/*! 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 60 In Chapter \(12,\) we analyzed s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 60

In Chapter \(12,\) we analyzed strength data for a sample of female high school athletes. When we predict the maximum number of pounds the athlete can bench press using the number of times she can do a 60 -pound bench press \(\left(\mathrm{BP}_{-} 60\right)\), we get \(r^{2}=0.643 .\) When we add the number of times an athlete can perform a 200 -pound leg press \(\left(\mathrm{LP}_{-} 200\right)\) to the model, we get \(\hat{y}=60.6+1.33\left(\mathrm{BP}_{-} 60\right)+0.21\left(\mathrm{LP}_{-} 200\right)\) and \(R^{2}=0.656\)

Short Answer

Expert verified
Adding \(LP_{-200}\) to the model slightly increases \(R^2\) from 0.643 to 0.656, improving explanatory power.

Step by step solution

01

Understand the given data

We have two regression models related to the bench press performance of female athletes. The first model uses only the variable \(BP_{-60}\) with an \(r^2\) value of 0.643. The second model adds another variable \(LP_{-200}\), resulting in an equation \(\hat{y} = 60.6 + 1.33(BP_{-60}) + 0.21(LP_{-200})\) and an \(R^2\) value of 0.656.
02

Analyze the change in model parameters

In the initial model, \(r^2 = 0.643\) indicates that 64.3% of the variability in the maximum bench press weight can be explained by \(BP_{-60}\) alone. Adding another variable \(LP_{-200}\) adjusts the \(R^2\) to 0.656, suggesting a slight increase in explanatory power to 65.6%.
03

Interpret the equation coefficients

The regression equation \(\hat{y} = 60.6 + 1.33(BP_{-60}) + 0.21(LP_{-200})\) indicates the predicted maximum weight for an athlete's bench press adjusts by 1.33 pounds for each additional lift of the 60-pound bench press, and by 0.21 pounds for each additional lift of the 200-pound leg press. The intercept of 60.6 pounds is the predicted bench press weight when both \(BP_{-60}\) and \(LP_{-200}\) are zero.

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 91影视!

Key Concepts

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

Coefficient Interpretation
When we delve into regression analysis, interpreting the coefficients is pivotal. In the equation \( \hat{y} = 60.6 + 1.33(BP_{-60}) + 0.21(LP_{-200}) \), each coefficient has a specific role. The number \( 1.33 \) associated with \( BP_{-60} \) tells us that for every additional repetition of the 60-pound bench press the athlete completes, we expect the maximum bench press capability to increase by 1.33 pounds. Likewise, the \( 0.21 \) coefficient for \( LP_{-200} \) suggests a 0.21 pound increase in the predicted bench press for each extra repetition of the 200-pound leg press.
The intercept, \( 60.6 \), reveals the estimated bench press weight when both \( BP_{-60} \) and \( LP_{-200} \) repetitions are zero. Although in real-world scenarios this might not make much practical sense, it is mathematically meaningful as it anchors the regression line.
  • Effect of \( BP_{-60} \): +1.33 pounds per repetition
  • Effect of \( LP_{-200} \): +0.21 pounds per repetition
  • Intercept: 60.6 pounds when repetitions are zero
Model Comparison
Model comparison in regression helps understand how well different sets of predictors explain the variability of the outcome. The initial model with just \( BP_{-60} \) yielded an \( r^2 \) value of \( 0.643 \), signifying that 64.3% of the variation in the bench press weight is explained by this single variable.
Adding \( LP_{-200} \) as an additional predictor altered our model to achieve an \( R^2 \) of \( 0.656 \). This denotes a modest enhancement in the model's explanatory power鈥攏ow accounting for 65.6% of the variability.
The slight rise in \( R^2 \) hints at the marginally improved fitting of the model when \( LP_{-200} \) is incorporated.
  • Initial Model (\( BP_{-60} \) only): \( r^2 = 0.643 \)
  • Enhanced Model (with \( LP_{-200} \)): \( R^2 = 0.656 \)
  • Increased explanatory power with additional variables
Explanatory Power
Explanatory power in regression describes the proportion of the variance in the dependent variable captured by the model. In simpler terms, it tells us how effective our model is in explaining the observed data. Initially, using \( BP_{-60} \) alone gave us an \( r^2 \) of \( 0.643 \). Therefore, it explained 64.3% of the variance in the bench press performance based solely on the 60-pound repetitions.
After including \( LP_{-200} \) into the model, our \( R^2 \) reached \( 0.656 \). This increment, although small, is significant. It shows that including another variable enhances the model鈥檚 capability to account for more variance, even if just slightly. The remaining 34.4% variance depicts unpredictable factors or noise not captured by the model.
  • \( r^2 = 0.643 \): Variance explained by \( BP_{-60} \) alone
  • \( R^2 = 0.656 \): Variance explained with \( LP_{-200} \) included
  • Reflects how well the model captures outcome variability

