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 surgical practice wants to obtain a 99% prediction interval estimate for the mean monthly income contributed by    

    surgeons that have 18 new patients. Compute this prediction interval:

  

		(25.4330 , 31.8540)
		(18.2330 , 57.4201)
		(16.6710 , 51.914)
		(26.2130 , 29.8540)

Answer 1

n=	12
bo=	17.7
b1=	1.1200
sxy =√MSE=	5.8040
Sxx=(n-1)sx^2=	980.917
x̅ =	25.083

predcited value at X=18:	37.86
std error prediction interval=	s*√(1+1/n+(x0-x̅)2/Sxx)	=	6.1819
for 99 % CI value of t=					3.1693
margin of error E=t*std error                            =			19.59
lower prediction bound=sample mean-margin of error =		18.23
Upper prediction bound=sample mean+margin of error=		57.42

from above:

prediction interval (18.2330 , 57.4201)