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Chapter 12: Problem 106

An investigation was carried out to study the relationship between speed (ft/s) and stride rate (number of steps taken/s) among female marathon runners. Resulting summary quantities included \(n=11, \Sigma(\) speed \()=205.4, \Sigma(\text { speed })^{2}\) \(=3880.08, \quad \Sigma(\) rate \()=35.16, \quad \Sigma(\text { rate })^{2}\) \(=112.681\), and \(\Sigma(\) speed \()(\) rate \()=660.130 .\) a. Calculate the equation of the least squares line that you would use to predict stride rate from speed. b. Calculate the equation of the least squares line that you would use to predict speed from stride rate. c. Calculate the coefficient of determination for the regression of stride rate on speed of part (a) and for the regression of speed on stride rate of part (b). How are these related? d. How is the product of the two slope estimates related to the value calculated in (c)?

Short Answer

Expert verified
a) Stride rate = 2.8586 + 0.0181 脳 speed. b) Speed = -36.3542 + 17.2178 脳 rate. c) R虏 varies per regression but relates identically in paired linear relations. d) Product of slopes indicates relation strength.

Step by step solution

01

Compute Means

For both variables, speed and stride rate, calculate the means:Mean of speed, \( \bar{x} \):\[ \bar{x} = \frac{\Sigma(\text{speed})}{n} = \frac{205.4}{11} = 18.6727 \text{ ft/s} \]Mean of rate, \( \bar{y} \):\[ \bar{y} = \frac{\Sigma(\text{rate})}{n} = \frac{35.16}{11} = 3.1964 \text{ steps/s} \]
02

Calculate Slope for Predicting Stride Rate from Speed

The slope \( b_1 \) for predicting stride rate from speed is calculated using the formula:\[ b_1 = \frac{\Sigma(xy) - n\bar{x}\bar{y}}{\Sigma(x^2) - n\bar{x}^2} \]Substitute the known values:\(\Sigma(xy) = 660.130, \Sigma(x^2) = 3880.08\)\[ b_1 = \frac{660.130 - 11 \times 18.6727 \times 3.1964}{3880.08 - 11 \times (18.6727)^2} = \frac{660.130 - 657.636}{137.777} = \frac{2.494}{137.777} = 0.0181 \]
03

Calculate Intercept for Predicting Stride Rate from Speed

The intercept \( a \) is calculated using:\[ a = \bar{y} - b_1 \bar{x} \]Substitute the known values:\[ a = 3.1964 - 0.0181 \times 18.6727 = 3.1964 - 0.3378 = 2.8586 \]
04

Equation for Predicting Stride Rate from Speed

The least squares line for predicting stride rate from speed is:\[ \text{rate} = 2.8586 + 0.0181 \times \text{speed} \]
05

Calculate Slope for Predicting Speed from Stride Rate

The slope \( m_1 \) for predicting speed from stride rate is:\[ m_1 = \frac{\Sigma(xy) - n\bar{x}\bar{y}}{\Sigma(y^2) - n\bar{y}^2} \]Given \(\Sigma(y^2) = 112.681\)\[ m_1 = \frac{660.130 - 11 \times 18.6727 \times 3.1964}{112.681 - 11 \times (3.1964)^2} = \frac{2.494}{0.1449} = 17.2178 \]
06

Calculate Intercept for Predicting Speed from Stride Rate

The intercept \( c \) is given by:\[ c = \bar{x} - m_1 \bar{y} \]\[ c = 18.6727 - 17.2178 \times 3.1964 = 18.6727 - 55.0269 = -36.3542 \]
07

Equation for Predicting Speed from Stride Rate

The least squares equation for predicting speed from stride rate is:\[ \text{speed} = -36.3542 + 17.2178 \times \text{rate} \]
08

Calculate Coefficient of Determination for Stride Rate on Speed

The coefficient of determination \( R^2 \) is related to the square of the correlation coefficient \( r \). We won't directly find \( R^2 \) without additional data, but given dependence, note that:\[ R^2_{xy} = (b_1 \times m_1)^2 \]
09

Relationship Between Slopes and Coefficient of Determination

The product of the two slopes is related to the correlation coefficient:\[ b_1 \times m_1 = 0.0181 \times 17.2178 \approx 0.311\]And it connects to \( R^2 \) because each model's fit is the square of linear relation strength. Given squared correlation \( r^2 \), both regression \( R^2 \) are equal and symmetric for a single paired data set.

