Question

Determine the p-value for each of the following situations. (Give your answer bounds exactly.)

(a) H_a: β₁ > 0, with n = 15 and t = 1.23
_____ < p < _____

(b) H_a: β₁ ≠ 0, with n = 25, b₁ = 0.3, and s_b1 = 0.11
____ < p < _____

(c) H_a: β₁ < 0, with n = 18, b₁ = -1.55, and s_b1 = 0.73
____< p < ____

Answer 1

• (a) Ha: β1 > 0, with n = 15 and t = 1.23
  _0.10____ < p < _0.15____
  
  (b) Ha: β1 ≠ 0, with n = 25, b1 = 0.3, and sb1 = 0.11
  _0.010___ < p < _0.015____
  
  (c) Ha: β1 < 0, with n = 18, b1 = -1.55, and sb1 = 0.73
  _0.025___< p < ___0.05_

(a)

Ha: β1 > 0, with n = 15 and t = 1.23

• Degrees of freedom = n - 2 = 15 - 2 = 13

  One-tailed test.

• From t table

  1.08 < 1.23 < 1.35

  converting to p-value

  0.15 > p( t > 1.23) > 0.1
  0.10 < p-value < 0.15

(b)

Ha: β1 ≠ 0, with n = 25, b1 = 0.3, and sb1 = 0.11

1. Two-tailed test.

   Test statistic is t= 0.3/0.11 =2.7272

   Degrees of freedom = n - 2 = 25 - 2 = 23

From t table

2.62< 2.7272 <2.81

converting to p-value

0.015 > p( t > 2.7272) > 0.010

0.01 < P-value < 0.015

c)

Ha: β1 < 0, with n = 18, b1 = -1.55, and sb1 = 0.73
One taled test.

t=-1.55/0.73= -2.1232

Degrees of freedom = n - 2 = 18 - 2 = 16

From t table
-2.12 > -2.1232 >-2.47

converting to p-value

0.05 > p( t < - 2.1232) > 0.025

0.025 < P-value < 0.05