Question

In: Statistics and Probability

The residuals for 15 consecutive time periods from a simple linear regression with one independent variable...

The residuals for 15 consecutive time periods from a simple linear regression with one independent variable are given in the following table.

Time_Period   Residual
1   +4
2   -6
3   -1
4   -5
5   +3
6   +6
7   -3
8   +7
9   +7
10   -3
11   +2
12   +3
13   0
14   -5
15   -7

B) Compute the​ Durbin-Watson statistic. At the 0.05 level of​significance, is there evidence of positive autocorrelation among the​ residuals?

The​ Durbin-Watson statistic is D=

​(Round to three decimal places as​ needed.)

What are the critical values for this test?

dL=   

dU=     

(Round to three decimal places as needed)

At the 0.05 level of​ significance, is there evidence of positive autocorrelation among the​ residuals?

A.Yes comma because the value of Upper D is less than nbspYes, because the value of D is less than d Subscript Upper LdL.

B.​No, because the value of D is greater than d Subscript Upper LdL.

C.​Yes, because the value of D is less than d Subscript Upper UdU.

D.No comma because the value of Upper D is greater than nbspNo, because the value of D is greater than d Subscript Upper UdU.

E.The test is inconclusive.

c. Based on​ (a) and​ (b), what conclusion can you reach about the autocorrelation of the​ residuals?

A.There appears to be strong negative autocorrelation among the residuals.

B.There appears to be positive autocorrelation among the residuals.

C.There appears to be positive and negative autocorrelation among the residuals.

D.There does not appear to be autocorrelation among the residuals.

Solutions

Expert Solution

t e (et- e_t-1)^2 e_t^2
1 4 16
2 -6 100 36
3 -1 25 1
4 -5 16 25
5 3 64 9
6 6 9 36
7 -3 81 9
8 7 100 49
9 7 0 49
10 -3 100 9
11 2 25 4
12 3 1 9
13 0 9 0
14 -5 25 25
15 -7 4 49
559 326

d = 559/326

= 1.715

k = 1

n = 15

dL = 1.08

dU = 1.36

d = 1.715 > dU(1.36)

hence there is no statistical evidence that the error terms are positively autocorrelated.

option D) is correct

c)

4-d = 4-1.715 = 2.285

since 4-d > dU

there is no statistical evidence that the error terms are negatively autocorrelated.

D.There does not appear to be autocorrelation among the residuals.


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