Answer
Here we have n = 11, x̅ = 1.173, s = 0.461, df = 10
For a 90% confidence level, using Table C we have:
t = 1.812
$s_{x̅} = \frac{0.461}{\sqrt 11} = 0.14$
Lower Bound = 1.173 - (1.812)(0.14) = 0.92
Upper Bound = 1.173 + (1.812)(0.14) = 1.42
We can use the interval for inference as the given data falls in this interval except 2.0.
Work Step by Step
Given above
