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Chapter 13: Problem 1

Predicting weight For a study of female college athletes, the prediction equation relating \(y=\) total body weight (in pounds) to \(x_{1}=\) height (in inches) and \(x_{2}=\) percent body fat is \(\hat{y}=-121+3.50 x_{1}+1.35 x_{2}\) a. Find the predicted total body weight for a female athlete at the mean values of 66 and 18 for \(x_{1}\) and \(x_{2}\). b. An athlete with \(x_{1}=66\) and \(x_{2}=18\) has actual weight \(y=115\) pounds. Find the residual and interpret it.

Short Answer

Expert verified
Predicted weight is 134.3 pounds; residual is -19.3 pounds, meaning the athlete weighs 19.3 pounds less than predicted.

Step by step solution

01

Apply the Prediction Equation

To find the predicted total body weight, use the prediction equation \( \hat{y} = -121 + 3.50x_{1} + 1.35x_{2} \). Substitute the mean values given for \( x_{1} = 66 \) and \( x_{2} = 18 \) into the equation:\[\hat{y} = -121 + 3.50 \times 66 + 1.35 \times 18\]
02

Perform the Calculations

Calculate each component of the prediction equation:1. \( 3.50 \times 66 = 231 \)2. \( 1.35 \times 18 = 24.3 \)Substitute these values back into the equation:\[\hat{y} = -121 + 231 + 24.3\]
03

Solve for the Predicted Weight

Add the calculated values to find the predicted weight:\[\hat{y} = -121 + 231 + 24.3 = 134.3\]So, the predicted total body weight for the athlete is 134.3 pounds.
04

Calculate the Residual

The residual is the difference between the actual weight and the predicted weight. Given that the actual weight \( y \) is 115 pounds, the residual \( e \) is calculated as:\[e = y - \hat{y} = 115 - 134.3\]
05

Interpret the Residual

Perform the calculation:\[e = 115 - 134.3 = -19.3\]The negative residual of -19.3 indicates that the athlete's actual weight is 19.3 pounds less than the weight predicted by the equation.

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Key Concepts

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

Prediction Equation
The prediction equation in linear regression is a mathematical formula that helps to estimate the dependent variable based on one or more independent variables. It is key to making informed predictions in statistical modeling. In our exercise, the equation is \[\hat{y} = -121 + 3.50x_{1} + 1.35x_{2}\]where \(\hat{y}\) is the predicted total body weight, \(x_{1}\) is the height in inches, and \(x_{2}\) is the percent body fat.

To use the prediction equation:
  • Substitute the given values for each variable. In our case, these are the average values: \(x_{1} = 66\) and \(x_{2} = 18\).
  • Calculate the equation step-by-step by multiplying and then adding each component.
  • The result gives you the predicted value of the dependent variable鈥攊n this instance, the total body weight.

The prediction equation forms the basis of many predictive analytics, playing a crucial role in decision making processes.
Residuals
Residuals in linear regression are the differences between the observed values and the values predicted by the model. They show how much the prediction deviates from the actual data. This is invaluable for assessing the accuracy and performance of a prediction model.In our example, the actual weight of the athlete is 115 pounds, while the predicted weight is 134.3 pounds. Thus, the residual is calculated as:\[e = y - \hat{y} = 115 - 134.3 = -19.3\]A negative residual like our -19.3 indicates that the actual value is less than the predicted value. This further helps in interpreting the model's performance:
  • A large absolute residual value suggests a significant deviation from the prediction.
  • Positive residuals indicate underestimation by the model, while negative ones indicate overestimation.
Residual analysis can highlight potential biases in the prediction equation.
Statistical Modeling
Statistical modeling is a broad concept that encompasses various techniques used to understand and predict data patterns. Linear regression, as discussed here, is a fundamental statistical modeling technique that is widely used due to its simplicity and interpretability. With linear regression, relationships between variables are explored through equations like our prediction equation. This technique has several merits:
  • It helps identify and quantify relationships among variables.
  • It can be used for forecasting by predicting an outcome variable based on one or more predictor variables.
  • The output can be easily interpreted to understand the strength and direction of relationships.
Overall, statistical modeling is crucial in turning data into actionable insights by validating hypotheses through mathematical formulations.

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Most popular questions from this chapter

Comparable number of bedrooms and house size effects In Example \(2,\) the prediction equation between \(y=\) selling price and \(x_{1}=\) house size and \(x_{2}=\) number of bedrooms was \(\hat{y}=60,102+63.0 x_{1}+15,170 x_{2}\) a. For fixed number of bedrooms, how much is the house selling price predicted to increase for each square foot increase in house size? Why? b. For a fixed house size of 2000 square feet, how does the predicted selling price change for two, three, and four bedrooms?

Predicting restaurant revenue An Italian restaurant keeps monthly records of its total revenue, expenditure on advertising, prices of its own menu items, and the prices of its competitors' menu items.a. Specify notation and formulate a multiple regression equation for predicting the monthly revenue using the available data. Explain how to interpret the parameters in the equation. b. State the null hypothesis that you would test if you want to analyze whether advertising is helpful, for the given prices of items in the restaurant's own menu and the prices of its competitors' menu items. c. State the null hypothesis that you would test if you want to analyze whether at least one of the predictors has some effect on monthly revenue.

Controlling has an effect The slope of \(x_{1}\) is not the same for multiple linear regression of \(y\) on \(x_{1}\) and \(x_{2}\) as compared to simple linear regression of \(y\) on \(x_{1},\) where \(x_{1}\) is the only predictor. Explain why you would expect this to be true. Does the statement change when \(x_{1}\) and \(x_{2}\) are uncorrelated?

A chain restaurant that specializes in selling pizza wants to analyze how \(y=\) sales for a customer (the total amount spent by a customer on food and beverage, in pounds) depends on the location of the restaurant, which is classified as inner city, suburbia, or at an interstate exit. a. Construct indicator variables \(x_{1}\) for inner city and \(x_{2}\) for suburbia so you can include location in a regression equation for predicting the sales. b. For part a, suppose \(\hat{y}=6.9+1.2 x_{1}+0.5 x_{2} .\) Find the difference between the estimated mean sales at inner-city locations and at interstate exits.

When we use multiple regression, what is the purpose of performing a residual analysis? Why is it better to work with standardized residuals than unstandardized residuals to detect outliers?
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