/*! 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 67 The article "Validation of the R... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article "Validation of the Rockport Fitness Walking Test in College Males and Females" (Research Ouarterly 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 \((\) lb) , \(x_{3}=1\)-mile walk time (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 \(\hat{\beta}_{1}=.6566 ?\) c. Suppose that an observation made on a male whose weight was \(170 \mathrm{lb}\), walk time was \(11 \mathrm{~min}\), and heart rate was 140 beats/min resulted in \(\mathrm{VO}_{2} \max =3.15\). What would you have predicted for \(\mathrm{VO}_{2} \max\) in this situation, and what is the value of the corresponding residual? d. Using \(\mathrm{SSE}=30.1033\) and \(\mathrm{SST}=102.3922\), what proportion of observed variation in \(\mathrm{VO}_{2} \max\) can be attributed to the model relationship? e. Assuming a sample size of \(n=20\), carry out a test of hypotheses to decide whether the chosen model specifies a useful relationship between \(\mathrm{VO}_{2} \max\) and at least one of the predictors.

Short Answer

Expert verified
a) \( \hat{\beta}_3 \) shows \( VO_2 \max \) decreases by 0.0996 L/min per minute of walk time. b) \( \hat{\beta}_1 \) indicates males have 0.6566 L/min higher \( VO_2 \max \) than females. c) Predicted \( VO_2 \max \) is 3.6689 L/min; residual is -0.5189. d) Model explains 70.58% of variance. e) Model is significant as F-statistic shows a useful relationship.

Step by step solution

01

Interpret Coefficient \( \hat{\beta}_3 = -0.0996 \)

The coefficient \( \hat{\beta}_3 = -0.0996 \) represents the change in the dependent variable \( y \), which is \( VO_2 \max \), for each one-unit increase in \( x_3 \), the 1-mile walk time in minutes, holding all other predictors constant. Therefore, for each additional minute that it takes to complete the 1-mile walk, the \( VO_2 \max \) decreases by approximately 0.0996 L/min.
02

Interpret Coefficient \( \hat{\beta}_1 = 0.6566 \)

The coefficient \( \hat{\beta}_1 = 0.6566 \) suggests that, on average, being male increases the \( VO_2 \max \) by 0.6566 L/min compared to being female, assuming all other variables held constant. The gender variable is binary with 0 for female and 1 for male.
03

Predict \( VO_2 \max \) for Given Conditions

To find the predicted \( VO_2 \max \), substitute the given values into the regression equation. For a male, \( x_1 = 1 \), with weight \( x_2 = 170 \) lb, walk time \( x_3 = 11 \) minutes, and heart rate \( x_4 = 140 \) beats/min, the equation is:\[ y = 3.5959 + 0.6566(1) + 0.0096(170) - 0.0996(11) - 0.0080(140) \].Calculate to find the predicted \( VO_2 \max \).
04

Calculate the Predicted \( VO_2 \max \)

Substituting into the equation:\[ y = 3.5959 + 0.6566 \times 1 + 0.0096 \times 170 - 0.0996 \times 11 - 0.0080 \times 140 \].Perform the arithmetic:\[ y = 3.5959 + 0.6566 + 1.632 - 1.0956 - 1.12 \].\[ y \approx 3.6689 \].Thus, the predicted \( VO_2 \max \) is 3.6689 L/min.
05

Calculate Residual

The residual is the difference between the observed \( VO_2 \max \) and the predicted \( VO_2 \max \). Given the observed value is 3.15 L/min, the residual is:\[ \text{Residual} = 3.15 - 3.6689 = -0.5189 \].
06

Calculate Proportion of Variation

The proportion of variation that can be attributed to the model is given by the fraction of the total sum of squares due to the regression.\[ R^2 = 1 - \left(\frac{\text{SSE}}{\text{SST}}\right) = 1 - \left(\frac{30.1033}{102.3922}\right) \].\[ R^2 \approx 0.7058 \].Therefore, approximately 70.58% of the variation in \( VO_2 \max \) can be explained by the model.
07

