Question

You may need to use the appropriate appendix table or technology to answer this question.

Given are five observations for two variables, x and y. (Round your answers to two decimal places.)

xi	1	2	3	4	5
yi	3	7	6	11	14

(a)

Use s_ŷ* = s

1 n + (x* − x)2 Σ(xi − x)2

to estimate the standard deviation of

ŷ*

when

x = 4.

(b)

Use

ŷ* ± t_α/2s_ŷ*

to develop a 95% confidence interval for the expected value of y when

x = 4.

to

(c)

Use s_pred = s

1 + 1 n + (x* − x)2 Σ(xi − x)2

to estimate the standard deviation of an individual value of y when

x = 4.

(d)

Use

ŷ* ± t_α/2s_pred

to develop a 95% prediction interval for y when

x = 4.

Answer 1

SSE =Syy-(Sxy)2/Sxx=

7.2000

	s2 =SSE/(n-2)=	2.4000
std error σ              =	=se =√s2=	1.5492

predicted value at X=4 is:2.6*4+0.4=

10.800

a)

std error of CI=s*√(1/n+(x0-x̅)2/Sxx)=

0.85

b)

for 95 % CI value of t=		3.182
margin of error E=t*std error=		2.700
lower confidence bound=xo-E=		8.10
Upper confidence bound=xo+E=		13.50

95% Confidence interval =(8.10 , 13.50)

c)

std error of PI=s*√(1+1/n+(x0-x̅)2/Sxx)=

1.77

d)

for 95 % CI value of t=		3.182
margin of error E=t*std error=		5.621
lower confidence bound=xo-E=		5.18
Upper confidence bound=xo+E=		16.42

95% prediction interval =(5.18 , 16.42)