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Chapter 12: Problem 19

Athletes and others suffering the same type of knee injury often require anterior and posterior ligament reconstruction. In order to determine the proper length of bone grafts, experiments were done using three imaging techniques, and these results were compared to the actual length required. A summary of the results of a simple linear regression analysis for each of these three methods is given in the following table. \({ }^{15}\) $$ \begin{array}{llrlc} \multicolumn{4}{c} {\text { Coefficient of }} \\ \text { Imaging Technique } & \text { Determination, } r^{2} & \text { Intercept } & \text { Slope } & p \text { -value } \\ \hline \text { Radiographs } & 0.80 & -3.75 & 1.031 & <0.0001 \\ \text { Standard MRI } & 0.43 & 20.29 & 0.497 & 0.011 \\ \text { 3-Dimensional MRI } & 0.65 & 1.80 & 0.977 & <0.0001 \end{array} $$ a. What can you say about the significance of each of the three regression analyses? b. How would you rank the effectiveness of the three regression analyses? What is the basis of your decision? c. How do the values of \(r^{2}\) and the \(p\) -values compare in determining the best predictor of actual graft lengths of ligament required?

Short Answer

Expert verified
Answer: The Radiographs imaging technique is the best predictor for the actual graft lengths, as it has the highest coefficient of determination (\(r^2\) = 0.80) among the three techniques. A higher \(r^2\) value indicates a better fit of the regression model, which means it can explain a larger proportion of the variation in the actual graft lengths.

Step by step solution

01

Understanding the Significance of Each Regression Analysis

To understand the significance of each regression analysis, we look at the p-value. The p-value helps us determine whether we should reject or accept the null hypothesis. A small p-value (typically 鈮� 0.05) indicates strong evidence against the null hypothesis, so you would reject the null hypothesis. In our case, all three imaging techniques have p-values below 0.05, which means that the regression analyses are significant. Radiographs: p-value < 0.0001 Standard MRI: p-value = 0.011 3-Dimensional MRI: p-value < 0.0001
02

Ranking the Effectiveness of the Regression Analyses

We can rank the effectiveness of the three imaging techniques by comparing their coefficients of determination, \(r^2\). A higher \(r^2\) value means a better fit of the regression model, i.e., it explains a larger proportion of the variation in the data. Here are the \(r^2\) values: Radiographs: \(r^2\) = 0.80 Standard MRI: \(r^2\) = 0.43 3-Dimensional MRI: \(r^2\) = 0.65 Based on these values, we can rank the effectiveness of the three regression analyses as follows: 1. Radiographs (\(r^2\) = 0.80) 2. 3-Dimensional MRI (\(r^2\) = 0.65) 3. Standard MRI (\(r^2\) = 0.43) The basis of this decision is that Radiographs has the highest coefficient of determination, which indicates that it provides the best fit for the regression model among the three imaging techniques.
03

Comparing \(r^2\) and p-values for Predicting Actual Graft Lengths

The values of \(r^2\) and p-values both provide information about the regression model's significance and fit: 1. \(r^2\): This value represents the proportion of the variation in the dependent variable that can be explained by the regression model. A higher \(r^2\) indicates a better fit and allows us to predict the dependent variable more accurately. 2. p-value: This value helps us determine the significance of the regression model by testing the null hypothesis that the relationship between the dependent and independent variables is purely due to chance. A low p-value (鈮� 0.05) indicates strong evidence against the null hypothesis and suggests that the regression model is significant. Comparing the two: - If we are interested in finding the best predictor for actual graft lengths, we should focus more on the \(r^2\) value, as it directly represents the proportion of the variation in the actual graft lengths that can be explained by the regression model. - The p-value, on the other hand, mainly helps us determine whether the regression model is significant or not. Though all the imaging techniques have p-values below 0.05, we should rank their effectiveness based on their \(r^2\) values to determine the best predictor of actual graft lengths.

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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, denoted as \( r^2 \), is a crucial metric in regression analysis. It tells us how well the regression model explains the variability of the dependent variable. In simple terms, it shows how well the data fits the model.

  • Values range from 0 to 1. A value closer to 1 means a better fit.
  • If \( r^2 = 0.80 \), it means 80% of the variation in the dependent variable is explained by the model.
  • Think of \( r^2 \) as the percentage of variance captured by your model.
In the context of knee ligament graft length prediction, a higher \( r^2 \) implies that the imaging technique used allows for more accurate predictions based on the available data. This makes \( r^2 \) an excellent tool for comparing different methods to see which one provides better predictive power for graft lengths.

For example, the Radiographs technique has an \( r^2 \) of 0.80, indicating it provides the best fit among the three techniques analyzed. This suggests that Radiographs is most effective at predicting actual graft lengths compared to both Standard and 3-Dimensional MRI.
p-value
The p-value is an essential concept in statistical testing. It helps you decide whether to reject the null hypothesis, which essentially states that there is no effect or no relationship in the population.

