The mean water temperature downstream from a discharge pipe at a power plant cooling towershould be no more than 100°C. Past experience suggests that the standard deviation oftemperature is 2°C. The water temperature measured on nine randomly chosen days, yields anaverage temperature of 98°F2.a) Is there evidence that the water temperature is acceptable at α-0.05?b) What is the P-value for this test?c) What is the probability of accepting the ul hypothesis at a 0.05 if the water has a true meantemperature of 104°F?

Question
Asked Mar 4, 2019
The mean water temperature downstream from a discharge pipe at a power plant cooling tower
should be no more than 100°C. Past experience suggests that the standard deviation of
temperature is 2°C. The water temperature measured on nine randomly chosen days, yields an
average temperature of 98°F
2.
a) Is there evidence that the water temperature is acceptable at α-0.05?
b) What is the P-value for this test?
c) What is the probability of accepting the ul hypothesis at a 0.05 if the water has a true mean
temperature of 104°F?
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The mean water temperature downstream from a discharge pipe at a power plant cooling tower should be no more than 100°C. Past experience suggests that the standard deviation of temperature is 2°C. The water temperature measured on nine randomly chosen days, yields an average temperature of 98°F 2. a) Is there evidence that the water temperature is acceptable at α-0.05? b) What is the P-value for this test? c) What is the probability of accepting the ul hypothesis at a 0.05 if the water has a true mean temperature of 104°F?

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Expert Answer

Step 1

Given that water temperature measured on 9 randomly chosen days is normally distributed with mean 98oF  and standard deviation of temperature is given as 2oF. We need to test if water temperature is acceptable.

Null hypothesis : mean water temperature = 100oF

alternative hypothesis : mean water temperature < 100oF

Here we know population standard deviation so we use a Z test. Test statistic is given by below mentioned formula . y substituting values we get test statistic = -3.

referring to Z table the Z score for probability 0.0505 is -1.64 and Z score for probability 0.4945 is -1.65. So the Z score for probability 0.95 is average of -1.64 and -1.65 = -1.645. At 0.05 significance level the critical Z score is -1.645.

Comparing to test statistic Z score is way less than critical Z score so we reject the null hypothesis.

So we can say at 0.05 significance level the water temperature is less than 100. So the water temperature is acceptable.

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Step 2

The p value for Z score -3  is 0.0013 with reference to the Z table shown below.

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Step 3

Given that water has true mean temperature of 104oF. we need to find the probability of accepting the null hypothesis.

we accept the nul hypothesis is  p value is greater than 0.05. So the critical score has to be greater than -1.645. The temperature with respect to Z...

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Hypothesis Testing

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