Question

# QUESTION 1 For a combination of α (level of significance) = 0.05, n (sample size) =...

QUESTION 1

1. For a combination of α (level of significance) = 0.05, n (sample size) = 25, k (number of independent variables in the model) = 1 and D (Durbin-Watson statistic) = 3.30 , what statistical decision should be made when testing the null hypothesis of no negative autocorrelation?

 a. Neither reject nor not reject the null hypothesis. b. Do not reject the null hypothesis. c. Accept the null hypothesis. d. Reject the null hypothesis

1 points

QUESTION 2

1. When testing the null hypothesis of no negative autocorrelation, you reject the null hypothesis if

 a. the Durbin-Watson statistic is greater than 4-dL b. the Durbin-Watson statistic is less than 4-dU c. the Durbin-Watson statistic is less than 4-dL d. the Durbin-Watson statistic is between 4-dU and 4-dL

1 points

QUESTION 3

1. A sample of 12 observations collected in a multiple regression study on two independent variables (X1 and X2) shows the following partial results. SST = 12321 and SSE = 3516. If SSR(X1) = 4496 and SSR(X2) = 5378, and you would like to test the null hypothesis that variable X1 does not significantly improve the model after variable X2 has been included, what would be the critical value of F at the 1% level of significance? Use our textbook statistical table to answer the question.

1 points

QUESTION 4

1. A sample of 12 observations collected in a multiple regression study on two independent variables (X1 and X2) shows the following partial results. SST = 12321 and SSE = 3516. If SSR(X1) = 4496 and SSR(X2) = 5378, and you would like to test the null hypothesis that variable X1 does not significantly improve the model after variable X2 has been included, what would be the critical value of F at the 5% level of significance? Use our textbook statistical table to answer the question.

1 points

QUESTION 5

1. A sample of 12 observations collected in a multiple regression study on two independent variables (X1 and X2) shows the following partial results. SST = 12321 and SSE = 3516. If SSR(X1) = 4496 and SSR(X2) = 5378, what would be the value of SSR(X2|X1)?

1 points

QUESTION 6

1. Assume you have two independent variables and you found that the coefficient of determination between the two independent variables is 0.95. Based on this information, there no reason to suspect the existence of collinearity. Use the Variance Inflationary Factor (VIF) to answer the question.

True

False

1 points

QUESTION 7

1. Assume you have two independent variables and you found that the coefficient of determination between the two independent variables is 0.40. Based on this information, what is the Variance Inflationary Factor (VIF)? Round your final result to two decimal places.

QUESTION 8

1. For a sample of 30 Australian cities, a sociologist studies the crime rate in each city (crime per 100,000 residents) as a function of its poverty rate (in %) and its average income (in \$1,000). A portion of the regression results shows that the coefficients for poverty and average income are 54.22 and 5.10, respectively. Based on this information, what is the upper critical value used to test the null hypothesis that X1 has no significant effect on Y at the 1% level of significance? Answer this question using our textbook statistical table. Note: Y = crime rate in each city (crime per 100,000 residents); X1 = poverty rate (in %); and X2 = average income (in \$1,000).

Solution :

1) The decision rule for Durbin Watson test, when testing for negative autocorrelation is as follows :

If (4 - D) < D(L,α) then we reject the null hypothesis of no negative autocorrelation.

If (4 - D) > D(U,α) then we fail to reject the null hypothesis of no negative autocorrelation.

If D(L,α) < (4 - D) < D(U)  then the test is inconclusive.

(Where, D is the value of the test statistic, D(L,α) and D(U,α) are critical values.)

For n = 25 and k = 1, the critical values at α = 0.05 will be as follows :

D(L,α) = 1.288, D(U) = 1.454

And we have, D = 3.30

Hence, (4 - D) = (4 - 3.30) = 0.70

0.70 < 1.288

Since, (4 - D) < D(L,α) therefore, we shall reject the null hypothesis at 0.05 significance level.

Option (d) is correct.

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