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Chapter 13: Problem 1

Cardiorespiratory fitness is widely recognized as a major component of overall physical well-being. Direct measurement of maximal oxygen uptake (VO \(\mathrm{VO}_{2}\) max \()\) 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 $$ \begin{aligned} &y=\mathrm{VO}_{2} \max (\mathrm{L} / \mathrm{min}) \quad x_{1}=\text { weight }(\mathrm{kg}) \\ &x_{2}=\text { age }(\mathrm{yr}) \\ &x_{3}=\text { time necessary to walk } 1 \text { mile (min) } \\ &x_{4}=\text { heart rate at the end of the walk (beats/min) } \\ &\text { Here is one possible model, for male students, consistent } \\ &\text { with the information given in the article "Validation of } \\ &\text { the Rockport Fitness Walking Test in College Males } \\ &\text { and Females" (Research Quarterly for Exercise and } \\ &\text { Sport, } 1994: 152-158): \\ &Y=5.0+.01 x_{1}-.05 x_{2}-.13 x_{3}-.01 x_{4}+\epsilon \\ &\sigma=.4 \end{aligned} $$ 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 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
尾1: VO2 max increases by 0.01 L/min per kg; 尾3: decreases by 0.13 L/min per min of walk. Expected VO2 max is 1.8 L/min. Probability that it's between 1.00 and 2.60 is 95.44%.

Step by step solution

01

Interpret 尾1

The coefficient \( \beta_1 \) (0.01) represents the change in \( \mathrm{VO}_2 \max \) for a one-unit increase in weight (\( x_1 \)) while holding all other variables constant. This implies that for each additional kilogram of weight, \( \mathrm{VO}_2 \max \) is expected to increase by 0.01 \( \mathrm{L/min} \).
02

Interpret 尾3

The coefficient \( \beta_3 \) (-0.13) represents the change in \( \mathrm{VO}_2 \max \) for a one-unit increase in time to walk one mile (\( x_3 \)). It suggests that \( \mathrm{VO}_2 \max \) decreases by 0.13 \( \mathrm{L/min} \) for each additional minute taken to walk one mile.
03

Substitute Values into Equation

To find the expected value of \( \mathrm{VO}_2 \max \), substitute the values: weight (76 kg), age (20 years), walk time (12 min), and heart rate (140 bpm) into the model: \[ Y = 5.0 + 0.01(76) - 0.05(20) - 0.13(12) - 0.01(140) \]
04

Calculate Expected VO2 Max

Calculate: \[ Y = 5.0 + 0.76 - 1.0 - 1.56 - 1.4 = 1.8 \] The expected \( \mathrm{VO}_2 \max \) for the given conditions is 1.8 \( \mathrm{L/min} \).
05

Understand the Normal Distribution

Since \( \epsilon \) is a normally distributed error term with mean 0 and standard deviation \( \sigma = 0.4 \), \( Y \) follows a normal distribution with mean 1.8 and standard deviation 0.4.
06

Calculate Probability Range

To find the probability that \( \mathrm{VO}_2 \max \) is between 1.00 and 2.60, use the standard normal distribution. Calculate the z-scores: \[ z_1 = \frac{1.00 - 1.8}{0.4} = -2 \] \[ z_2 = \frac{2.60 - 1.8}{0.4} = 2 \] Using standard normal tables, \( P(-2 \leq Z \leq 2) = 0.9772 - 0.0228 = 0.9544 \).
07

Conclusion

The probability that \( \mathrm{VO}_2 \max \) will be between 1.00 and 2.60 is approximately 95.44%.

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Key Concepts

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

VO2 Max Prediction
Predicting \(\mathrm{VO}_2 \max\) is essential because it offers a way to estimate cardiorespiratory fitness without directly measuring oxygen uptake. Direct measurements are costly and labor-intensive. Therefore, modeling \(\mathrm{VO}_2 \max\) based on more easily obtained variables (like weight, age, time to walk a mile, and ending heart rate) is practical. In the given model, each coefficient represents the change in \(\mathrm{VO}_2 \max\) per unit change of that variable while others are kept constant. Understanding these coefficients is crucial:
  • \(\beta_1\) (weight): A coefficient of 0.01 indicating \(\mathrm{VO}_2 \max\) increases by 0.01 \(\mathrm{L/min}\) for every additional kilogram.
  • \(\beta_3\) (walk time): A coefficient of -0.13, meaning \(\mathrm{VO}_2 \max\) decreases by 0.13 \(\mathrm{L/min}\) for every extra minute taken to walk a mile.
These coefficients help to formulate a prediction equation that is directly applicable to individuals going through cardiorespiratory fitness assessments.
Cardiorespiratory Fitness
Cardiorespiratory fitness refers to the efficiency with which the body supplies oxygen to muscles during prolonged physical activities. It's a clear indicator of overall health. A fit cardiovascular and respiratory system means that the heart, lungs, and arteries effectively supply oxygenated blood during exercise. Analyzing cardiorespiratory fitness involves assessing the \(\mathrm{VO}_2 \max\), as it is the upper limit of oxygen utilization during intense exercise. The higher the \(\mathrm{VO}_2 \max\), the more fit an individual is naturally perceived to be.

