## Statistics (12th Edition)

$t = \frac{2}{\sqrt 2/\sqrt 6} = 2 / 0.577 = 3.464$ df = n - 1 = 6 - 1 = 5 $t_{α/2} = 0.025$ From student's t distribution table, we have: $t_{0.025} = 2.571$ Hence rejection region will contain values larger than 2.571 Since t > 2.571, we reject Null hypothesis