/*! 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 36 Cardiorespiratory fitness is wid... [FREE SOLUTION] | 91Ó°ÊÓ

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Cardiorespiratory fitness is widely recognized as a major component of overall physical well-being. Direct measurement of maximal oxygen uptake \(\left(\mathrm{VO}_{2} \max \right)\) is the single best measure of such fitness, but direct measurement is timeconsuming and expensive. It is therefore desirable to have a prediction equation for \(\mathrm{VO}_{2} \max\) in terms of easily obtained quantities. Consider the variables $$ \begin{aligned} y &=\mathrm{VO}_{2} \max (\mathrm{L} / \mathrm{min}) \quad x_{1}=\text { weight }(\mathrm{kg}) \\ x_{2} &=\text { age }(\mathrm{yr}) \\ x_{3} &=\text { time necessary to walk } 1 \text { mile }(\mathrm{min}) \\ x_{4} &=\text { heart rate at the end of the walk (beats/min) } \end{aligned} $$ Here is one possible model, for male students, consistent with the information given in the article "Validation of the Rockport Fitness Walking Test in College Males and Females" (Research Quarterly for Exercise and Sport, 1994: 152-158): $$ \begin{aligned} &Y=5.0+.01 x_{1}-.05 x_{2}-.13 x_{3}-.01 x_{4}+\epsilon \\ &\sigma=.4 \end{aligned} $$ a. Interpret \(\beta_{1}\) and \(\beta_{3}\). b. What is the expected value of \(\mathrm{VO}_{2} \max\) when weight is \(76 \mathrm{~kg}\), age is \(20 \mathrm{yr}\), walk time is \(12 \mathrm{~min}\), and heart rate is \(140 \mathrm{~b} / \mathrm{m}\) ? c. What is the probability that \(\mathrm{VO}_{2} \max\) will be between \(1.00\) and \(2.60\) for a single observation made when the values of the predictors are as stated in part (b)?

Short Answer

Expert verified
\(\beta_1\) predicts weight effect, \(\beta_3\) predicts walk time impact. Expected \(\mathrm{VO}_2 \max\) is 1.8 L/min; probability range 95.45%.

Step by step solution

01

Interpretation of Coefficient \(\beta_1\)

The coefficient \(\beta_1 = 0.01\) in the prediction equation indicates that for each additional kilogram of weight, the predicted \(\mathrm{VO}_2 \max\) increases by 0.01 L/min, assuming all other variables are held constant. This suggests that heavier individuals are predicted to have a slightly higher maximal oxygen uptake when other factors do not change.
02

Interpretation of Coefficient \(\beta_3\)

The coefficient \(\beta_3 = -0.13\) implies that for each additional minute taken to walk 1 mile, the predicted \(\mathrm{VO}_2 \max\) decreases by 0.13 L/min, assuming the other variables remain constant. This decrease reflects the impact of longer walking times on cardiorespiratory fitness assessments.
03

Calculate Expected \(\mathrm{VO}_2 \max\)

Using the prediction equation: \[\begin{align*}Y &= 5.0 + 0.01 \cdot 76 - 0.05 \cdot 20 - 0.13 \cdot 12 - 0.01 \cdot 140 \Y &= 5.0 + 0.76 - 1.0 - 1.56 - 1.4 \Y &= 1.8\end{align*}\]The expected value of \(\mathrm{VO}_2 \max\) is 1.8 L/min for the given values of the predictors.
04

Probability Calculation for \(\mathrm{VO}_2 \max\) Range

For part (c), we need to calculate the probability that \(\mathrm{VO}_2 \max\) lies between 1.00 and 2.60. Given the normal distribution assumption with \(\sigma = 0.4\), we find the Z-scores for the two endpoints:\[Z_1 = \frac{1.00 - 1.8}{0.4} = -2.0 \Z_2 = \frac{2.60 - 1.8}{0.4} = 2.0\]Using a standard normal distribution table or calculator, \\(P(-2.0 < Z < 2.0) \approx 0.9545\). Thus, the probability that \(\mathrm{VO}_2 \max\) is between 1.00 and 2.60 is approximately 95.45%.

