Question

In a study of long bones in a variety of mammalian species, a regression analysis of y = humerus bone circumference (cm) on x = femur bone circumference was performed. From their examination of the data the investigators felt that the assumptions of the linear regression model were satisfied. Summary quantities from this analysis yield: 7 = 76.031 n = 186 y=-2.772 + 1.0428x Sxx = 355634.92 Se = 6.701 Obtain a 95% prediction interval for a single humerus bone circumference when the femur bone circumference is 40 cm. Interpret your result. What is the lower limit for the CI?

Answer 1

x0             =	40
predcited value at X=40:	38.94
std error prediction interval=	s*√(1+1/n+(x0-x̅)2/Sxx)	=	6.7312
for 95 % CI value of t=					1.9730
margin of error E=t*std error                            =			13.28
lower prediction bound=sample mean-margin of error =		25.6594
Upper prediction bound=sample mean+margin of error=		52.2206

lower limit of the CI =25.6594

upper limit of the CI = 52.2206