Improvements in fitness often lead to lowered risk of heart disease, improved lung function, and better endurance. It's important to emphasize the use of predictive models for \(\mathrm{VO}_2 \max\), as they enable health and fitness professionals to assess cardiorespiratory fitness without intensive lab tests.
Normal Distribution
The concept of normal distribution is vital when predicting \(\mathrm{VO}_2 \max\) using regression analysis. In this context, the \(\epsilon\) (error term) in the predictive equation follows a normal distribution with a mean of 0 and standard deviation \(\sigma = 0.4\). This characteristic suggests that residuals (or deviations from the anticipated values) are symmetrically distributed around the mean.

When assessing the predictive model, the expected outcome, such as predicted \(\mathrm{VO}_2 \max\), can also be interpreted as normally distributed. This distribution is characterized by its mean and standard deviation, assisting in probability calculations. For instance, the probability that \(\mathrm{VO}_2 \max\) is between specified ranges can be calculated using z-scores and standard normal distribution tables.

Understanding this probability helps to provide confidence in the predictive capabilities of the model and ensures that predictions about an individual's cardiorespiratory fitness are statistically robust.

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Most popular questions from this chapter

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 \(\mu_{\gamma \cdot 75}\), 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 \(\Sigma y_{i}^{2}=8386.43, \Sigma y_{i}=210.70, \Sigma x_{i} y_{i}=17,002.00\), and \(\Sigma x_{i}^{2} y_{i}=1,419,780\), compute SSE \(\left[=\Sigma y_{i}^{2}-\right.\) \(\left.\hat{\beta}_{0} \Sigma y_{i}-\hat{\beta}_{1} \Sigma x_{i} y_{i}-\hat{\beta}_{2} \Sigma x_{i}^{2} y_{i}\right]\) 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 \(s_{\hat{\beta}_{2}}=.00226\), test \(H_{0}: \beta_{2}=0\) versus \(H_{2}: \beta_{2} \neq 0\) at level \(.01\), and interpret the result.

46\. A regression analysis carried out to relate \(y=\) repair time for a water filtration system (hr) to \(x_{1}=\) elapsed time since the previous service (months) and \(x_{2}=\) type of repair ( 1 if electrical and 0 if mechanical) yielded the following model based on \(n=12\) observations: \(y=.950+.400 x_{1}+1.250 x_{2}\). In addition, SST \(=12.72\), \(\mathrm{SSE}=2.09\), and \(s_{\hat{\beta}_{2}}=.312\). a. Does there appear to be a useful linear relationship between repair time and the two model predictors? Carry out a test of the appropriate hypotheses using a significance level of \(.05\). b. Given that elapsed time since the last service remains in the model, does type of repair provide useful information about repair time? State and test the appropriate hypotheses using a significance level of .01. c. Calculate and interpret a \(95 \% \mathrm{CI}\) for \(\beta_{2}\). d. The estimated standard deviation of a prediction for repair time when elapsed time is 6 months and the repair is electrical is \(.192\). Predict repair time under these circumstances by calculating a \(99 \%\) prediction interval. Does the interval suggest that the estimated model will give an accurate prediction? Why or why not?

a. Show that \(\sum_{i=1}^{n} e_{i}=0\) when the \(e_{i}\) 's are the residuals from a simple linear regression. b. Are the residuals from a simple linear regression independent of one another, positively correlated, or negatively correlated? Explain. c. Show that \(\sum_{i=1}^{n} x_{i} e_{i}=0\) for the residuals from a simple linear regression. (This result along with part (a) shows that there are two linear restrictions on the \(e_{i}^{\text {'s, resulting }}\) in a loss of 2 df when the squared residuals are used to estimate \(\sigma^{2}\).) d. Is it true that \(\Sigma_{i=1}^{n} e_{i}^{*}=0\) ? Give a proof or a counter example.