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Key Concepts

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

Understanding Cardiorespiratory Fitness
Cardiorespiratory fitness (CRF) is essential for assessing an individual's overall physical health. It reflects how well the heart, lungs, and muscles work together to supply oxygen during intense physical activity. CRF is commonly measured by the maximal oxygen uptake, known as \((\mathrm{VO}_{2} \max\)). This measurement demonstrates the maximum amount of oxygen the body can utilize during exercise.
\(\mathrm{VO}_{2} \max\) is viewed as an excellent measure of cardiovascular fitness and aerobic endurance. High levels indicate a strong cardiovascular system, which is vital for athletic performance and general health. Improving CRF can reduce risks of chronic diseases such as heart disease and can enhance overall life quality.
Due to its significant role in fitness and health, accurately estimating \((\mathrm{VO}_{2} \max)\) is crucial, especially for trainers and healthcare professionals. But since direct measurement is expensive and resource-intensive, regression analysis helps by providing an equation to predict \((\mathrm{VO}_{2} \max)\). This approach can significantly simplify evaluations using easily obtainable variables, offering convenience and cost-effectiveness.
Crafting a Prediction Equation
A prediction equation serves as a mathematical model that estimates an unknown variable based on known data. In the context of cardiorespiratory fitness, this equation estimates \((\mathrm{VO}_{2} \max)\) from variables like weight, age, walking time, and heart rate.
The provided prediction equation is: \[Y = 5.0 + 0.01x_1 - 0.05x_2 - 0.13x_3 - 0.01x_4 + \epsilon\] Here, \(x_1\) is weight, \(x_2\) is age, \(x_3\) is walking time for 1 mile, and \(x_4\) is heart rate. Each term represents the impact of these variables on \((\mathrm{VO}_{2} \max)\).
The coefficients in front of each variable indicate their relationship with \((\mathrm{VO}_{2} \max)\):
  • For weight, an increase of 1 kg results in a 0.01 increase in \((\mathrm{VO}_{2} \max)\).
  • Increased age tends to decrease \((\mathrm{VO}_{2} \max)\) by 0.05 per year.
  • Longer walking times reduce \((\mathrm{VO}_{2} \max)\) by 0.13 per minute.
  • A higher heart rate leads to a slight reduction, decreasing \((\mathrm{VO}_{2} \max)\) by 0.01 per beat/min.
Using such equations facilitates a quick, efficient estimate of fitness, aiding decisions about training and health interventions.
Model Interpretation in Regression Analysis
Model interpretation involves understanding what the coefficients and parameters of a prediction equation imply about the relationships between variables. It provides insight into how each predictor in the equation influences the response variable.
In the example equation, the coefficient \(\beta_1 = 0.01\) signifies a positive relationship between weight and \((\mathrm{VO}_{2} \max)\). This means that, holding other variables constant, increases in weight lead to small increases in predicted oxygen uptake. On the other hand, the coefficient \(\beta_3 = -0.13\) indicates a negative association between walking time and \((\mathrm{VO}_{2} \max)\).
The constant term (5.0 in this case) represents the expected value of \((\mathrm{VO}_{2} \max)\) when all predictors are zero, though this may not always have a practical interpretation. Meanwhile, \((\epsilon)\) stands for the random error, accounting for variation not explained by the model. Together, these details help us interpret and apply the regression model more effectively in real-world settings.
Performing Probability Calculations
Probability calculations in regression analysis are used to make predictions about how likely a particular outcome is. For example, in this context, how likely the \((\mathrm{VO}_{2} \max)\) will fall within a specific range.
Given the prediction results from the regression model, we can calculate probabilities assuming normal distribution of the data. With a standard deviation \(\sigma = 0.4\), the equation allows for a probability model to estimate the likelihood of the \((\mathrm{VO}_{2} \max)\) being between any two values.
By transforming the limits of the range into Z-scores:
  • For a value of 1.00, \((Z_1 = \frac{1.00 - 1.8}{0.4} = -2.0)\).
  • For a value of 2.60, \((Z_2 = \frac{2.60 - 1.8}{0.4} = 2.0)\).
These Z-scores can then be used with normal distribution tables to find that the probability \((P(-2.0 < Z < 2.0))\) of \((\mathrm{VO}_{2} \max)\) lying within this range is approximately 95.45%. Probability calculations hence allow us to understand the variability and reliability of the predicted outcomes.

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Most popular questions from this chapter

The article "Readability of Liquid Crystal Displays: A Response Surface" (Human Factors, 1983: 185-190) used a multiple regression model with four independent variables to study accuracy in reading liquid crystal displays. The variables were \(y=\) error percentage for subjects reading a four-digit liquid crystal display \(x_{1}=\) level of backlight (ranging from 0 to \(122 \mathrm{~cd} / \mathrm{m}^{2}\) ) \(x_{2}=\) character subtense (ranging from \(.025^{\circ}\) to \(1.34^{\circ}\) ) \(x_{3}=\) viewing angle (ranging from \(0^{\circ}\) to \(60^{\circ}\) ) \(x_{4}=\) level of ambient light (ranging from 20 to 1500 lux) The model fit to data was \(Y=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\beta_{3} x_{3}+\) \(\beta_{4} x_{4}+\epsilon\). The resulting estimated coefficients were \(\hat{\beta}_{0}=1.52, \hat{\beta}_{1}=.02, \hat{\beta}_{2}=-1.40, \hat{\beta}_{3}=.02\), and \(\hat{\beta}_{4}=-.0006\). a. Calculate an estimate of expected error percentage when \(x_{1}=10, x_{2}=5, x_{3}=50\), and \(x_{4}=100 .\) b. Estimate the mean error percentage associated with a backlight level of 20 , character subtense of \(.5\), viewing angle of 10 , and ambient light level of 30 . c. What is the estimated expected change in error percentage when the level of ambient light is increased by 1 unit while all other variables are fixed at the values given in part (a)? Answer for a 100-unit increase in ambient light level. d. Explain why the answers in part (c) do not depend on the fixed values of \(x_{1}, x_{2}\), and \(x_{3}\). Under what conditions would there be such a dependence? e. The estimated model was based on \(n=30\) observations, with SST \(=39.2\) and SSE \(=20.0\). Calculate and interpret the coefficient of multiple determination, and then carry out the model utility test using \(\alpha=.05\).

