Question 9:
A car dealer wants to find the relationships of car prices related to 3 variables: “Year”, “Mileage” and “Horsepower”. Here is the Excel Output results with 30 observations:
Coefficient |
Standard Error |
t-stat |
p-value |
||
Intercept |
2501198.34 |
931631 |
2.684 |
0.012 |
|
Year |
-1205.23 |
462.62 |
-2.60 |
0.014 |
|
Mileage |
-0.850 |
0.23 |
-3.62 |
0.001 |
|
Horsepower |
56.67 |
30.63 |
1.95 |
0.04 |
What is the 99% confidence interval for the slope of Horsepower?
(a) (56.67-2.756*30.63, 56.67+2.756*30.63)
(b) (56.67-2.771*30.63, 56.67+2.771*30.63)
(c) (56.67-2.763*30.63, 56.67+2.763*30.63)
(d) (56.67-2.779*30.63, 56.67+2.779*30.63)
Solution: Here we have given the Excel output for 30 observations.
We have to find the 99% confidence interval for the slope of Horsepower.
Confidence interval for slope= (Coefficient Critical value*Standard error)
Coefficient = 56.67, Standard error =30.63, n=30, =0.99
Degrees of freedom = n-2 = 30- 2 =28 ( In regression analysis t test n-2 degrees of freedom use)
Critical value for 28 degrees of freedom and 0.01 region =2.763----------(from t table)
So the option is c
(c) (56.67-2.763*30.63, 56.67+2.763*30.63)
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