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

The least squares prediction equation provides predicted values \(\hat{y}\) with the strongest possible correlation with \(y,\) out of all possible prediction equations of that form. Based on this property, explain why the multiple correlation \(R\) cannot decrease when you add a variable to a multiple regression model.

In the model \(\mu_{y}=\alpha+\beta_{1} x_{1}+\beta_{2} x_{2},\) suppose that \(x_{2}\) is an indicator variable for gender, equaling 1 for females and 0 for males. a. We set \(x_{2}=0\) if we want a predicted mean without knowing gender. b. The slope effect of \(x_{1}\) is \(\beta_{1}\) for males and \(\beta_{2}\) for females. c. \(\beta_{2}\) is the difference between the population mean of \(y\) for females and for males. d. \(\beta_{2}\) is the difference between the population mean of \(y\) for females and males, for all those subjects having \(x_{1}\) fixed, such as \(x_{1}=10\)

In a study (reported in New York Times, February 15,1991 ) on the effects of AZT in slowing the development of AIDS symptoms, 338 veterans whose immune systems were beginning to falter after infection with the AIDS virus were randomly assigned either to receive AZT immediately or to wait until their T cells showed severe immune weakness. The study classified the veterans' race, whether they received AZT immediately, and whether they developed AIDS symptoms during the three-year study. Let \(x_{1}\) denote whether used \(\mathrm{AZT}(1=\) yes, \(0=\) no ) and let \(x_{2}\) denote race \((1=\) white \(, 0=\) black \()\). A logistic regression analysis for the probability of developing AIDS symptoms gave the prediction equation $$\hat{p}=\frac{e^{-1.074-0.720 x_{1}+0.056 x_{2}}}{1+e^{-1.074-0.720 x_{1}+0.056 x_{2}}} .$$ a. Interpret the sign of the effect of AZT. b. Show how to interpret the AZT effect for a particular race by comparing the estimated probability of AIDS symptoms for black veterans with and without immediate AZT use. c. The se value was 0.279 for the AZT use effect. Does AZT use have a significant effect, at the 0.05 significance level? Show all steps of a test to justify your answer.

Multiple regression is used to model \(y=\) annual income using \(x_{1}=\) number of years of education and \(x_{2}=\) number of years employed in current job. a. It is possible that the coefficient of \(x_{2}\) is positive in a bivariate regression but negative in multiple regression. b. It is possible that the correlation between \(y\) and \(x_{1}\) is 0.30 and the multiple correlation between \(y\) and \(x_{1}\) and \(x_{2}\) is 0.26 . c. If the \(F\) statistic for \(\mathrm{H}_{0}: \beta_{1}=\beta_{2}=0\) has a \(\mathrm{P}\) -value \(=0.001,\) then we can conclude that both predictors have an effect on annual income. d. If \(\beta_{2}=0,\) then annual income is independent of \(x_{2}\) in bivariate regression.

For each of the following statements, indicate whether it is true or false. If false, explain why it is false. In regression analysis: a. The estimated coefficient of \(x_{1}\) can be positive in the bivariate model but negative in a multiple regression model. b. When a model is refitted after \(y=\) income is changed from dollars to euros, \(R^{2},\) the correlation between \(y\) and \(x_{1}\), the \(F\) statistics and \(t\) statistics will not change. c. If \(r^{2}=0.6\) between \(y\) and \(x_{1}\) and if \(r^{2}=0.6\) between \(y\) and \(x_{2}\), then for the multiple regression model with both predictors \(R^{2}=1.2\) d. The multiple correlation between \(y\) and \(\hat{y}\) can equal -0.40 .
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation
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.