Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most 150 degrees Fahrenheit, there will be no negative effects on the river ecosystem. To investigate whether the power plant is in compliance with regulations that prohibit a mean discharge water temperature above 150 degrees, a scientist will take 50 water samples at randomly selected times and will record the water temperature of each sample. She will then use a Z statistic ?=, "-./0 1√3 4 to decide between the hypothesis H0: ?=150 and H1: ?>150 where ? is the true mean temperature of the discharged water. Assume that ? is known to be 10 degrees.
(a) Explain why a Z statistic is appropriate in this setting?
(b) Suppose that the Critical Value Decision Rule is: Reject H0 if
Z ≥ 1.75. What significance level does this rule correspond to?
Round to the nearest percentage point.
(c) Suppose that the Critical Value Decision Rule is: Reject H0 if
Z ≥ 1.75. What is the rejection region for this test in terms of ?
"? (In other words: for what values of ? " will you reject H0?)
(d) Suppose that the sample mean from the 50 water samples is ? "=152.1 degrees. If H0 is rejected if Z ≥ 1.75 then what is the appropriate conclusion for this test? Write your conclusion and justify it.
(e) Suppose that the true value of ? is 153 degrees (i.e., this
means that the alternative hypothesis is actually true) and that
you reject H0 if Z ≥ 1.75. What is the power of this test?
(a) Since sample size is large and is known
z statistic is appropriate
(b) from z table , the area from 1.75 and beyond is 0.0401
significance level is 4.01%
(c) we know that
when z = 1.75 , , n = 50
X'' = 152.47
(d)
= 1.48
since calculated value of z (test statistic) is less than 1.75
we fail to reject H0
There is not enough evidence to conclude that mean temp of discharged water more than 150 degrees
(e) Power = P(rejecting H0 I H1 true)
=
=
= P(Z > -0.37) = 0.6443
power = 0.6443
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