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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's 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 \(H_{a}: \mu>150,\) 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 actual 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. What would be the conclusion of the test if \(H_{0}\) is rejected when \(z \geq 1.8\) and \(\bar{x}=152.4\) ? What type of error might have been made in reaching this conclusion?

Step by step solution

01

Explain why use of the z statistic is appropriate

The z statistic is appropriate in this setting because of the following reasons: 1. The sample size is large (n=50), allowing usage of Central Limit Theorem. 2. The population standard deviation (\(\sigma\)) is known and the population is assumed normal.

02

Describe Type I and II errors

Type I error occurs when the null hypothesis (H0) is true but is rejected. In this context, it means concluding that the mean temperature of the discharged water is more than \(150^{\circ} \mathrm{F}\) when in fact it isn't. Type II error occurs when the null hypothesis is false but is not rejected. In this context, it means failing to conclude that the mean temperature of the discharged water is more than \(150^{\circ} \mathrm{F}\) when in fact it is.

03

Calculate the value of alpha

The value of alpha can be calculated using the z-table or a calculator that can compute probabilities for the standard normal distribution. For \(z=1.8\), reading off a standard normal table, the area to the left is approximately 0.9641. Therefore, the area (or probability) to the right (which is \(\alpha\)) is \(1-0.9641=0.0359\).

04

Draw the Sampling Distribution

When drawing the sampling distribution, the x-axis represents the temperatures, the curve represents the distribution and shading would represent the Type II Error (β). The region to the right of the z-value of 1.8 under the curve of the sampling distribution would be shaded. The mean of this distribution is \(μ=153\).

05

Compute the value of beta when \(\mu=153\)

Beta (\(β\)) is the probability of a Type II error occurring. Calculation of \(β\) involves finding the z score of 1.8 in a new distribution where the mean (\(μ\)) is 153. This new z score is given by \((1.8 - (153 - 150) / (10/√50)) = 0.4\). Checking the z-table for this z score, we get the area as 0.6554. This means that \(β = 1 - 0.6554 = 0.3446\).

06

Compute the value of beta when \(\mu=160\)

Here, the calculation of \(β\) involves finding the z score of 1.8 under a distribution where \(μ=160\). The new z score will be \((1.8 - (160 - 150) / (10/√50)) = -3.58\). Checking the z-table, the area under the curve for this z-score approximately 0. So, \(β \approx 1\).

07

Conclusion of the test if \(H_{0}\) is rejected when \(z \geq 1.8\) and \(\bar{x}=152.4\)

First, find the observed z-score using \(\bar{x}=152.4\), \(μ=150\), \(\sigma=10\), and \(n=50\). The z-score comes out to be \((152.4 -150)/(10/√50)=1.7\). Since \(z=1.7\) which is less than 1.8, the null hypothesis would not get rejected. If it was wrongly rejected, a type I error could have occurred.

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Key Concepts

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

Type I and II Errors

Understanding the concepts of Type I and Type II errors is crucial in hypothesis testing. A Type I error happens when you reject a null hypothesis that is actually true. In simpler terms, you might be saying there's an effect or a change when there really isn't.
For example, in the context of the power plant's water temperature exercise, a Type I error would be concluding that the average water temperature is above 150°F, even though it truly is not. This could lead to unnecessary changes or penalties.
A Type II error occurs when you fail to reject a false null hypothesis. It's like saying nothing is wrong when there is indeed an issue. In the same exercise, if the true water temperature is above 150°F but you conclude it's not, this is a Type II error. The consequence here could be that potential harm to the river ecosystem goes unnoticed. Recognizing these errors helps in making better decisions based on hypothesis testing.

Z-Statistic

The z-statistic is a crucial tool in statistics, especially when dealing with hypothesis testing. It helps us understand how far away our sample mean is from the population mean.

• This statistic is particularly useful when the sample size is large (typically n > 30) and when the population standard deviation is known.
• In the context of the power plant problem, it helps us determine whether the temperature of discharged water significantly deviates from the regulatory limit of 150°F.

The formula to calculate the z-statistic is:\[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]Here, \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is known as the population standard deviation, and \(n\) is the sample size. Using this formula, you can compare whether the results you've observed are within the typical range of variance, or if they're extreme enough to suggest rejecting the null hypothesis.

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental principle in statistics. It says that the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution, provided the sample size is large enough (usually n > 30).
In the water temperature scenario, the scientist takes 50 samples, which is considered sufficiently large. Therefore, she can assume the sampling distribution of the sample mean temperature is normal even if the actual temperature readings are not.
This normality is what allows the use of the z-statistic to make decisions about the water temperature data. The CLT is vital because it simplifies many aspects of inferential statistics, enabling us to estimate probabilities and make predictions using the normal distribution.

Standard Normal Distribution

The standard normal distribution is a special form of the normal distribution. It's a symmetrical, bell-shaped curve with a mean of 0 and a standard deviation of 1.
In statistical testing, the standard normal distribution is used as a reference to determine how far a result is from what is expected under the null hypothesis. This helps in interpreting z-scores, where each point on the distribution shows how many standard deviations away from the mean a result lies.

• For example, in hypothesis testing of the power plant's discharged water temperature, once the z-statistic is calculated, you can use the standard normal distribution to find probabilities and critical values.
• This has practical importance in assessing whether to reject the null hypothesis based on how extreme the test statistic (z-score) falls on this curve.

Understanding this distribution allows scientists to make decisions with estimates of error based on where observed results fall in comparison to this well-defined pattern.