Answer
We will not reject the Null hypothesis. Hence, there is insufficient evidence to conclude that the part is not being manufactured according to the specifications.
Work Step by Step
Here we have μ = 1.3825, n = 10, x̅ = 1.3826, s = 0.0003 and α = 0.05
Using the Classical approach:
Using Table VI, we have: $t_{0.025} = 2.262$
$E = 2.262 \times \frac{0.0003}{\sqrt 10} = 0.0002 $
Lower Bound = x̅ - E = 1.3826 - 0.0002 = 1.3824
Upper Bound = x̅ + E = 1.3826 + 0.0002 = 1.3828
Since 1.3825 is in the confidence interval, we will not reject the Null hypothesis. Hence, there is insufficient evidence to conclude that the part is not being manufactured according to the specifications.
