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Chapter 14: Problem 20

This exercise requires the use of a computer package. The paper "Habitat Selection by Black Bears in an Intensively Logged Boreal Forrest" (Canadian Journal of Zoology [2008]: \(1307-1316\) ) gave the accompanying data on \(n=11\) female black bears. $$ \begin{array}{ccc} \begin{array}{c} \text { Age } \\ \text { (years) } \end{array} & \begin{array}{c} \text { Weight } \\ (\mathrm{kg}) \end{array} & \begin{array}{c} \text { Home-Range } \\ \text { Size }\left(\mathrm{km}^{2}\right) \end{array} \\ \hline 10.5 & 54 & 43.1 \\ 6.5 & 40 & 46.6 \\ 28.5 & 62 & 57.4 \\ 6.5 & 55 & 35.6 \\ 7.5 & 56 & 62.1 \\ 6.5 & 62 & 33.9 \\ 5.5 & 42 & 39.6 \\ 7.5 & 40 & 32.2 \\ 11.5 & 59 & 57.2 \\ 9.5 & 51 & 24.4 \\ 5.5 & 50 & 68.7 \\ \hline \end{array} $$ a. Fit a multiple regression model to describe the relationship between \(y=\) home-range size and the predictors \(x_{1}=\) age and \(x_{2}=\) weight. b. Construct a normal probability plot of the 11 standardized residuals. Based on the plot, does it seem reasonable to regard the random deviation distribution as approximately normal? Explain. c. If appropriate, carry out a model utility test with a significance level of \(.05\) to determine if the predictors age and weight are useful for predicting homerange size.

Short Answer

Expert verified
The steps for solving this problem involve fitting a multiple regression model, examining residuals through a normal probability plot, and if suitable, performing a model utility test. The final results will vary depending on the data analysis software and the data used. The interpretation of the results will give insight into whether age and weight are useful predictors for home-range size of the bears

Step by step solution

01

Fit a multiple regression model.

To describe the relationship between home-range size (dependent variable) and predictors age and weight (independent variables), fit a multiple regression model using a statistical software package. These software typically require inputting your data and specifying your dependent and independent variables, then they provide the regression equation which would have the format like \(y = a + b_{1}x_{1} + b_{2}x_{2} + e\), where \(y\) is the home-range size, \(x_1\) and \(x_2\) represent age and weight respectively, \(a\) is the y-intercept, \(b_1\) and \(b_2\) are the regression coefficients and \(e\) is the error term.
02

Construct a normal probability plot of the standardized residuals.

Use the same software package to generate a normal probability plot of the residuals. Standardized residuals are the deviation between observed and predicted home-range sizes, normalized by the standard deviation. These residuals should follow a normal distribution if the assumption of homoscedasticity holds true. This step is crucial for verifying the appropriateness of the regression model.
03

Interpret the normal probability plot.

Examine the normal probability plot. If the points lie along (or roughly along) a straight, 45掳 line, this suggests that the residuals are normally distributed and thus the assumption of normality holds. If the points significantly deviate from this line, the distribution of residuals might not be normal.
04

Perform a model utility test.

If the normality assumption is plausible, perform a model utility test (ANOVA F test) to determine whether the predictor variables age and weight are useful for predicting home range size. This test checks if the regression model provides a significantly better fit than a model with no predictors. The null hypothesis states that all the regression slope coefficients are equal to zero (meaning, none of the predictors contribute to model significantly). If the p-value obtained from this test is less than the significance level (0.05), reject the null hypothesis, and conclude that at least one predictor is useful for predicting home range size.

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Key Concepts

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

Understanding Standardized Residuals
In multiple regression analysis, understanding residuals is crucial because they tell us about the fit of the model. A residual is the difference between an observed value and the value predicted by the regression model. However, raw residuals do not offer a complete picture since they can vary greatly in magnitude and can be influenced by the scale of the variables.

This is where standardized residuals come into play. Standardized residuals are raw residuals divided by their estimated standard deviation, essentially normalizing them. This makes it easier to identify outliers and understand if the variance is consistent across the range of predictions (homoscedasticity). A value of a standardized residual that falls outside the range of approximately -2 to 2 may be considered unusual and could indicate a point that does not fit the model well.