  • A p-value less than or equal to 0.05 typically indicates strong evidence against the null hypothesis.
  • It tells you how likely it is to obtain the observed results if the null hypothesis were true.
  • Lower p-values suggest that the associations or effects observed are statistically significant.
When evaluating regression analyses for predicting knee ligament graft lengths, a low p-value for each imaging technique (Radiographs, Standard MRI, and 3-Dimensional MRI) signifies that the regression results are statistically significant.

For instance, both Radiographs and 3-Dimensional MRI have p-values of less than 0.0001, which indicates very strong evidence that these models are significant. Even the Standard MRI, with a p-value of 0.011, confirms that its results aren't due to random chance, thereby making its findings significant as well.
Linear Regression
Linear regression is a straightforward yet powerful statistical method. It models the relationship between two variables by fitting a linear equation to observed data.

  • The linear equation is of the form \( y = mx + c \).
  • Here, \( y \) is the dependent variable, \( x \) is the independent variable, \( m \) is the slope, and \( c \) is the intercept.
  • The slope \( m \) indicates how much \( y \) changes for a unit change in \( x \).
In the case of predicting knee ligament graft lengths, the aim of using linear regression is to find how accurately we can predict the required graft length using different imaging techniques.
Radiographs show a slope of 1.031, which suggests a nearly one-to-one relationship in predicting actual graft lengths. By using linear regression, medical professionals can understand how changes in graft length measurements from different imaging techniques might predict the actual lengths needed, helping in effective decision-making and planning for surgeries.

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Most popular questions from this chapter

You are given these data: $$ \begin{array}{l|rrrrrrr} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 2 & 2 & 3 & 4 & 4 \end{array} $$ a. Plot the data points. Based on your graph, what will be the sign of the sample correlation coefficient? b. Calculate \(r\) and \(r^{2}\) and interpret their values.

A marketing research \end{tabular}experiment was conducted to study the relationship between the length of time necessary for a buyer to reach a decision and the number of alternative package designs of a product presented. Brand names were eliminated from the packages and the buyers made their selections using the manufacturer's product descriptions on the packages as the only buying guide. The length of time necessary to reach a decision was recorded for 15 participants in the marketing research study. $$ \begin{array}{c|c|c|c} \text { Length of Decision } & & & \\ \text { Time, } y \text { (sec) } & 5,8,8,7,9 & 7,9,8,9,10 & 10,11,10,12,9 \\ \hline \text { Number of } & & & & \\ \text { Alternatives, } x & 2 & & 3 & & 4 \end{array} $$ a. Find the least-squares line appropriate for these data. b. Plot the points and graph the line as a check on your calculations. c. Calculate \(s^{2}\). d. Do the data present sufficient evidence to indicate that the length of decision time is linearly related to the number of alternative package designs? (Test at the \(\alpha=.05\) level of significance.) e. Find the approximate \(p\) -value for the test and interpret its value. f. If they are available, examine the diagnostic plots to check the validity of the regression assumptions. g. Estimate the average length of time necessary to reach a decision when three alternatives are presented, using a \(95 \%\) confidence interval.

Graph the line corresponding to the equation \(y=2 x+1\) by graphing the points corresponding to \(x=0,1,\) and 2 . Give the \(y\) -intercept and slope for the line.

An experiment was conducted to determine the effect of various levels of phosphorus on the inorganic phosphorus levels in a particular plant. The data in the table represent the levels of inorganic phosphorus in micromoles ( \(\mu\) mol) per gram dry weight of Sudan grass roots grown in the greenhouse for 28 days, in the absence of zinc. Use the MINI TAB output to answer the questions. $$ \begin{array}{l} \text { Phosphorus Applied, } x \text { Phosphorus in Plant, } y \\ \hline .50 \mu \mathrm{mol} & 204 \\ & 195 \\ & 247 \\ & 245 \\ & \\ .25 \mu \mathrm{mol} & 159 \\ & 127 \\ & 95 \\ & 144 \\ .10 \mu \mathrm{mol} & 128 \\ & 192 \\ & 84 \\ & 71 \end{array} $$ a. Plot the data. Do the data appear to exhibit a linear relationship? b. Find the least-squares line relating the plant phosphorus levels \(y\) to the amount of phosphorus applied to the soil \(x\). Graph the least-squares line as a check on your answer. c. Do the data provide sufficient evidence to indicate that the amount of phosphorus present in the plant is linearly related to the amount of phosphorus applied to the soil? d. Estimate the mean amount of phosphorus in the plant if \(.20 \mu \mathrm{mol}\) of phosphorus is applied to the soil, in the absence of zinc. Use a \(90 \%\) confidence interval.

What diagnostic plot can you use to determine whether the incorrect model has been used? What should the plot look like if the correct model has been used?
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