Answer
We reject the Null hypothesis.
Work Step by Step
Here we have α = 0.05 and n = 16. df = 16 - 1 = 15
Using the classical approach, we have:
$t = \frac{\bar d - µ_{d}}{s_{d}/\sqrt n}$
$ = \frac{1.14 - 0}{1.75/\sqrt 16} = \frac{1.14}{0.4375} = 2.606$
Using Student's t distribution table, we have:
t = 1.753
Hence the rejection region will be all values greater than 1.753
Since 2.606 > 1.753 , we will reject the Null Hypothesis.
