The mean water temperature downstream from a power plant cooling tower discharge pipe should be no more than 100 degrees F. Past experience has indicated that the standard deviation of temperature is 2 degrees F. The water temperature is measured on nine randomly chosen days, and the average temperature is found to be 98 degrees F. 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 null hypothesis at α = 0.05 if the water has a true mean temperature of 104 degrees F?

A) Yes. At a significance level of 0.05, there is enough evidence to support the claim that the mean water temperature is significantly below 100 °F.

B) P-value = 0.001

C) The probability of not rejecting the null hypothesis at α = 0.05 if the true mean is 104 °F is P(M>98.9)=1. This means that is almost impossible to reject the null hypothesis μ≤100 given that the true mean is 104.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean water temperature is significantly below 100 °F.

Then, the null and alternative hypothesis are:

The significance level is 0.05.

The sample has a size n=9.

The sample mean is M=98.

The standard deviation of the population is known and has a value of σ=2.

We can calculate the standard error as:

Then, we can calculate the z-statistic as:

This test is a left-tailed test, so the P-value for this test is calculated as:

As the P-value (0.001) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the mean water temperature is significantly below 100 °F.

The critical value for this left-tailed test is zc=-1.645.

The null hypothesis would be accepted if the test statistic is higher than zc=-1.645. For a normal distribution with the parameters of the null hypothesis (μ=100, σ=2), this would correspond to a value of:

This means that if we get a sample of size n=9 and mean bigger than 98.9, we will failed to reject the null hypothesis.

If the true mean is 104 °F, the probability of getting a sample mean over 98.9 for this sample size can be calculated as:

The probability of not rejecting the null hypothesis at α = 0.05 if the true mean is 104 °F is P(M>98.9)=1. This means that is almost impossible to reject the null hypothesis μ≤100 given that the true mean is 104.

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step-by-step explanation:

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)