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Key Concepts

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

Least Squares Method
The Least Squares Method is a popular approach for finding the best-fit line through a scatter plot of data points. This method minimizes the sum of the squares of the vertical distances of the points from the line. Simply put, it tries to draw a straight line that is as close as possible to all points in the dataset.
  • To achieve this, we first calculate the slope and intercept of the line. This involves some fundamental calculations using data values such as means and summations.
  • The slope indicates the rate of change between the independent and dependent variables, while the intercept represents the expected value of the dependent variable when the independent variable is zero.
The utility of this method lies in its ability to simplify complex data into a model that can predict or explain the relationship between variables.
Coefficient of Determination
The Coefficient of Determination, denoted as \(R^2\), is a statistical measure that shows the proportion of variance for a dependent variable that's explained by an independent variable in a regression model.
  • It provides insight into how well the model fits the data; the closer \(R^2\) is to 1, the better the model explains the data variance.
  • In our context, \(R^2\) reflects how changes in speed can explain variations in stride rate, or vice versa.
An important aspect of \(R^2\) is its symmetry in a single data set allowing it to relate the slopes from reciprocal relationships (e.g., speed on stride rate and stride rate on speed). Its calculation requires the square of the correlation coefficient, highlighting the role of both the strength and direction of a linear relationship.
Correlation Coefficient
The Correlation Coefficient, denoted by \(r\), measures the degree to which two variables move in relation to each other. This value ranges from -1 to 1, where:
  • -1 indicates a perfectly inverse relationship
  • 0 indicates no linear relationship
  • 1 indicates a perfectly direct relationship
For the regression of stride rate on speed, \(r\) helps us understand if increases in speed result in predictable increases (or decreases) in stride rate, and vice versa. Knowing \(r\) gives us insights into the strength and direction of the relationship, which is pivotal for interpreting \(R^2\) and its implications for real-world predictions.
Slope and Intercept Calculations
Calculating the slope and intercept is fundamental to forming the least squares regression line. Let's break down how it is done: **Slope Calculation**
  • The slope illustrates the relation between variables - how much the dependent variable changes for a unit change in the independent variable.
  • To compute the slope in our problem, we use the formula to incorporate sums of products and squares of the data points. This provides a ratio that defines the line angle.
**Intercept Calculation**
  • The intercept is computed by modifying the average of the dependent variable with the product of the slope and average of the independent variable.
  • It establishes the starting point for the regression line at the independent variable value of zero.
These calculations transform raw data into a predictive model, helping to interpret relationships in context, such as how speed affects stride rate in athletes.

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Most popular questions from this chapter

A regression analysis carried out to relate \(y=\) repair time for a water filtration system ( \(\mathrm{hr}\) ) 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\% 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?

Verify Property 2 of the correlation coefficient, the value of \(r\) is independent of the units in which \(x\) and \(y\) are measured; that is, if \(x_{i}^{\prime}=a x_{i}+c\) and \(y_{i}^{\prime}=b y_{i}+d, a>0, b>0\), then \(r\) for the \(\left(x_{i}^{\prime}, y_{i}^{\prime}\right)\) pairs is the same as \(r\) for the \(\left(x_{i}, y_{i}\right)\) pairs.

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鈥�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)?

The article "Objective Measurement of the Stretchability of Mozzarella Cheese" \((J\). Texture Stud., 1992: 185-194) reported on an experiment to investigate how the behavior of mozzarella cheese varied with temperature. Consider the accompanying data on \(x=\) temperature and \(y=\) elongation (\%) at failure of the cheese. [Note: The researchers were Italian and used real mozzarella cheese, not the poor cousin widely available in the United States.] $$ \begin{array}{r|rrrrrrr} x & 59 & 63 & 68 & 72 & 74 & 78 & 83 \\ \hline y & 118 & 182 & 247 & 208 & 197 & 135 & 132 \end{array} $$ a. Construct a scatter plot in which the axes intersect at \((0,0)\). Mark \(0,20,40,60,80\), and 100 on the horizontal axis and \(0,50,100,150,200\), and 250 on the vertical axis. b. Construct a scatter plot in which the axes intersect at \((55,100)\), as was done in the cited article. Does this plot seem preferable to the one in part (a)? Explain your reasoning. c. What do the plots of parts (a) and (b) suggest about the nature of the relationship between the two variables?

The article "Analysis of the Modeling Methodologies for Predicting the Strength of Air-Jet Spun Yarns" (Textile Res. \(J ., 1997: 39-44\) ) reported on a study carried out to relate yarn tenacity \((y\), in \(\mathrm{g} /\) tex) to yarn count ( \(x_{1}\), in tex), percentage polyester \(\left(x_{2}\right)\), first nozzle pressure \(\left(x_{3}\right.\), in \(\left.\mathrm{kg} / \mathrm{cm}^{2}\right)\), and second nozzle pressure \(\left(x_{4}\right.\), in \(\left.\mathrm{kg} / \mathrm{cm}^{2}\right) .\) The estimate of the constant term in the corresponding multiple regression equation was \(6.121\). The estimated coefficients for the four predictors were \(-.082, .113, .256\), and \(-.219\), respectively, and the coefficient of multiple determination was .946. Assume that \(n=25\). a. State and test the appropriate hypotheses to decide whether the fitted model specifies a useful linear relationship between the dependent variable and at least one of the four model predictors. b. Calculate the value of adjusted \(R^{2}\) and comment. c. Calculate a \(99 \%\) confidence interval for true mean yarn tenacity when yarn count is 16.5, yarn contains \(50 \%\) polyester, first nozzle pressure is 3 , and second nozzle pressure is 5 if the estimated standard deviation of predicted tenacity under these circumstances is \(.350\).
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