Life Expectancy An actuary wanted to develop a model to predict how long individuals will live. After consulting a number of physicians, she collected the age at death ( y ), the average number of hours of exercise per week ( x ₁), the cholesterol level ( x ₂), and the number of points that the individual's blood pressure exceeded the recommended value ( x ₃). A random sample of 40 individuals was selected. The computer output of the multiple regression model is shown below. THE REGRESSION EQUATION IS y = 55.8 + 1.79 x ₁ - 0.021 x ₂ - 0.061 x ₃

S = 9.47 R - Sq = 22.5%
ANALYSIS OF VARIANCE {Life Expectancy Narrative} Is there enough evidence at the 5% significance level to infer that the model is useful in predicting length of life?

Question

Life Expectancy An actuary wanted to develop a model to predict how long individuals will live. After consulting a number of physicians, she collected the age at death ( y ), the average number of hours of exercise per week ( x ₁), the cholesterol level ( x ₂), and the number of points that the individual's blood pressure exceeded the recommended value ( x ₃). A random sample of 40 individuals was selected. The computer output of the multiple regression model is shown below. THE REGRESSION EQUATION IS y = 55.8 + 1.79 x ₁ - 0.021 x ₂ - 0.061 x ₃

S = 9.47 R - Sq = 22.5%
ANALYSIS OF VARIANCE

{Life Expectancy Narrative} Is there enough evidence at the 5% significance level to infer that the model is useful in predicting length of life?

Quizplus · Accepted Answer

The Answer of Life Expectancy An actuary wanted to develop a model to predict how long individuals will live. After consulting a number of physicians, she collected the age at death ( y ), the average number of hours of exercise per week ( x ₁), the cholesterol level ( x ₂), and the number of points that the individual's blood pressure exceeded the recommended value ( x ₃). A random sample of 40 individuals was selected. The computer output of the multiple regression model is shown below. THE REGRESSION EQUATION IS y = 55.8 + 1.79 x ₁ - 0.021 x ₂ - 0.061 x ₃ S = 9.47 R - Sq = 22.5% ANALYSIS OF VARIANCE {Life Expectancy Narrative} Is there enough evidence at the 5% significance level to infer that the model is useful in predicting length of life?

Life Expectancy an Actuary Wanted to Develop a Model to Predict