Question

Date	DOW			RUSSELL
8-Apr-20	23,719.37			1,246.73
1-Apr-20	22,653.86			1,139.17
25-Mar-20	21,917.16			1,153.10
18-Mar-20	20,704.91			1,096.54
11-Mar-20	21,237.38			1,106.51
4-Mar-20	25,018.16			1,350.90
26-Feb-20	25,917.41			1,486.08
19-Feb-20	27,081.36			1,571.90
12-Feb-20	29,232.19			1,683.52
5-Feb-20	29,276.34			1,677.51
29-Jan-20	28,807.63			1,656.77
22-Jan-20	28,722.85			1,658.31
15-Jan-20	29,196.04			1,685.90
8-Jan-20	28,939.67			1,675.74
1-Jan-20	28,583.68			1,658.31

a. Create a scatter plot. Say what you see.

b.  Use the data to develop an estimated regression equation showing how your team data is related to DOW, the Dow Jones industrial average. What is the estimated regression model (y-mx+b, slope & intercept)?
Let x represent the DOW indexes.

c.  How much of the variation in the sample values of your team data does the model estimated in part (b) explain?
Round your answer to two decimal places.

d. Suppose that the closing price for the DOW is 29,000. Estimate the closing price for your data

e. Preform a hypothesis test for the model (F test) with an significance of 0.05. State your conclusion

Answer 1

Ʃx =	391008.01
Ʃy =	21846.99
Ʃxy =	580536532.1717
Ʃx² =	10340354614.3419
Ʃy² =	32656195.8231
Sample size, n =	15
x̅ = Ʃx/n = 391008.01/15 =	26067.20067
y̅ = Ʃy/n = 21846.99/15 =	1456.466
SSxx = Ʃx² - (Ʃx)²/n = 10340354614.3419 - (391008.01)²/15 =	147870355.4
SSyy = Ʃy² - (Ʃy)²/n = 32656195.8231 - (21846.99)²/15 =	836797.6858
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 580536532.1717 - (391008.01)(21846.99)/15 =	11046659.88

a) Scatter plot:

b)

Slope, b = SSxy/SSxx = 11046659.87904/147870355.3979 = 0.074705

y-intercept, a = y̅ -b* x̅ = 1456.466 - (0.07471)*26067.20067 = -490.8851

Regression equation :

ŷ = (0.0747) x - 490.8851

c)

Coefficient of determination, r² = (SSxy)²/(SSxx*SSyy)

= (11046659.87904)²/(147870355.3979*836797.68576) = 0.9862

98.62% variation in y is explained by the least squares model.

d)

Predicted value of y at x = 29000

ŷ = -490.8851 + (0.0747) * 29000 = 1675.5609

e)

Null and alternative hypothesis:

Ho: β₁ = 0 ; Ha: β₁ ≠ 0

SSE = SSyy -SSxy²/SSxx = 836797.68576 - (11046659.87904)²/147870355.3979 = 11556.5875

SSR = SSxy²/SSxx = (11046659.87904)²/147870355.3979 = 825241.0982

Test statistic:

F = SSR/(SSE/(n-2)) = 825241.0982/(11556.5875/13) = 928.3133

Critical value F crit = F.INV.RT(0.05, 1, 13) = 4.667

p-value = F.DIST.RT(928.3133, 1, 13) = 0.0000

Conclusion:

p-value < α Reject the null hypothesis.