Question

Please solve for (a)-(d)
below. The data in the accompanying table represent the rate of return of a certain company stock for 11 months, compared with the rate of return of a certain index of 500 stocks. Both are in percent. Complete parts (a) through Click the icon to view the data table. (a) Treating the rate of return of the index as the explanatory variable, x, use technology to determine the estimates of Bo and B1 The estimate of Bo is a (Round to four decimal places as needed.) The estimate of B, is (Round to four decimal places as needed.) (6) Assuming the residuals are normally distributed, test whether a linear relation exists between the rate of return of the index, x, and the rate of return for the company stock, y, at the a = 0.10 level of significance. Choose the correct answer below. State the null and alternative hypotheses

Answer 1

S.No х Y (y-7) 1 2 4.23 3.35 3.28 5.09 (x-x)(y-7) 11.7318 16.6022 3 4 -1.78 -3.2 1.29 0.54 2.88 2.69 un 6 7 8 9 3.58 1.48 -4.4 -0.86 7.41 -4.83 -2.38 2.37 -4.27 -3.77 9.01 (x-2)2 22.7269 15.1109 1.5444 7.0901 3.3389 16.9519 4.0694 14.9207 0.1042 31.1668 8.6596 125.6838 6.06 18.24 0.08 4.25 3.50 43.44 31.91 10.23 2.41 25.90 21.06 167.0727 SSY 0.3468 -5.4876 3.4187 27.1366 - 11.3958 12.3572 -0.5005 28.4110 13.5044 96.1248 -6.12 10 11 Total Mean -3.48 -5.91 -0.537 0.82 SSX SXY sample size n- 11 125.684 -0.5373 y = 0.8191 SOE (X-X)? Swy 2(X-X)(Y-Y)= slope B. = S/S intercepta ßo - 7-B.X= Least square line equation: 9 =1.23+0.7648*x 96.125 0.76481 Co 1.2300

a)

ß0 =1.2300

ß1 =0.7648

b)

correlation r='Sxy/(√Sxx*Syy) =		0.663
test stat t=	r*(√(n-2)/(1-r2))=	2.659
P value    =	0.026	(from excel:tdist(2.659,9,2)

O A. Reject Ho. There is sufficient evidence to conclude that a linear relation exists between the rate of return of the index and the rate of return of the company stock.

c)

std error of slope =se(β1) =s/√Sxx=

0.2876

for 95 % CI value of t=		2.262	(from excel:tinv(0.05,9)
margin of error E=t*std error =		0.6506
lower bound=b1+E=		0.1142
Upper bound=b1+E=		1.4154

d)

predicted val=1.23+3.15*0.7648=

3.639