Question

Based on a sample of n = 20, the least-squares method was used to develop the following prediction line: Y_i = 5 + 3X_i

In addition,

S_YX = 1.0

$\bar{X}$ = 2

$\bg_white \fn_jvn \sum_{i = 1}^{n} (X_{I}-\bar{X})^2 = 20$

a) Construct a 95% confidence interval estimate of the population mean response for X = 2.

b) Construct a 95% prediction interval of an individual response for x = 2.

Answer 1

n=	20
bo=	5
b1=	3.0000
sxy =√MSE=	1.0000
Sxx=(n-1)sx^2=	20
x̅ =	2

a)

x   =	2
predcited value at X=2:		11.0000
std error confidence interval=		s*√(1/n+(x0-x̅)2/Sxx)		=	0.2236
for 95 % CI value of t=					2.1010
margin of error E=t*std error                            =					0.4698
lower confidence bound=sample mean-margin of error =					10.5302
Upper confidence bound=sample mean+margin of error=					11.4698

95% confidence interval =(10.5302 , 11.4698) (please use decimals as required in question)

b)

x0             =	2
predcited value at X=2:		11.00
std error prediction interval=		s*√(1+1/n+(x0-x̅)2/Sxx)		=	1.0247
for 95 % CI value of t=					2.1010
margin of error E=t*std error                            =					2.15
lower prediction bound=sample mean-margin of error =					8.8471
Upper prediction bound=sample mean+margin of error=					13.1529

95% prediction interval =(8.8471 , 13.1529)