When you look at a solution involving standardized residuals, you should think of them as a diagnostic tool. They can be plotted to assess the normality assumption of the regression model's errors 鈥� a key to ensuring accurate conclusions from the analysis.
Deciphering the Normal Probability Plot
The normal probability plot is a graphical tool used to assess whether a set of data follows a normal distribution. When standardized residuals from a multiple regression model are plotted on this graph, the resulting points should form a straight line if the residuals are normally distributed.

This is vital as many statistical tests rely on the assumption of normality. When you perform this step in a software package, pay close attention to the shape of the plot. If the standardized residuals closely follow a straight line, this suggests that the original data do not grossly violate the normality assumption, and the use of regression analysis is appropriate.

If the points deviate systematically from the line, this may suggest a potential problem with non-normality. In that case, it might require more in-depth statistical techniques like transformation of the data or non-parametric tests. Understanding how to interpret this plot is important for verifying the validity of your regression analysis.
Performing a Model Utility Test
The model utility test, often conducted via an Analysis of Variance (ANOVA) F-test in regression analysis, serves to determine whether your model provides a significant improvement over a baseline model with no predictors. It tests the hypothesis that all regression coefficients are zero, which would mean that none of the predictors are contributing to explaining the variability in the dependent variable.

A p-value is derived from this test, and if it's lower than the significance level (commonly set at 0.05), you reject the null hypothesis. This indicates that at least one of the predictors has a significant relation with the dependent variable, affirming the 'utility' or 'usefulness' of your model.

In the context of the exercise, if the p-value is below 0.05 after you run the model utility test, you would conclude that factors like the age and weight of the bears are indeed significant in predicting the size of their home ranges. This step not only aids in confirming the value of your predictors but also in adding scientific credibility to your regression model.

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Most popular questions from this chapter

The following statement appeared in the article "Dimensions of Adjustment Among College Women" (Journal of College Student Development [19981: 364): Regression analyses indicated that academic adjustment and race made independent contributions to academic achievement, as measured by current GPA. Suppose $$ \begin{aligned} y &=\text { current GPA } \\ x_{1} &=\text { academic adjustment score } \\ x_{2} &=\text { race (with white }=0, \text { other }=1) \end{aligned} $$ What multiple regression model is suggested by the statement? Did you include an interaction term in the model? Why or why not?

According to "Assessing the Validity of the Post-Materialism Index" (American Political Science Review [1999]: \(649-664\) ), one may be able to predict an individual's level of support for ecology based on demographic and ideological characteristics. The multiple regression model proposed by the authors was $$ \begin{aligned} y=& 3.60-.01 x_{1}+.01 x_{2}-.07 x_{3}+.12 x_{6}+.02 x_{5} \\ &-.04 x_{6}-.01 x_{7}-.04 x_{8}-.02 x_{9}+e \end{aligned} $$ where the variables are defined as follows: \(y=\) ecology score (higher values indicate a greater concern for ecology) \(x_{1}=\) age times 10 \(x_{2}=\) income (in thousands of dollars) \(x_{3}=\) gender \((1=\) male, \(0=\) female \()\) \(x_{j}=\operatorname{race}(1=\) white, \(0=\) nonwhite \()\) \(x_{5}=\) education (in years) \(x_{6}=\) ideology \((4=\) conservative, \(3=\) right of center, \(2=\) middle of the road, \(1=\) left of \(\begin{aligned}\text { center, and } 0=\text { liberal }) \\ x_{7}=& \text { social class }(4=\text { upper, } 3=\text { upper middle, }\\\ 2=\text { middle, } 1=\text { lower middle, and } \\\0=\text { lower }) \end{aligned}\) \(x_{8}=\) postmaterialist ( 1 if postmaterialist, 0 otherwise) \(x_{9}=\) materialist ( 1 if materialist, 0 otherwise) a. Suppose you knew a person with the following characteristics: a 25 -year- old, white female with a college degree (16 years of education), who has a $$\$ 32,000$$ -peryear job, is from the upper middle class, and considers herself left of center, but who is neither a materialist nor a postmaterialist. Predict her ecology score. b. If the woman described in Part (a) were Hispanic rather than white, how would the prediction change? c. Given that the other variables are the same, what is the estimated mean difference in ecology score for men and women? d. How would you interpret the coefficient of \(x_{2}\) ? e. Comment on the numerical coding of the ideology and social class variables. Can you suggest a better way of incorporating these two variables into the model?