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 paper. \begin{tabular}{l|cccccccc} \(\mathrm{HR}\) & \(43.5\) & \(44.0\) & \(44.0\) & \(44.5\) & \(44.0\) & \(45.0\) & \(48.0\) & \(49.0\) \\ \hline \(\mathrm{VO}_{2}\) & \(22.0\) & \(21.0\) & \(22.0\) & \(21.5\) & \(25.5\) & \(24.5\) & \(30.0\) & \(28.0\) \end{tabular} \begin{tabular}{l|llllllll} \(\mathrm{HR}\) & \(49.5\) & \(51.0\) & \(54.5\) & \(57.5\) & \(57.7\) & \(61.0\) & \(63.0\) & \(72.0\) \\ \hline \(\mathrm{VO}_{2}\) & \(32.0\) & \(29.0\) & \(38.5\) & \(30.5\) & \(57.0\) & \(40.0\) & \(58.0\) & \(72.0\) \end{tabular} 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 ability of ecologists to identify regions of greatest species richness could have an impact on the preservation of genetic diversity, a major objective of the World Conservation Strategy. The article "Prediction of Rarities from Habitat Variables: Coastal Plain Plants on Nova Scotian Lakeshores" (Ecology, 1992: 1852-1859) used a sample of \(n=37\) lakes to obtain the estimated regression equation $$ \begin{aligned} y=3.89+.033 x_{1}+.024 x_{2}+.023 x_{3} \\ &-.0080 x_{4}-.13 x_{5}-.72 x_{6} \end{aligned} $$ where \(y=\) species richness, \(x_{1}=\) watershed area, \(x_{2}=\) shore width, \(x_{3}=\) poor drainage \((\%), x_{4}=\) water color (total color units), \(x_{5}=\) sand (\%), and \(x_{6}=\) alkalinity. The coefficient of multiple determination was reported as \(R^{2}=.83\). Carry out a test of model utility.

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 \(n=6 \mathrm{ob}\) servations was $$ y=-113.0937+3.3684 x-.01780 x^{2} $$ a. Estimate \(\mu_{\ldots, 5}\), 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 \(\sum y_{i}^{2}=8386.43, \Sigma y_{i}=210.70, \Sigma x_{i} y_{i}=17,002.00\), and \(\sum x_{i}^{2} y_{i}=1,419,780\), compute SSE \(\left[=\sum y_{i}^{2}-\right.\) \(\left.\hat{\beta}_{0} \Sigma y_{i}-\hat{\beta}_{1} \Sigma x_{i} y_{i}-\hat{\beta}_{2} \sum x_{i}^{2} y_{i}\right], s^{2}\), 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 \(s_{\beta_{2}}=.00226\), test \(H_{0}: \beta_{2}=0\) versus \(H_{a}: \beta_{2} \neq 0\) at level \(.01\), and interpret the result.

High-alumina refractory castables have been extensively investigated in recent years because of their significant advantages over other refractory brick of the same classlower production and application costs, versatility, and performance at high temperatures. The accompanying data on \(x=\) viscosity \((\mathrm{MPa} \cdot \mathrm{s})\) and \(y=\) free-flow \((\%)\) was read from a graph in the article "Processing of Zero-Cement Self-Flow Alumina Castables" (The Amer. Ceramic Soc. Bull., 1998: 60-66): \begin{tabular}{c|ccccccc} \(x\) & 351 & 367 & 373 & 400 & 402 & 456 & 484 \\ \hline\(y\) & 81 & 83 & 79 & 75 & 70 & 43 & 22 \end{tabular} The authors of the cited paper related these two variables using a quadratic regression model. The estimated regression function is \(y=-295.96+2.1885 x-.0031662 x^{2}\). a. Compute the predicted values and residuals, and then SSE and \(s^{2}\). b. Compute and interpret the coefficient of multiple determination. c. The estimated SD of \(\hat{\beta}_{2}\) is \(s_{\hat{\beta}_{3}}=.0004835\). Does the quadratic predictor belong in the regression model? d. The estimated SD of \(\hat{\beta}_{1}\) is \(.4050\). Use this and the information in (c) to obtain joint CIs for the linear and quadratic regression coefficients with a joint confidence level of (at least) \(95 \%\). e. The estimated SD of \(\hat{\mu}_{y, 200}\) is 1.198. Calculate a 95\% CI for true average free-flow when viscosity \(=400\) and also a \(95 \%\) PI for free- flow resulting from a single observation made when viscosity \(=400\), and compare the intervals.

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