SUMMARY OUTPUT cov (X,Y) -2487708.147 Regression Statistics Sx 6572.761629 Multi

Question

SUMMARY OUTPUT










cov (X,Y) -2487708.147


Regression Statistics

Sx 6572.761629


Multiple R 0.793745367

Sy 476.837435


R Square






Adjusted R Square 0.62625652






Standard Error






Observations 100















ANOVA







df SS MS F Significance F


Regression 1 14182026.33 14182026.33


Residual 98 8327993.674 84979.52728




Total 99 22510020











Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 80.0% Upper 80.0%
Intercept 16887.9327 163.0279 103.5892 6.0666E-102 16564.4091 17211.4563 16677.5860 17098.27946
Odometer -0.05758 -0.06643 -0.04874 -0.063335537 -0.05183295

1 SUMMARY OUTPUT cov (X,Y)-2487708.1 6572.76163 476.837435 Regression Statistics 4 Multiple R 0.79374537 5 R Square 6 Adjusted R S 0.62625652 7 Standard Error 8 Observations 100 10 ANOVA df MS Significance F 1 14182026.33 14182026.3 98 8327993.674 84979.5273 9922510020 12 Regression 13 Residual 14 Total 15 16 17 Intercept 18 Odometer er 80.0% Upper 80.0% 163.0279 103.5892 6.067E-10216564.4091 17211.4563 16677.5860 17098.2795 0.06643-0.04874 -0.0633355 0.051833 Coefficients Standard Error tStat P-value Lower 95% Upper 95% Low 16887.9327 -0.05758

Explanation / Answer

Solution

Let x = odometer reading and y = used car price.

The regression equation is: y = + x.

Least square estimates: cap and cap

Given cap = 16887.9327 (intercept) and cap = - 0.05758 (odometer)

Part (1) Value of R2 = 0.793745372 = 0.6300 (rounded to 4 decimal places) ANSWER

Part (2) Standard error of the estimate (cap) = 0.004455 ANSWER

Details of computations:

100(1 - )% Confidence Interval (CI) for = cap ± {SE(b) x tn – 2,/2}, where n = number of observations and

tn – 2,/2 = upper (/2)% point of t-distribution with degrees of freedom = n – 2.

Given 95% CI for = (- 0.06643, - 0.04874), n = 100

t98, 0.025 = 1.9845 [using Excel Function]

So, - 0.04874 – (cap) = {SE(cap) x 1.9845

i.e., - 0.04874 – (- 0.05758) = {SE(cap) x 1.9845

Or, SE(cap) = 0.00884/1.9845 = 0.004455

Part (3)Test statistic for testing slope is non-zero: t = (cap - 0)/SE(cap) which is as tn – 2

Value of t = - 0.05758/0.004455 = - 12.9249 ANSWER

Part (4) p-value of t

Since t ~ t98, p-value = P(t98 > |t-value|) = P(t98 > | - 12.9249| ) = 6.76E-23 ANSWER [using Excel Function]

Part (5) Conclusion at 5% level of significance

Since p-value is much less than 0.05, there is enough evidence to conclude that slope is non-zero

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