Question

Total 10 Points Assignment 4 1. Consider the following equation where we report estimates along with their standard errors in brackets: sleep = 3683.25 -0.148totwrk - 11.13educ +2.20age (112.28) (0.017) (5.88) (1.45) n = 706, R2 = .113

Answer 1

For the estimate of totwrk:
Coefficient = -0.148
Standard error = 0.017

df = n - p - 1 = 706 - 3 - 1 = 702

t-critical value at alpha = 0.1 is 1.647
t-critical value at alpha = 0.01 is 2.583

Confidence interval at 90%:

CI: (Coeff. - SE*t, Coeff + SE*t))
CI: (-0.148 - 1.647*0.017, -0.148 + 1.647*0.017)
CI: (-0.148 - 0.028, -0.148 + 0.028)
CI: (-0.176, -0.12)

As the 90% CI does not contain the value of 0, we reject the null hypothesis at 10% level of significance. Hence, the coefficient is significantly different than zero.

Confidence interval at 99%:

CI: (Coeff. - SE*t, Coeff + SE*t))
CI: (-0.148 - 2.583*0.017, -0.148 + 2.583*0.017)
CI: (-0.148 - 0.044, -0.148 + 0.044)
CI: (-0.192, -0.104)

As the 99% CI does not contain the value of 0, we reject the null hypothesis at 1% level of significance. Hence, the coefficient is significantly different than zero.