data Preparation

First, organize the given data into two lists for the independent variable \(x\) (rock surface area) and the dependent variable \(y\) (colony density). This will facilitate performing calculations. We have: \[x = [50, 71, 55, 50, 33, 58, 79, 26, 69, 44, 37, 70, 20, 45, 49]\] \[y = [152, 1929, 48, 22, 2, 5, 35, 7, 269, 38, 171, 13, 43, 185, 25]\]
02

Fit the Linear Regression Model

Use statistical software or manual calculation to fit a linear regression model of the form \( y = \beta_0 + \beta_1 x \). Calculate the slope \(\beta_1\) and intercept \(\beta_0\). These represent the estimated change in \(y\) for a 1-unit change in \(x\) and the \(y\)-intercept respectively.
03

Predict Colony Density

Using the regression equation from the previous step, substitute \( x = 70 \) and \( x = 71 \) to find predicted colony densities \( \hat{y}_{70} \) and \( \hat{y}_{71} \). These values will help in calculating the residuals.
04

Calculate Residuals

Residuals are the differences between observed and predicted values. Calculate the residuals \( e_{70} = y_{actual,70} - \hat{y}_{70} \) and \( e_{71} = y_{actual,71} - \hat{y}_{71} \). Compare the magnitude of these residuals to assess model accuracy for these predictions.
05

Coefficient of Determination

Calculate \( R^2 \), the coefficient of determination, which indicates the proportion of the variance in \( y \) that is predictable from \( x \). This is done by squaring the correlation coefficient \( r \) obtained from the regressed data.
06

Impact of Extreme Values

Exclude the extreme observation(s) (e.g., \( y = 1929 \)) from the dataset and refit the regression model. Recalculate the regression parameters \( \beta_0 \) and \( \beta_1 \), and compare with those from Step 2. Assess changes in the slope, intercept, \( R^2 \), and standard error \( s \) to understand the impact of the extreme values on model fit and accuracy.

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Key Concepts

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

Coefficient of Determination
The coefficient of determination, often represented as \( R^2 \), is a key metric in linear regression analysis. It measures the proportion of variance in the dependent variable, here the colony density, that can be predicted from the independent variable, which is the rock surface area in this context.

An \( R^2 \) value ranges from 0 to 1, where 0 indicates that the model does not explain any of the variability in the response data around its mean. On the other hand, an \( R^2 \) value of 1 means the model explains all the variability of the response data around its mean.

In practical terms:
  • Higher \( R^2 \) values indicate a better fit of the model to the observed data.
  • It shows how well the relation between the variables is captured by the model.
  • It helps in understanding how much of the variation in colony density is due to changes in rock surface area.
For this exercise, calculating \( R^2 \) involves squaring the correlation coefficient obtained from our regression analysis. The closer \( R^2 \) is to 1, the better the model predicts the data.
Residual Analysis
Residual analysis is an important part of assessing linear regression models. A residual is the difference between an observed value \( y \) and the predicted value \( \hat{y} \) from the regression model. By breaking down these residuals, we gain insight into how well our model fits the data.

Residuals are calculated as:
  • For each data point: \( e = y - \hat{y} \)
  • Where \( y \) is the actual observed value and \( \hat{y} \) is the predicted value.
Analyzing residuals can reveal:
  • If the residuals are randomly dispersed around zero, it suggests a good fit.
  • Patterns or systematic structures in residuals, indicating potential issues with the model.
  • Outliers or unusual observations that may affect the model's accuracy.
In our exercise, comparing the residuals for predictions at surface areas 70 and 71 helps validate the accuracy and reliability of the regression model in those specific cases. A smaller residual indicates a more accurate prediction by the model.
Extreme Values Impact
Extreme values, or outliers, are observations that lie far from the other entries in the dataset. They can significantly impact the results of a linear regression analysis.

In our exercise, the observation with \( y = 1929 \) is considered extreme, given its stark contrast with other \( y \)-values. Here's how extreme values can affect linear regression outcomes:
  • They can skew the slope \( \beta_1 \) and intercept \( \beta_0 \) of the regression line, leading to a model that doesn't accurately reflect the majority of the data points.
  • \( R^2 \), the coefficient of determination, could be misleadingly increased or decreased.
  • Standard error \( s \) might show more variability, suggesting less precision in the regression estimates.
To assess the impact, removing such an extreme point and recalculating the regression line might provide a clearer picture of the "true" relationship. By noting changes in \( \beta_0 \), \( \beta_1 \), \( R^2 \), and \( s \), one can determine the influence of extreme values on the overall fit and predictions of the regression model.

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Most popular questions from this chapter

The efficiency ratio for a steel specimen immersed in a phosphating tank is the weight of the phosphate coating divided by the metal loss (both in \(\mathrm{mg} / \mathrm{ft}^{2}\) ). The article "Statistical Process Control of a Phosphate Coating Line" (Wire J. Intl., May 1997: 78-81) gave the accompanying data on tank temperature \((x)\) and efficiency ratio \((y)\). $$ \begin{array}{l|rrrrrrr} \text { Temp. } & 170 & 172 & 173 & 174 & 174 & 175 & 176 \\ \hline \text { Ratio } & .84 & 1.31 & 1.42 & 1.03 & 1.07 & 1.08 & 1.04 \\ \text { Temp. } & 177 & 180 & 180 & 180 & 180 & 180 & 181 \\ \hline \text { Ratio } & 1.80 & 1.45 & 1.60 & 1.61 & 2.13 & 2.15 & .84 \\ \text { Temp. } & 181 & 182 & 182 & 182 & 182 & 184 & 184 \\ \hline \text { Ratio } & 1.43 & .90 & 1.81 & 1.94 & 2.68 & 1.49 & 2.52 \\ \text { Temp. } & 185 & 186 & 188 & & & & \\ \hline \text { Ratio } & 3.00 & 1.87 & 3.08 & & & & \end{array} $$ a. Construct stem-and-leaf displays of both temperature and efficiency ratio, and comment on interesting features. b. Is the value of efficiency ratio completely and uniquely determined by tank temperature? Explain your reasoning. c. Construct a scatterplot of the data. Does it appear that efficiency ratio could be very well predicted by the value of temperature? Explain your reasoning.

