Question

Exhibit 15-6
Below you are given a partial Excel output based on a sample of 16 observations.

ANOVA
	df	SS	MS	F
Regression		4,853	2,426.5
Residual			485.3
	Coefficients	Standard Error
Intercept	12.924	4.425
x1	-3.682	2.630
x2	45.216	12.560

1. Refer to Exhibit 15-6. The estimated regression equation is

                a.             y = β0 + β1x1 + β2x2 + ε

                b.             E(y) = β0 + β1x1 + β2x2

                c.             y hat = 12.924 - 3.682 x1 + 45.216 x2

                d.             y hat = 4.425 + 2.63 x1 + 12.56 x2

2. Refer to Exhibit 15-6. The interpretation of the coefficient of x1 is that

                a.             a one unit change in x1 will lead to a 3.682 unit decrease in y

                b.             a one unit increase in x1 will lead to a 3.682 unit decrease in y when all other variables are held constant

                c.             a one unit increase in x1 will lead to a 3.682 unit decrease in x2 when all other variables are held constant

                d.             It is impossible to interpret the coefficient.

3. Refer to Exhibit 15-6. We want to test whether the parameter β1 is significant. The test statistic equals

                a.             -1.4

                b.             1.4

                c.             3.6

                d.             5

4. Refer to Exhibit 15-6. The t value obtained from the table which is used to test an individual parameter at the 1% level is

                a.             2.65

                b.             2.921

                c.             2.977

                d.             3.012

Answer 1

1.

We can clearly see the intercept is 12.924, coefficient of x1 is - 3.682 and coefficient of x2 is 45.216

So, correct option is option c)

2.

Since coefficient of x1 is negative so,

a one unit increase in x1 will lead to a 3.682 unit decrease in y when all other variables are held constant.

correct option is option b)

3.

Test statistic =-3.682/2.630 =-1.4

correct option is option a)

4.

Sample size =16

Dof =16-2-1=13

So, t value from table =t_/2=0.005,13 = 3.012

correct option is option d)

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