Answer
Here we have n = 16, x̅ = 56.3, s = 77.3
$H_{o}: μ = 130$
$H_{a}: μ > 130$
$s_{x̅} = \frac{77.3}{\sqrt 16} = -3.81$
$t = \frac{56.3 - 130}{19.325} = 2.913$
Using technology, we have p = 0.999 at 95% confidence level.
Since P > 0.05, we will fail to reject the Null Hypothesis.
Hence we do not have sufficient evidence that on average the E. coli levels in these swimming areas were unsafe.
Work Step by Step
As above