When coastal power stations take in large quantities of cooling water, it is inevitable that a number of fish are drawn in with the water. Various methods have been designed to screen out the fish. The article "Multiple Regression Analysis for Forecasting Critical Fish Influxes at Power Station Intakes" (Journal of Applied Ecology [1983]: 33-42) examined intake fish catch at an English power plant and several other variables thought to affect fish intake: $$ \begin{aligned} y &=\text { fish intake (number of fish) } \\ x_{1} &=\text { water temperature }\left({ }^{\circ} \mathrm{C}\right) \\ x_{2} &=\text { number of pumps running } \\ x_{3} &=\text { sea state }(\text { values } 0,1,2, \text { or } 3) \\ x_{4} &=\text { speed (knots) } \end{aligned} $$ Part of the data given in the article were used to obtain the estimated regression equation $$ \hat{y}=92-2.18 x_{1}-19.20 x_{2}-9.38 x_{3}+2.32 x_{4} $$ (based on \(n=26\) ). SSRegr \(=1486.9\) and SSResid \(=\) \(2230.2\) were also calculated. a. Interpret the values of \(b_{1}\) and \(b_{4}\). b. What proportion of observed variation in fish intake can be explained by the model relationship? c. Estimate the value of \(\sigma\). d. Calculate adjusted \(R^{2}\). How does it compare to \(R^{2}\) itself?

This exercise requires the use of a computer package. The authors of the article "Absolute Versus per Unit Body Length Speed of Prey as an Estimator of Vulnerability to Predation" (Animal Behaviour [1999]: 347 - 352) found that the speed of a prey (twips/s) and the length of a prey (twips \(\times 100\) ) are good predictors of the time (s) required to catch the prey. (A twip is a measure of distance used by programmers.) Data were collected in an experiment in which subjects were asked to "catch" an animal of prey moving across his or her computer screen by clicking on it with the mouse. The investigators varied the length of the prey and the speed with which the prey moved across the screen. The following data are consistent with summary values and a graph given in the article. Each value represents the average catch time over all subjects. The order of the various speed-length combinations was randomized for each subject. $$ \begin{array}{ccc} \text { Prey Length } & \text { Prey Speed } & \text { Catch Time } \\ \hline 7 & 20 & 1.10 \\ 6 & 20 & 1.20 \\ 5 & 20 & 1.23 \\ 4 & 20 & 1.40 \\ 3 & 20 & 1.50 \\ 3 & 40 & 1.40 \\ 4 & 40 & 1.36 \\ 6 & 40 & 1.30 \\ 7 & 40 & 1.28 \\ 7 & 80 & 1.40 \\ 6 & 60 & 1.38 \\ 5 & 80 & 1.40 \\ 7 & 100 & 1.43 \\ 6 & 100 & 1.43 \\ 7 & 120 & 1.70 \\ 5 & 80 & 1.50 \\ 3 & 80 & 1.40 \\ 6 & 100 & 1.50 \\ 3 & 120 & 1.90 \\ \hline \end{array} $$ a. Fit a multiple regression model for predicting catch time using prey length and speed as predictors. b. Predict the catch time for an animal of prey whose length is 6 and whose speed is 50 . c. Is the multiple regression model useful for predicting catch time? Test the relevant hypotheses using \(\alpha=.05\). d. The authors of the article suggest that a simple linear regression model with the single predictor \(x=\frac{\text { length }}{\text { speed }}\) might be a better model for predicting catch time. Calculate the \(x\) values and use them to fit this linear regression model. e. Which of the two models considered (the multiple regression model from Part (a) or the simple linear regression model from Part (d)) would you recommend for predicting catch time? Justify your choice.

The authors of the paper "Weight-Bearing Activity during Youth Is a More Important Factor for Peak Bone Mass than Calcium Intake" (Journal of Bone and Mineral Density [1994]: \(1089-1096\) ) used a multiple regression model to describe the relationship between $$ \begin{aligned} &\begin{aligned} y &=\text { bone mineral density }\left(\mathrm{g} / \mathrm{cm}^{3}\right) \\\ x_{1} &=\text { body weight }(\mathrm{kg}) \\ x_{2} &=\text { a measure of weight-bearing activity, with } \end{aligned}\\\ &\text { higher values indicating greater activity } \end{aligned} $$ a. The authors concluded that both body weight and weight-bearing activity were important predictors of bone mineral density and that there was no significant interaction between body weight and weightbearing activity. What multiple regression function is consistent with this description? b. The value of the coefficient of body weight in the multiple regression function given in the paper is \(0.587 .\) Interpret this value.
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