91Ó°ÊÓ

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 \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above \(150^{\circ} \mathrm{F}\), 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 $$ z=\frac{\bar{x}-150}{\frac{\sigma}{\sqrt{n}}} $$ to decide between the hypotheses \(H_{0}: \mu=150\) and \(H_{a^{2}}\) \(\mu>150\), where \(\mu\) is the mean temperature of discharged water. Assume that \(\sigma\) is known to be 10 . a. Explain why use of the \(z\) statistic is appropriate in this setting. b. Describe Type I and Type II errors in this context. c. The rejection of \(H_{0}\) when \(z \geq 1.8\) corresponds to what value of \(\alpha ?\) (That is, what is the area under the \(z\) curve to the right of \(1.8 ?\) ) d. Suppose that the true value for \(\mu\) is 153 and that \(H_{0}\) is to be rejected if \(z \geq 1.8 .\) Draw a sketch (similar to that of Figure \(10.5\) ) of the sampling distribution of \(\bar{x}\), and shade the region that would represent \(\beta\), the probability of making a Type II error. e. For the hypotheses and test procedure described, compute the value of \(\beta\) when \(\mu=153\). f. For the hypotheses and test procedure described, what is the value of \(\beta\) if \(\mu=160 ?\) g. If \(H_{0}\) is rejected when \(z \geq 1.8\) and \(\bar{x}=152.8\), what is the appropriate conclusion? What type of error might have been made in reaching this conclusion?

Step by step solution

01

- Explaining use of z statistic

The z statistic is suitable in this case due to the Central Limit Theorem (CLT), which states that: if we draw a large enough sample, the distribution of the sample mean will be approximately normally distributed, independent of the population distribution, providing that the population's standard deviation is known and the sample size is large enough (N > 30). Here, the sample size is 50, which is > 30.

02

- Describing Type I and Type II errors

Type I Error: The null hypothesis \(H_0\) is true (i.e., the mean temperature of discharged water is 150) but is erroneously rejected. This means an incorrect claim would be made that the plant is violating the regulations. \n Type II Error: The alternate hypothesis \(H_a\) is true (i.e., the mean temperature of discharged water is >150) and yet the null hypothesis is not rejected. The error here is failing to detect a violation of regulations.

03

- Calculating value of alpha

The rejection area to the right of z = 1.8 on a z-distribution table corresponds to the significance level \(\alpha\). Looking up a Z-table for the value of 1.8, we get 0.9641. Therefore, \(\alpha = 1 - 0.9641 = 0.0359\).

04

- Sketching sampling distribution

The sampling distribution of \(\bar{x}\) would be a bell shaped curve centered at 150 (assuming \(H_0\) is true), with a standard deviation equal to \(\frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{50}}\). For \(H_{a^{2}}\) (with \(\mu\) = 153), the curve is shifted to the right. The region to the right of z = 1.8 under the curve for \(H_{a^{2}}\) represents \(\beta\), the probability of making a Type II error.

05

- Computing value of beta when μ = 153

We first compute the corresponding z-score: \(z = \frac{153 - 150}{\frac{10}{\sqrt{50}}} = 2.12 \). Since the critical z = 1.8, we are interested in the area to the right of z = 2.12 under the standard normal curve. Looking up a Z-table, this corresponds to 1 - 0.9830 = 0.0170.

06

- Computing value of beta when μ = 160

The z-score is \(z = \frac{160 - 150}{\frac{10}{\sqrt{50}}} = 7.07 \). As 7.07 is quite large, it's safe to assume that the value of \(\beta\) will be nearly zero.

07

- Drawing conclusion

Since z = 1.8 is the decision boundary, and the computed z for \(\bar{x} = 152.8\) is \(z = \frac{152.8 - 150}{\frac{10}{\sqrt{50}}} = 1.988\) which is greater than 1.8, we reject \(H_0\). This suggests that the power plant is not complying with the regulations. However, we may have committed a Type I error, implying that although we concluded that the plant does not comply, in reality, it does.

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Key Concepts

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

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental statistical principle that makes hypothesis testing possible for many types of data. It tells us that when you take a sufficiently large number of samples from any distribution, the distribution of the sample mean will approach a normal distribution. This is true even if the original data itself is not normally distributed. In our water temperature example, we have a sample size of 50, which is large enough to apply the CLT.
This is important because the CLT allows us to confidently use the z statistic (a test statistic for normal distributions) without worrying about the distribution of the temperatures we're sampling. As a rule of thumb, a sample size greater than 30 is usually considered adequate for the CLT to hold. That's why the scientist in this exercise can use it with the sample of 50 water temperatures.

Type I Error

A Type I Error occurs when we wrongly reject a true null hypothesis. In the context of our water discharge problem, a Type I Error would mean concluding that the mean temperature of the discharged water exceeds 150°F when it actually doesn’t. Essentially, it would suggest a violation of environmental regulations when none exists.
The consequences of a Type I Error can be significant as it may lead to unnecessary corrective actions or penalties against the power plant. This error is defined by the significance level \(\alpha\), which in this case, is 0.0359 (from our calculation for the rejection region). This means there is a 3.59% chance of making a Type I Error if in fact, the temperature does not exceed the set limit.

Type II Error

A Type II Error happens when we fail to reject a false null hypothesis. In simpler words, it means that in our exercise, the mean discharged water temperature is actually greater than 150°F, but our test fails to detect this fact. So, the plant is non-compliant with regulations, yet it goes unnoticed.
The probability of a Type II Error is denoted by the symbol \(\beta\). For a situation where the true mean temperature is 153°F, we calculated \(\beta\) to be 0.0170. This indicates a 1.70% chance of failing to notice that the mean discharge temperature exceeds 150°F. It's crucial for balancing the chances of making Type I and II errors, and often depends on the true mean's distance from the null hypothesis, sample size, and \(\alpha\).

Z Statistic

The Z statistic is a powerful tool in hypothesis testing, particularly when dealing with large samples or when the population standard deviation is known. It measures the number of standard deviations a data point is from the population mean.
In this exercise, the Z statistic formula used is \[\ z = \frac{\bar{x} - 150}{\frac{\sigma}{\sqrt{n}}} \] where \(\bar{x}\) is the sample mean, 150°F is the hypothesized population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.By comparing this Z statistic to a critical value (like 1.8 in our example), the scientist determines whether to reject or fail to reject the null hypothesis \((H_0)\). If the Z score is greater than 1.8, \((H_0)\) is rejected, suggesting the plant may not be in compliance. This method relies heavily on the properties of the normal distribution, aided by the Central Limit Theorem.