Question

Suppose you are interested in buying a new Lincoln Navigator or Town Car. You are standing on the sales lot looking at a model with different options. The list price is on the vehicle. As a salesperson approaches, you wonder what the dealer invoice price is for this model with its options. The following data are based on a random selection o these cars of different models and options. Let y be the dealer invoice in thousands of dollars) for the given vehicle. 32.4 34.3 36.1 44.0 47.8 L 30.4 31.8 32.0 42.1 42.2 (a) Verify that Ex = 194.6, Ly = 178.5, Ex2 = 7750.3, Ey2 = 6512.65, Exy = 7100.46, and r2 0.974. Ey Ex2 Exy (b) Use a 10% level of significance to test the claim that p > 0. (Use 2 decimal places.) critical t Conclusion O Reject the null hypothesis, there is sufficient evidence that p > 0. Reject the null hypothesis, there is insufficient evidence that p > 0. Fail to reject the null hypothesis, there is insufficient evidence that p > 0. Fail to reject the null hypothesis, there is sufficient evidence that p > 0. (c) Verify that se 1.5417, a 1.9029, and b 0.8684. Sa b (d) Find the predicted dealer invoice when the list price is x = 44 (thousand dollars). (Use 2 decimal places.) Find a 95% confidence interval for y when x = 44 (thousand dollars). (Use 2 decimal place.) lower limit upper limit Use a 10% level of significance to test the claim that ß > 0. (Use 2 decimal places.) critical t Conclusion O Reject the null hypothesis, there is sufficient evidence that 8 > 0. Reject the null hypothesis, there is insufficient evidence that 8 > 0. O Fail to reject the null hypothesis, there is insufficient evidence that B > 0. Fail to reject the null hypothesis, there is sufficient evidence that B > 0.

Answer 1

S.NO X XY 1 2 3 32.4 34.3 36.1 ΔΑ 47.8 194.6 Υ 30.4 31.8 32 42.1 42.2 178.5 Y, 924.1600 1011.2400 1024.0000 1772.4100 1780.8400 6512.6500 ΣΥ 1049.7600 1176.4900 1303.2100 1936.0000 2284.8400 7750.3000 ΣΧ 984.96 1090.74 1155.20 1852.40 2017.16 7100.4600 4 5 Total Mean 38.920 35.7000 ΣΧΥ

	n=	5.0000
	X̅=ΣX/n	38.92
	Y̅=ΣY/n	35.70
	sx=(√(Σx2-(Σx)2/n)/(n-1))=	6.64
	sy=(√(Σy2-(Σy)2/n)/(n-1))=	5.9203
	Cov=sxy=(ΣXY-(ΣXΣY)/n)/(n-1)=	38.3100
	r=Cov/(Sx*Sy)=	0.9742
	slope= β̂1 =r*Sy/Sx=	0.8684
	intercept= β̂0 ='y̅-β1x̅=	1.9029

a)

ΣX =	194.600
ΣY=	178.500
ΣX2 =	7750.300
ΣY2 =	6512.650
ΣXY =	7100.460
r =	0.974

b)

t=r*(√(n-2)/(1-r2))=	7.48
critical t =	1.64

Reject the null hypothesis, there is sufficient evidence that p > 0.

c)

Se =√(SSE/(n-2))=	1.5417
a=	1.9029
b=	0.8684

d)

predicted val=1.903+44*0.868=

40.11

standard error of PI=s*√(1+1/n+(x0-x̅)2/Sxx)=		1.7888
for 95 % CI value of t =	3.182
margin of error E=t*std error   =		5.693
lower confidence bound=xo-E =		34.42
Upper confidence bound=xo+E=		45.80

t=r*(√(n-2)/(1-r2))=	7.48
critical t =	1.64

O Reject the null hypothesis, there is sufficient evidence that B > 0.