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The mean water temperature downstream from a discharge pipe at a power plant cooling tower should be no more than \(100^{\circ} \mathrm{F}\). Past experience has indicated that the standard deviation of temperature is \(2^{\circ} \mathrm{F}\). The water temperature is measured on nine randomly chosen days, and the average temperature is found to be \(98^{\circ} \mathrm{F}\). (a) Is there evidence that the water temperature is acceptable at \(\alpha=0.05 ?\) (b) What is the \(P\) -value for this test? (c) What is the probability of accepting the null hypothesis at \(\alpha=0.05\) if the water has a true mean temperature of \(104^{\circ} \mathrm{F} ?\)

Short Answer

Expert verified
(a) Yes, the water temperature is acceptable. (b) P-value is 0.0013. (c) Probability of accepting null if true mean is 104°F is 0.

Step by step solution

01

Define the Hypotheses

We need to set up our null and alternative hypotheses for the test. The null hypothesis (\(H_0\)) states that the mean water temperature is \( \mu = 100^{\circ} \mathrm{F} \). The alternative hypothesis (\(H_a\)) suggests that the mean water temperature is less than \(100^{\circ} \mathrm{F}\): - \(H_0: \mu = 100^{\circ} \mathrm{F}\) - \(H_a: \mu < 100^{\circ} \mathrm{F}\)
02

Calculate the Test Statistic

Given that the sample mean \( \bar{x} = 98^{\circ} \mathrm{F}\), population standard deviation \( \sigma = 2^{\circ} \mathrm{F} \), and sample size \( n = 9 \), we use the formula for the test statistic: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{98 - 100}{\frac{2}{\sqrt{9}}} = \frac{98 - 100}{0.667} = -3 \]
03

Determine the Critical Value and Compare

With a significance level of \( \alpha = 0.05\) for a left-tailed test, we look up the critical value for \( z \) in a standard normal distribution table. The critical value is \( z = -1.645\). Since our calculated \( z = -3 \) is less than \( -1.645 \), we reject the null hypothesis.
04

Calculate the P-value for the Test Statistic

The p-value is the probability of obtaining a test statistic as extreme as, or more extreme than, the value observed, assuming the null hypothesis is true. Here, with \( z = -3 \), the p-value is found from the standard normal table as approximately 0.0013.
05

Calculate the Probability of Type II Error

To find the probability of not rejecting the null hypothesis when the true mean is \( \mu = 104^{\circ} \mathrm{F}\), we calculate type II error: The test statistic under \( \mu = 104^{\circ} \mathrm{F}\) is: \[ z = \frac{104 - 100}{\frac{2}{\sqrt{9}}} = \frac{4}{0.667} = 6 \] We need to find the probability that a \( z \)-score for \( \bar{x} \) falls to the right of the critical value \( -1.645 \) given this new true mean. Since \( z = 6 \) falls far to the right of \( -1.645 \), the probability is practically 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis
In hypothesis testing, the null hypothesis (\(H_0\)) is a key concept that represents a default position that suggests "no effect" or "no difference."
If you imagine a court trial, the null hypothesis is like assuming the defendant is innocent until proven guilty.
In our exercise, the null hypothesis is that the mean water temperature downstream from a discharge pipe at a power plant cooling tower is \(100^{\circ} \mathrm{F}\).
This claim is what we test against the alternative, which suggests a change or difference, in this case, that the mean temperature is less than \(100^{\circ} \mathrm{F}\).

Understanding the null hypothesis is crucial for the framework of statistical testing:
  • It acts as the hypothesis to be tested and possibly rejected, based on evidence suggesting a different scenario.
  • The null hypothesis provides a baseline or standard with which the alternative hypotheses are compared. Rejection of \(H_0\) suggests that the alternative hypothesis might be true.
Hypothesis testing, therefore, pivots around stating, testing, and interpreting the results of the null hypothesis against the given data.
P-value
The p-value is a statistical measurement that helps you determine the strength of your results when testing a hypothesis.
It's the probability of observing data as extreme, or even more extreme, under the assumption that the null hypothesis is true.
Smaller p-values often mean stronger evidence against the null hypothesis, leading to its rejection.

Consider this scenario: You calculated a p-value of 0.0013, as in our exercise.
  • Because 0.0013 is less than the significance level (usually \(\alpha = 0.05\), an accepted threshold for testing), it suggests that the observed results are not likely due to random chance alone.
  • This small p-value provides evidence strong enough to reject the null hypothesis, supporting the alternative that the mean temperature is indeed less than \(100^{\circ} \mathrm{F}\).
Keep in mind that:
  • A very low p-value (below \(\alpha\)) typically indicates a significant result.
  • Conversely, a high p-value suggests there is not sufficient evidence to reject the null hypothesis.
  • Despite its power, the p-value does not measure the size of the effect or the importance of the result.
Being familiar with how p-values function helps you interpret whether your test results are significant in a given context.
Type II Error
In statistical hypothesis testing, a Type II Error, also known as a "false negative," occurs when the hypothesis test fails to reject the null hypothesis even though it is false.
This error can be thought of as missing the detection of an effect or difference when one truly exists.

Let's connect this to our example. You calculated the probability of a Type II Error when the true mean temperature was actually \(104^{\circ} \mathrm{F}\).
  • Given this scenario, a Type II Error would mean claiming the water temperature is acceptable (failing to reject \(H_0\)) when it is, in fact, too high (\(\mu = 104^{\circ} \mathrm{F}\)).
  • Optimizing power and reducing the risk of a Type II Error generally involves increasing the sample size, thus acquiring more information about the population.
Recognizing the implications of Type II Errors is essential:
  • It helps you understand the trade-offs between different types of errors when designing tests.
  • Balancing between Type I (false positive) and Type II (false negative) errors is crucial in making informed decisions based on test outcomes.
  • Practical significance and context can also influence how each error type affects decisions in real-world scenarios.
Understanding Type II Errors ensures you make more effective interpretations and decisions in hypothesis testing.
Standard Deviation
Standard deviation is a measure used to quantify the amount of variation or dispersion in a set of data values.
It shows how much individual data points differ from the mean of the dataset, presenting a picture of the data's spread.

In our exercise, the standard deviation of water temperature was given as \(2^{\circ} \mathrm{F}\).
  • This means that on average, the temperature readings deviate by \(2^{\circ} \mathrm{F}\) from the mean temperature.
  • A smaller standard deviation indicates that the data points are closer to the mean, while a larger standard deviation suggests more spread out data.
Knowing the standard deviation allows you to better understand and describe your data:
  • It is essential for calculating test statistics in hypothesis testing, as seen in the determination of the z-score in this scenario.
  • Standard deviation plays a critical role in constructing confidence intervals and prediction intervals, offering insights into the precision of your estimates.
The standard deviation is a powerful statistical tool for exploring the variability within a dataset, forming the backbone for many analyses in research and data-driven decision-making.

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Most popular questions from this chapter

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