The Null Hypothesis - H0: β0 = 0, H0: β₁ = 0, H0: β₂ = 0, Β_i = 0

The Alternative Hypothesis - H1: β0 ≠ 0, H0: β₁ ≠ 0, H0: β₂ ≠ 0, Β_i ≠ 0      _{i =0, 1, 2, 3}

Reject H0 if |t | > t?¹ = 1.96153 or Reject H0 when P-value ≤ α = 0.05

Predictor         Coef          SE Coef            T       P          VIF

Constant       0.37794     0.01369        27.61   0.000

totexp     -0.00119745    0.00006058  -19.77  0.000  1.272

income     -0.00007625   0.00004302  -1.77    0.077  1.282

age          0.0016660      0.0003076    5.42     0.000  1.062

nk            0.029515        0.004765      6.19     0.000  1.005

 Inverse Cumulative Distribution Function

Student's t distribution with 1514 DF

P( X <= x )        x

      0.975  1.96153

Since the constant = 27.61 > 1.96153 (CV) and the P-Value is ≤ α = 0.05, H0 would be rejected as there is sufficient evidence.      

Since the totexp = -19.77 < 1.96153 (CV) and the P-Value is ≤ α = 0.05, there is evidence to suggest that H0 should be accepted according to the T value however the P-Value suggest otherwise but it is not significant.

Since the income = -1.77 < 1.96153 (CV) and the P-Value is ≤ α = 0.05, there is evidence to suggest that H0 should be accepted according to the T value however the P-Value suggest otherwise but it is not significant.

Since the age = 5.42 > 1.96153 (CV) and the P-Value is ≤ α = 0.05, H0 would be rejected as there is sufficient evidence.      

Since the nk = 6.19 > 1.96153 (CV) and the P-Value is ≤ α = 0.05, H0 would be rejected as there is sufficient evidence.