#### Answer

Reject Null hypothesis

#### Work Step by Step

$t = \frac{2}{\sqrt 2/\sqrt 6} = 2 / 0.577 = 3.464$
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