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 + 1.79x 1 − 0.021x 2 − 0.061x 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 + 1.79x 1 − 0.021x 2 − 0.061x 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 + 1.79x 1 − 0.021x 2 − 0.061x 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