If there is at least one \(x\) value at which more than one observation has been made, there is a formal test procedure for testing \(H_{0}: \mu_{\gamma-x}=\beta_{0}+\beta_{1} x\) for some values \(\beta_{0}, \beta_{1}\) (the true regression function is linear) versus \(H_{a}: H_{0}\) is not true (the true regression function is not linear) Suppose observations are made at \(x_{1}, x_{2}, \ldots, x_{c}\). Let \(Y_{11}, Y_{12}, \ldots, Y_{\mathrm{Ln}_{\mathrm{i}}}\) denote the \(n_{1}\) observations when \(x=x_{1} ; \ldots ; Y_{c 1}, Y_{c 2}, \ldots, Y_{c a}\) denote the \(n_{c}\) observations when \(x=x_{c^{-}}\)With \(n=\Sigma n_{j}\) (the total number of observations), SSE has \(n-2\) df. We break SSE into two pieces, SSPE (pure error) and SSLF (lack of fit), as follows: $$ \begin{aligned} \mathrm{SSPE} &=\sum_{i} \sum_{j}\left(Y_{i j}-\bar{Y}_{i-}\right)^{2} \\ &=\sum \sum Y_{i j}^{2}-\sum n_{i} \bar{Y}_{i-}^{2} \\ \mathrm{SSLF} &=\mathrm{SSE}-\mathrm{SSPE} \end{aligned} $$ The \(n_{i}\) observations at \(x_{i}\) contribute \(n_{i}-1\) df to SSPE, so the number of degrees of freedom for SSPE is \(\Sigma_{i}\left(n_{i}-1\right)=n-c\), and the degrees of freedom for SSLF is \(n-2-(n-c)=c-2\). Let MSPE \(=\operatorname{SSPE} /(n-c)\) and MSLF \(=S S L F /(c-2)\). Then it can be shown that whereas \(E(\mathrm{MSPE})=\sigma^{2}\) whether or not \(H_{0}\) is true, \(E(\mathrm{MSLF})=\sigma^{2}\) if \(H_{0}\) is true and \(E(\mathrm{MSLF})>\sigma^{2}\) if \(H_{0}\) is false. The test statistic is \(F=\) MSLF/MSPE, and the corresponding \(P\)-value is the area under the \(F_{c-2, n-c}\) curve to the right of \(f\). The following data comes from the article "Changes in Growth Hormone Status Related to Body Weight of Growing Cattle" (Growth, 1977: 241-247), with \(x=\) body weight and \(y=\) metabolic clearance rate/body weight. $$ \begin{array}{l|ccccccc} x & 110 & 110 & 110 & 230 & 230 & 230 & 360 \\ \hline y & 235 & 198 & 173 & 174 & 149 & 124 & 115 \\ x & 360 & 360 & 360 & 505 & 505 & 505 & 505 \\ \hline y & 130 & 102 & 95 & 122 & 112 & 98 & 96 \end{array} $$ a. Test \(H_{0}\) versus \(H_{a}\) at level 05 using the lack-of-fit test just described. b. Does a scatterplot of the data suggest that the relationship between \(x\) and \(y\) is linear? How does this compare with the result of part (a)? (A nonlinear regression function was used in the article.)

Utilization of sucrose as a carbon source for the production of chemicals is uneconomical. Beet molasses is a readily available and low-priced substitute. The article "Optimization of the Production of \(\beta\)-Carotene from Molasses by Blakeslea Trispora" \((J\). of Chem. Tech. and Biotech., 2002: 933-943) carried out a multiple regression analysis to relate the dependent variable \(y=\) amount of \(\beta\)-carotene \(\left(\mathrm{g} / \mathrm{dm}^{3}\right)\) to the three predictors amount of lineolic acid, amount of kerosene, and amount of antioxidant \(\left(\right.\) all \(\mathrm{g} / \mathrm{dm}^{3}\) ). a. Fitting the complete second-order model in the three predictors resulted in \(R^{2}=987\) and adjusted \(R^{2}=\) \(.974\), whereas fitting the first-order model gave \(R^{2}=\) .016. What would you conclude about the two models? b. For \(x_{1}=x_{2}=30, x_{3}=10\), a statistical software package reported that \(\hat{y}=.66573, s_{\hat{y}}=.01785\), based on the complete second-order model. Predict the amount of \(\beta\)-carotene that would result from a single experimental run with the designated values of the independent variables, and do so in a way that conveys information about precision and reliability.
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