/*! 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 78 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 time-consuming 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 \(\begin{aligned} y &=\mathrm{VO}_{2} \max (\mathrm{L} / \mathrm{min}) \quad x_{1}=\text { weight }(\mathrm{kg}) \\ x_{2} &=\text { age }(\mathrm{yr}) \\\ x_{3} &=\text { time necessary to walk } 1 \text { mile }(\mathrm{min}) \\\ x_{4} &=\text { heart rate at the end of the walk }(\text { beats } / \mathrm{min}) \end{aligned}\) 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" (Res. Q. Exercise Sport, 1994: 152–158): $$ \begin{aligned} &Y=5.0+.01 x_{1}-.05 x_{2}-.13 x_{3}-.01 x_{4}+\varepsilon \\ &\sigma=.4 \end{aligned} $$ 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 year, walk time is \(12 \mathrm{~min}\), and heart rate is 140 beats \(/ \mathrm{min}\) ? 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
a. \( \beta_1 = 0.01 \), \( \beta_3 = -0.13 \). b. Expected \( \mathrm{VO}_2 \max \) is 1.8 L/min. c. Probability is approximately 0.9545.

Step by step solution

01

Interpret Coefficients

Coefficients \( \beta_1 \) and \( \beta_3 \) represent the change in \( \mathrm{VO}_2 \max\) for a one-unit increase in the corresponding variables, keeping other variables constant. \( \beta_1 = 0.01 \): This implies that for every additional 1 kg of weight, \( \mathrm{VO}_2 \max\) increases by 0.01 L/min.\( \beta_3 = -0.13 \): This indicates that for every additional minute taken to walk 1 mile, \( \mathrm{VO}_2 \max\) decreases by 0.13 L/min.
02

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

Substitute the given values into the prediction equation:\[ Y = 5.0 + 0.01(76) - 0.05(20) - 0.13(12) - 0.01(140) + \varepsilon \]First, calculate each component:- \( 0.01 \times 76 = 0.76 \)- \( -0.05 \times 20 = -1.0 \)- \( -0.13 \times 12 = -1.56 \)- \( -0.01 \times 140 = -1.4 \)Add these values to the constant term:\[ Y = 5.0 + 0.76 - 1.0 - 1.56 - 1.4 = 1.8 + \varepsilon \]The expected value is \( 1.8 \mathrm{~L/min} \), ignoring the error term \( \varepsilon \).
03

Calculate the Probability of \( \mathrm{VO}_2 \max \) within Range

We want the probability that \( \mathrm{VO}_2 \max \) is between 1.00 and 2.60 L/min. The standard deviation \( \sigma = 0.4 \).Calculate the standard score (z-score) for each limit using the expected value of 1.8:For 1.00:\[ z_1 = \frac{1.00 - 1.8}{0.4} = -2.0 \]For 2.60:\[ z_2 = \frac{2.60 - 1.8}{0.4} = 2.0 \]Consult the standard normal distribution table or calculator:- Probability \( z > -2.0 \) is approximately 0.9772.- Probability \( z < 2.0 \) is approximately 0.9772.The probability of \( \mathrm{VO}_2 \max \) between 1.00 and 2.60 is:\[ P(z_1 < Z < z_2) = 0.9772 - (1 - 0.9772) = 0.9545 \]

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Key Concepts

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

Linear Regression
Linear regression is a statistical method used to model the relationship between a dependent variable and one or more independent variables. In this exercise, \(MODERN{Linear}{Regression}\) involves a dependent variable, \(\mathrm{VO}_2 \max\), which measures cardiorespiratory fitness, and independent variables such as weight, age, walking time, and heart rate. The goal is to understand how changes in these predictors affect \(\mathrm{VO}_2 \max\).

In a linear regression model, the equation is typically expressed as \(Y = \beta_0 + \beta_1 x_1 + \ldots + \beta_n x_n + \varepsilon\), where \(Y\) is the dependent variable, \(\beta_i\) are the regression coefficients, \(x_i\) are the independent variables, and \(\varepsilon\) is the error term.

Understanding how each independent variable influences the dependent variable, as shown in the coefficients \(\beta_1\) and \(\beta_3\), is crucial. Here, \(\beta_1 = 0.01\) implies that for every one kilogram increase in weight, \(\mathrm{VO}_2 \max\) increases by 0.01 L/min, holding other variables constant. Similarly, \(\beta_3 = -0.13\) implies that an additional minute taken to walk a mile results in a decrease of 0.13 L/min in \(\mathrm{VO}_2 \max\).

Linear regression provides a straightforward way to make predictions and draw insights from data, helping us forecast and understand outcomes such as fitness levels based on measurable inputs.
Predictive Models
Predictive models are tools used to forecast future outcomes based on historical data. In the context of our exercise, we're building a predictive model to estimate \(\mathrm{VO}_2 \max\) using easily obtained factors like weight, age, and walking performance.

Predictive models generally follow these steps:
  • Define the variables: Identify which variables can best predict the outcome.
  • Model training: Use statistical techniques, like linear regression, to "train" the model on existing data.
  • Validation: Test the model's predictive power on new data to ensure accuracy and reliability.

Good predictive models give us valuable insights and enable effective planning. In this case, by predicting \(\mathrm{VO}_2 \max\), we can make informed decisions about medical evaluations or fitness training programs without the need for direct and costly measurement.
Standard Normal Distribution
The standard normal distribution is a key statistical concept used to understand the distribution of values and to calculate probabilities. It's a normal distribution with a mean of 0 and a standard deviation of 1.

In this problem, we use the standard normal distribution to calculate the probability of \(\mathrm{VO}_2 \max\) being within a specific range. By transforming our predicted \(\mathrm{VO}_2 \max\) values into \MODERN{standard}{normal}{distribution}\ scores (z-scores), we can assess how likely certain values are.

For example, we derived z-scores for \(\mathrm{VO}_2 \max\) limits by comparing them to the expected value of 1.8 L/min. The z-scores range from -2.0 to 2.0, allowing us to determine that about 95.45% of observations will fall within this range, using symmetry and values from the z-table.

This method provides a simple way to quantify uncertainty and make data-driven decisions based on probabilities.
Regression Coefficients
Regression coefficients are fundamental in understanding how changes in independent variables affect the dependent variable. They tell us the size and direction of the relationship between variables.

In the given model, each predictor has its own coefficient:
  • \(\beta_1\) for weight implies a positive relationship: more weight leads to a higher \(\mathrm{VO}_2 \max\).
  • \(\beta_2\) for age and \(\beta_4\) for heart rate, although not fully explored in our point, also influence \(\mathrm{VO}_2 \max\) as modeled.
  • \(\beta_3\) for walk time indicates a negative relationship: longer times reduce \(\mathrm{VO}_2 \max\).

Each coefficient plays a unique role in shaping the model's predictions. By interpreting these coefficients, we gain deeper insights into the factors affecting outcomes, enabling targeted actions to improve fitness or perform risk assessments.

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Most popular questions from this chapter

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