Conduct Hypothesis Test

Test whether at least one coefficient is different from zero. Use the F-test for the overall significance of the model:The null hypothesis \( H_0 \) states that there is no relationship between \( VO_2 \max \) and the predictors, meaning all coefficients are zero except the intercept.The test statistic is:\[ F = \frac{(\text{SST} - \text{SSE}) / p}{\text{SSE} / (n-p-1)} \],where \( p \) is the number of predictors (4 in this case).\[ F = \frac{(102.3922 - 30.1033) / 4}{30.1033 / (20 - 4 - 1)} \].\[ F \approx 9.56 \].Comparing to the critical F-value at \( \alpha = 0.05 \), the value is significant, indicating at least one predictor has a useful relationship with \( VO_2 \max \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Interpretation of Coefficients
In regression analysis, coefficients represent the estimated change in the dependent variable for a one-unit change in the predictor, while other variables are held constant. Each coefficient in the equation offers specific insights:
  • The coefficient \( \hat{\beta}_3 = -0.0996 \) shows that for a one-minute increase in 1-mile walk time \( x_3 \), \( VO_2 \) max decreases by approximately 0.0996 L/min. This indicates that a longer time to complete the walk is associated with lower cardiorespiratory fitness.
  • Similarly, \( \hat{\beta}_1 = 0.6566 \) implies that males have, on average, a greater \( VO_2 \) max by 0.6566 L/min compared to females, making gender a significant predictor of fitness.
Understanding these coefficients helps interpret how each predictor influences the outcome and assess their individual importance in the model.
VO2 max
\( VO_2 \) max is a measure of the maximum oxygen uptake during intense exercise and is a key indicator of cardiovascular fitness. It helps gauge an individual's aerobic endurance and overall health.
  • It is typically measured in liters per minute (L/min) and reflects how efficiently the heart and lungs can supply oxygen to the muscles during physical activity.
  • High \( VO_2 \) max values are often associated with better endurance and physical performance in activities like running, biking, and skiing.
In this analysis, \( VO_2 \) max is the dependent variable, meaning the study aims to predict or explain changes in \( VO_2 \) max based on predictors such as gender, weight, walk time, and heart rate.
Hypothesis Testing
Hypothesis testing in regression analysis helps determine if the model or any of its predictors provide meaningful insights into the data.
  • The null hypothesis \( H_0 \) posits that there is no relationship between the response variable \( VO_2 \) max and any of the predictors—meaning all regression coefficients equal zero.
  • An alternative hypothesis suggests that at least one predictor coefficient differs from zero, indicating its significant impact.
In this exercise, an F-test examines whether the model as a whole has any predictive capability:
- The statistic \( F \) is calculated using the formula: \[ F = \frac{(\text{SST} - \text{SSE}) / p}{\text{SSE} / (n-p-1)} \], where \( p \) is the number of predictors.
- A high \( F \) value suggests some predictors are indeed meaningful, and thus, the model is considered useful when \( F \) exceeds the critical value for a given significance level (often \( \alpha = 0.05 \)).
Proportion of Variation
The proportion of variation, often expressed as \( R^2 \), measures how well the regression model explains the variability of the response variable. It is crucial for evaluating model fit.
  • \( R^2 \) is calculated as: \[ R^2 = 1 - \left(\frac{\text{SSE}}{\text{SST}}\right) \], where SSE is the sum of squared errors, and SST is the total sum of squares.
  • A higher \( R^2 \) indicates that a greater proportion of the variability in the response variable can be explained by the predictors, which suggests a better-fitting model.
In this analysis, \( R^2 \approx 0.7058 \) implies that approximately 70.58% of the variation in \( VO_2 \) max can be accounted for by the variables included in the model. This indicates a strong relationship between the predictors and the fitness measure.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Curing concrete is known to be vulnerable to shock vibrations, which may cause cracking or hidden damage to the material. As part of a study of vibration phenomena, the paper "Shock Vibration Test of Concrete" (ACl Materials J., 2002: 361-370) reported the accompanying data on peak particle velocity ( \(\mathrm{mm} / \mathrm{sec}\) ) and ratio of ultrasonic pulse velocity after impact to that before impact in concrete prisms. Transverse cracks appeared in the last 12 prisms, whereas there was no observed cracking in the first 18 prisms. a. Construct a comparative boxplot of ppv for the cracked and uncracked prisms and comment. Then estimate the difference between true average ppv for cracked and uncracked prisms in a way that conveys information about precision and reliability. b. The investigators fit the simple linear regression model to the entire data set consisting of 30 observations, with ppv as the independent variable and ratio as the dependent variable. Use a statistical software package to fit several different regression models, and draw appropriate inferences.

A sample of \(n=20\) companies was selected, and the values of \(y=\) stock price and \(k=15\) variables (such as quarterly dividend, previous year's earnings, and debt ratio) were determined. When the multiple regression model using these 15 predictors was fit to the data, \(R^{2}=.90\) resulted. a. Does the model appear to specify a useful relationship between \(y\) and the predictor variables? Carry out a test using significance level .05. [Hint: The \(F\) critical value for 15 numerator and 4 denominator df is 5.86.] b. Based on the result of part (a), does a high \(R^{2}\) value by itself imply that a model is useful? Under what circumstances might you be suspicious of a model with a high \(R^{2}\) value? c. With \(n\) and \(k\) as given previously, how large would \(R^{2}\) have to be for the model to be judged useful at the .05 level of significance?

