Question

A surgical practice reported the results of a regression analysis, designed to predict the monthly income the surgeons bring into the practice (Y) – measured in thousands of dollars. One independent variable used to predict monthly income for the surgical practice is the number of new patients (X) enrolled by each surgeon. The firm samples 12 surgeons and determines for each the number of new patients they have enrolled monthly and the amount in thousands of dollars they have brought into the practice. The data were used to fit a linear model. The results of the simple linear regression are provided below.

        Y = 17.7 + 1.12X; S_YX =$5.804; 2 – tailed p value = 0.000126 (for testing ß₁);                

           S_b1=0.185;    X = 25.083; SSX=Σ( X_i –X )²=980.917;   n=12 ;

    Suppose the general manager wants to obtain a 99% confidence interval estimate for the monthly income

    contribution to the practice by surgeons based on their number of new patients. This confidence interval

    estimate is:     

   

		(0.5327 , 1.7073)
		(1.3935 , 1.9465)
		(0.6079 , 1.6321)
		(1.3302 , 2.0498)

Answer 1

The answer is (0.5327, 1.7073).

b1	1.12
se	0.185
t	3.169273
Lower Limit	0.5327
Upper Limit	1.7073

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