A sample of \(n=500(x, y)\) pairs was collected and a test of \(H_{0}: \rho=0\) versus \(H_{\mathrm{a}}: \rho \neq 0\) was carried out. The resulting \(P\)-value was computed to be \(.00032\). a. What conclusion would be appropriate at level of significance .001? b. Does this small \(P\)-value indicate that there is a very strong linear relationship between \(x\) and \(y\) (a value of \(\rho\) that differs considerably from 0 )? Explain. c. Now suppose a sample of \(n=10,000(x, y)\) pairs resulted in \(r=.022\). Test \(H_{0}: \rho=0\) versus \(H_{\mathrm{a}}: \rho \neq 0\) at level .05. Is the result statistically significant? Comment on the practical significance of your analysis.

Magnetic resonance imaging (MRI) is well established as a tool for measuring blood velocities and volume flows. The article "Correlation Analysis of Stenotic Aortic Valve Flow Patterns Using Phase Contrast MRI,"鈥 referenced in Exercise 1.67, proposed using this methodology for determination of valve area in patients with aortic stenosis. The accompanying data on peak velocity \((\mathrm{m} / \mathrm{s})\) from scans of 23 patients in two different planes was read from a graph in the cited paper. $$ \begin{aligned} &\begin{array}{l|rrrrrrrr} \text { Level- } & .60 & .82 & .85 & .89 & .95 & 1.01 & 1.01 & 1.05 \\ \hline \text { Level- } & .50 & .68 & .76 & .64 & .68 & .86 & .79 & 1.03 \end{array}\\\ &\begin{array}{c|rrrrrrrr} \text { Level- } & 1.08 & 1.11 & 1.18 & 1.17 & 1.22 & 1.29 & 1.28 & 1.32 \\ \hline \text { Level-- } & .75 & .90 & .79 & .86 & .99 & .80 & 1.10 & 1.15 \end{array}\\\ &\begin{array}{l|lllllll} \text { Level- } & 1.37 & 1.53 & 1.55 & 1.85 & 1.93 & 1.93 & 2.14 \\ \hline \text { Level-- } & 1.04 & 1.16 & 1.28 & 1.39 & 1.57 & 1.39 & 1.32 \end{array} \end{aligned} $$ a. Does there appear to be a difference between true average velocity in the two different planes? Carry out an appropriate test of hypotheses (as did the authors of the article). b. The authors of the article also regressed level-velocity against level- velocity. The resulting estimated intercept and slope are .14701 and .65393, with corresponding estimated standard errors \(.07877\) and \(.05947\), coefficient of determination .852, and \(s=.110673\). The article included a comment that this regression showed evidence of a strong linear relationship but a regression slope well below 1. Do you agree?

The article "Quantitative Estimation of Clay Mineralogy in Fine-Grained Soils" (J. of Geotechnical and Geoenvironmental Engr., 2011: 997-1008) reported on various chemical properties of natural and artificial soils. Here are observations on \(x=\) cation exchange capacity (CEC, in meq/100 g) and \(y=\) specific surface area (SSA, in \(\mathrm{m}^{2} / \mathrm{g}\) ) of 20 natural soils. $$ \begin{array}{c|cccccccccc} x & 66 & 121 & 134 & 101 & 77 & 89 & 63 & 57 & 117 & 118 \\ \hline y & 175 & 324 & 460 & 288 & 205 & 210 & 295 & 161 & 314 & 265 \\ x & 76 & 125 & 75 & 71 & 133 & 104 & 76 & 96 & 58 & 109 \\ \hline y & 236 & 355 & 240 & 133 & 431 & 306 & 132 & 269 & 158 & 303 \end{array} $$ Minitab gave the following output in response to a request for \(r\) : Normal probability plots of \(x\) and \(y\) are quite straight. a. Carry out a test of hypotheses to see if there is a positive linear association in the population from which the sample data was selected. b. With \(n=20\), how small would the value of \(r\) have to be in order for the null hypothesis in the test of (a) to not be rejected at significance level .01? c. Calculate a confidence interval for \(\rho\) using a \(95 \%\) confidence level.

Bivariate data often arises from the use of two different techniques to measure the same quantity. As an example, the accompanying observations on \(x=\) hydrogen concentration (ppm) using a gas chromatography method and \(y=\) concentration using a new sensor method were read from a graph in the article "'A New Method to Measure the Diffusible Hydrogen Content in Steel Weldments Using a Polymer Electrolyte-Based Hydrogen Sensor" (Welding Res., July 1997: \(251 \mathrm{~s}-256 \mathrm{~s})\). $$ \begin{array}{c|cccccccccc} x & 47 & 62 & 65 & 70 & 70 & 78 & 95 & 100 & 114 & 118 \\ \hline y & 38 & 62 & 53 & 67 & 84 & 79 & 93 & 106 & 117 & 116 \\ x & 124 & 127 & 140 & 140 & 140 & 150 & 152 & 164 & 198 & 221 \\ \hline y & 127 & 114 & 134 & 139 & 142 & 170 & 149 & 154 & 200 & 215 \end{array} $$ Construct a scatterplot. Does there appear to be a very strong relationship between the two types of concentration measurements? Do the two methods appear to be measuring roughly the same quantity? Explain your reasoning.

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