Question

1.

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 of these cars of different models and options. Let y be the dealer invoice (in thousands of dollars) for the given vehicle.

x	31.0	34.3	36.1	44.0	47.8
y	30.2	31.4	32.0	42.1	42.2

(a) Verify that Σx = 193.2, Σy = 177.9, Σx² = 7661.54, Σy² = 6475.25, Σxy = 7037.98, and r ≈ 0.970.

Σx	=193.2
Σy	177.9
Σx2	=7661.54
Σy2	=6475.25
Σxy	=7037.98
r	=.97

(b) Use a 5% level of significance to test the claim that ρ > 0. (Use 2 decimal places.)

t
critical t

Conclusion

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

Reject the null hypothesis, there is insufficient evidence that ρ > 0.

Fail to reject the null hypothesis, there is insufficient evidence that ρ > 0.

Fail to reject the null hypothesis, there is sufficient evidence that ρ > 0.

(c) Verify that S_e ≈ 1.7005, a ≈ 3.3116, and b ≈ 0.8351.

Se	=1.7005
a	=3.3116
b	=.8351

(d) Find the predicted dealer invoice when the list price is x = 33 (thousand dollars). (Use 2 decimal places.)


(e) Find a 99% confidence interval for y when x = 33 (thousand dollars). (Use 2 decimal place.)

lower limit
upper limit

(f) Use a 5% level of significance to test the claim that β > 0. (Use 2 decimal places.)

t
critical t

Conclusion

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

Reject the null hypothesis, there is insufficient evidence that β > 0.

Fail to reject the null hypothesis, there is insufficient evidence that β > 0.

Fail to reject the null hypothesis, there is sufficient evidence that β > 0.

(g) Find a 99% confidence interval for β and interpret its meaning. (Use 3 decimal places.)

lower limit
upper limit

2. Let x be the number of different research programs, and let y be the mean number of patents per program. As in any business, a company can spread itself too thin. For example, too many research programs might lead to a decline in overall research productivity. The following data are for a collection of pharmaceutical companies and their research programs.

x	10	12	14	16	18	20
y	1.9	1.4	1.6	1.4	1.0	0.7

Complete parts (a) through (e), given Σx = 90, Σy = 8, Σx² = 1420, Σy² = 11.58, Σxy = 112.6, and

r ≈ −0.925.

(a) Draw a scatter diagram displaying the data.

I have the graph already completed.

(b) Verify the given sums Σx, Σy, Σx², Σy², Σxy, and the value of the sample correlation coefficient r. (Round your value for r to three decimal places.)

Σx =
Σy =
Σx2 =
Σy2 =
Σxy =
r =

(c) Find x, and y. Then find the equation of the least-squares line  = a + bx. (Round your answers for x and y to two decimal places. Round your answers for a and b to three decimal places.)

x	=
y	=
	=	+   x

(d) Find the value of the coefficient of determination r². What percentage of the variation in y can be explained by the corresponding variation in x and the least-squares line? What percentage is unexplained? (Round your answer for r² to three decimal places. Round your answers for the percentages to one decimal place.)

r2 =
explained	%
unexplained	%

(e) Suppose a pharmaceutical company has 13 different research programs. What does the least-squares equation forecast for y = mean number of patents per program? (Round your answer to two decimal places.)
patents per program

Answer 1

1)

S.No х y XY 1 2 3 31 34.3 36.1 44 47.8 193.2 30.2 31.4 32 42.1 42.2 177.9 X? 961.0000 1176.4900 1303.2100 1936.0000 2284.8400 7661.5400 YA? 912.0400 985.9600 1024.0000 1772.4100 1780.8400 6475.2500 936.20 1077.02 1155.20 1852.40 2017.16 7037.9800 4 5 Total Mean 38.640 35.5800 ΣΧΥ

	n=	5.0000
	X̅=ΣX/n	38.64
	Y̅=ΣY/n	35.58
	sx=(√(Σx2-(Σx)2/n)/(n-1))=	7.01
	sy=(√(Σy2-(Σy)2/n)/(n-1))=	6.0326
	Cov=sxy=(ΣXY-(ΣXΣY)/n)/(n-1)=	40.9810
	r=Cov/(Sx*Sy)=	0.9697
	slope= β̂1 =r*Sy/Sx=	0.8351
	intercept= β̂0 ='y̅-β1x̅=	3.3116

a)

ΣX =	193.200
ΣY=	177.900
ΣX2 =	7661.540
ΣY2 =	6475.250
ΣXY =	7037.980
r =	0.970

b)

t=r*(√(n-2)/(1-r2))=	6.88
critical t =	2.35

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

c)

Se =√(SSE/(n-2))=	1.7005
a=	3.3116
b=	0.8351

d)

predicted val=3.312+33*0.835=

30.87

e)

standard error of PI=s*√(1+1/n+(x0-x̅)2/Sxx)=		1.9845
for 99 % CI value of t =	5.841
margin of error E=t*std error   =		11.592
lower confidence bound=xo-E =		19.28
Upper confidence bound=xo+E=		42.46

f)

t=r*(√(n-2)/(1-r2))=	6.88
critical t =	2.35

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

g)

std error of slope =se(β1) =s/√Sxx=		0.1214
for 90 % CI value of t =	2.353
margin of error E=t*std error   =		0.286
lower confidence bound=xo-E =		0.549
Upper confidence bound=xo+E=		1.121