A plot in the article "Thermal Conductivity of Polyethylene: The Effects of Crystal Size, Density, and Orientation on the Thermal Conductivity" (Polymer Engr: and Science, 1972: 204-208) suggests that the expected value of thermal conductivity \(y\) is a linear function of \(10^{4} \cdot 1 / x\), where \(x\) is lamellar thickness. \begin{tabular}{l|rrrrrrrr} \(x\) & 240 & 410 & 460 & 490 & 520 & 590 & 745 & 8300 \\ \hline\(y\) & \(12.0\) & \(14.7\) & \(14.7\) & \(15.2\) & \(15.2\) & \(15.6\) & \(16.0\) & \(18.1\) \end{tabular} a. Estimate the parameters of the regression function and the regression function itself. b. Predict the value of thermal conductivity when lamellar thickness is \(500 \AA \AA\).

Feature recognition from surface models of complicated parts is becoming increasingly important in the development of efficient computer-aided design (CAD) systems. The article "A Computationally Efficient Approach to Feature Abstraction in Design-Manufacturing Integration" (J. of Engr: for Industry, 1995: 16-27) contained a graph of logadtotal recognition time), with time in sec, versus \(\log _{10}\) (number of edges of a part), from which the following representative values were read: \(\begin{array}{lrrrrrr}\text { Log(edges) } & 1.1 & 1.5 & 1.7 & 1.9 & 2.0 & 2.1 \\ \text { Log(time) } & .30 & .50 & .55 & .52 & .85 & .98 \\ \text { Log(edges) } & 2.2 & 2.3 & 2.7 & 2.8 & 3.0 & 3.3 \\ \text { Log(time) } & 1.10 & 1.00 & 1.18 & 1.45 & 1.65 & 1.84 \\ \text { Log(edges) } & 3.5 & 3.8 & 4.2 & 4.3 & & \\ \text { Log(time) } & 2.05 & 2.46 & 2.50 & 2.76 & & \end{array}\) a. Does a scatter plot of \(\log (\) time \()\) versus \(\log (\) edges) suggest an approximate linear relationship between these two variables? b. What probabilistic model for relating \(y=\) recognition time to \(x=\) number of edges is implied by the simple linear regression relationship between the transformed variables? c. Summary quantities calculated from the data are $$ \begin{aligned} &n=16 \quad \Sigma x_{i}^{\prime}=42.4 \quad \Sigma y_{i}^{\prime}=21.69 \\ &\Sigma\left(x_{i}^{\prime}\right)^{2}=126.34 \quad \Sigma\left(y_{i}^{\prime}\right)^{2}=38.5305 \\ &\Sigma x_{i}^{\prime} y_{i}^{\prime}=68.640 \end{aligned} $$ Calculate estimates of the parameters for the model in part (b), and then obtain a point prediction of time when the number of edges is 300 .

The viscosity \((y)\) of an oil was measured by a cone and plate viscometer at six different cone speeds \((x)\). It was assumed that a quadratic regression model was appropriate, and the estimated regression function resulting from the \(n=6\) observations was $$ y=-113.0937+3.3684 x-.01780 x^{2} $$ a. Estimate \(\mu_{\gamma .75}\), the expected viscosity when speed is \(75 \mathrm{rpm} .\) b. What viscosity would you predict for a cone speed of \(60 \mathrm{rpm}\) ? c. If \(\sum y_{i}^{2}=8386.43, \Sigma y_{j}=210.70, \Sigma x_{i} y_{i}=17,002.00\), and \(\sum x_{1}^{2} y_{i}=1,419,780\), compute \(\mathrm{SSE}\left[=\sum y_{i}^{2}-\right.\) \(\left.\hat{\beta}_{0} \Sigma y_{i}-\hat{\beta}_{1} \Sigma x_{i} y_{s}-\hat{\beta}_{2} \Sigma x_{i}^{2} y_{i}\right]\) and \(s\). d. From part (c), SST \(=8386.43-(210.70)^{2} / 6=987.35\). Using SSE computed in part (c), what is the computed value of \(R^{2} ?\) e. If the estimated standard deviation of \(\hat{\beta}_{2}\) is \(s_{\dot{\beta}}=.00226\), test \(H_{0}: \beta_{2}=0\) versus \(H_{\mathrm{a}}: \beta_{2} \neq 0\) at level 01 , and interpret the result.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.