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The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100 and the standard deviation is 2 . We wish to test \(H_{0}: \mu=100\) versus \(H_{1}: \mu \neq 100\) with a sample of \(n=9\) specimens. (a) If the acceptance region is defined as \(98.5 \leq \bar{x} \leq 101.5\), find the type I error probability \(\alpha\). (b) Find \(\beta\) for the case where the true mean heat evolved is 103 . (c) Find \(\beta\) for the case where the true mean heat evolved is 105. This value of \(\beta\) is smaller than the one found in part (b) above. Why?

Short Answer

Expert verified
(a) \(\alpha = 0.0244\); (b) \(\beta = 0.0122\); (c) \(\beta = 0\). \(\beta\) decreases with larger \(\mu\) because the distribution shifts further from \(H_0\).

Step by step solution

01

Define the Acceptance Region and Null Hypothesis

The acceptance region for the test is defined as \(98.5 \leq \bar{x} \leq 101.5\). The null hypothesis to be tested is \(H_0: \mu = 100\) versus the alternative hypothesis \(H_1: \mu eq 100\), where \(\mu\) is the true mean of the heat evolved.
02

Calculate the Type I Error Probability (\(\alpha\))

The type I error probability \(\alpha\) is the probability of rejecting the null hypothesis when it is true. Here, it is given by the probability that the sample mean falls outside the acceptance region under \(H_0\).Since \(\bar{x}\) is normally distributed with mean \(\mu = 100\) and standard deviation \(\sigma/\sqrt{n} = 2/3\), find\[\alpha = P(\bar{x} < 98.5 \text{ or } \bar{x} > 101.5 \mid \mu = 100).\]Find the z-scores:\(z_1 = \frac{98.5 - 100}{2/3} = -2.25\),\(z_2 = \frac{101.5 - 100}{2/3} = 2.25\).Using the standard normal distribution table, \(P(Z < -2.25) \approx 0.0122\) and \(P(Z > 2.25) \approx 0.0122\).Then, \(\alpha = 0.0122 + 0.0122 = 0.0244\).
03

Calculate \(\beta\) for \(\mu = 103\)

The type II error probability \(\beta\) is the probability of failing to reject the null hypothesis when the true mean is \(\mu = 103\). For \(\mu = 103\), \(\bar{x}\) is normally distributed with mean \(103\) and standard deviation \(2/3\). Thus, \[\beta = P(98.5 \leq \bar{x} \leq 101.5 \mid \mu = 103).\]Find the z-scores:\(z_1 = \frac{98.5 - 103}{2/3} = -6.75\),\(z_2 = \frac{101.5 - 103}{2/3} = -2.25\).Using the standard normal distribution table, \(P(Z < -6.75) \approx 0\) and \(P(Z < -2.25) \approx 0.0122\).Therefore, \(\beta = P(-6.75 < Z < -2.25) \approx 0.0122\).
04

Calculate \(\beta\) for \(\mu = 105\)

To find \(\beta\) when the true mean is \(\mu = 105\), calculate:\[\beta = P(98.5 \leq \bar{x} \leq 101.5 \mid \mu = 105).\]For \(\mu = 105\), \(\bar{x}\) is normally distributed with mean \(105\) and standard deviation \(2/3\). Find the z-scores:\(z_1 = \frac{98.5 - 105}{2/3} = -9.75\),\(z_2 = \frac{101.5 - 105}{2/3} = -5.25\).Standard normal distribution table gives:\(P(Z < -9.75) \approx 0\) and \(P(Z < -5.25) \approx 0\).Thus, \(\beta \approx 0 - 0 = 0\).
05

Conclusion: Why \(\beta\) decreases with increasing \(\mu\)

As \(\mu\) increases (moving further from \(H_0: \mu=100\)), the probability \(\beta\) of incorrectly accepting \(H_0\) when it is false, decreases. The sample mean distribution shifts further away from the acceptance region, leading to a reduced overlap with the region defined by \(H_0\). Therefore, \(\beta\) for \(\mu = 105\) is smaller than \(\beta\) for \(\mu = 103\).

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Key Concepts

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

Type I Error
A Type I error occurs when we reject a true null hypothesis. In the context of hypothesis testing for the cement mixture problem, it means we conclude that the mean heat evolved is different from 100 calories per gram when, in reality, it is exactly 100. This kind of mistake is significant because it suggests there's a statistically significant effect when there actually isn't one. In our exercise, the Type I error probability, denoted as \( \alpha \), was calculated by determining the likelihood that our sample mean falls outside the acceptance region \(98.5 \leq \bar{x} \leq 101.5\) when the true mean is indeed 100. The calculations involve using the standard normal distribution to find the probability that the sample mean is either below 98.5 or above 101.5, resulting in \( \alpha = 0.0244 \). This value represents our chance of making a Type I error in this scenario.
Type II Error
A Type II error is the probability of not rejecting the null hypothesis when it is, in fact, false. This means we wrongly accept that the mean heat is 100 calories per gram when it actually is not. In the given problem, we calculate Type II error probability, \( \beta \), for different true mean values (103 and 105).
  • For \( \mu = 103 \): The calculated \( \beta \) reflects that the sample mean falls within the acceptance region when it should not. Utilizing the standard normal distribution, \( \beta \) is found to be approximately 0.0122; indicating errors in detection steps when the real mean is slightly shifted from 100.
  • For \( \mu = 105 \): \( \beta \) dramatically reduces to 0 due to the mean's further deviation from 100. The sample mean's distribution does not overlap with the acceptance region anymore, explaining the absence of a Type II error (\( \beta = 0 \)).
These values illustrate how \( \beta \) changes as the true mean moves further away from 100, emphasizing the improvement in detecting false assumptions.
Normal Distribution
The normal distribution is a foundational concept in statistics, often known as the "bell curve" due to its shape. It is symmetrical around its mean. In the given exercise, the sample mean (\( \bar{x} \)) follows a normal distribution with a certain mean and standard deviation. This concept is critical because it allows us to use z-scores to determine probabilities, such as the probabilities involved in Type I and Type II errors.
  • When the true mean is 100, \( \bar{x} \) follows the normal distribution with mean 100 and a standard deviation of \( \frac{2}{3} \) (which is derived from the formula \( \sigma/\sqrt{n} \), where \( \sigma = 2 \) and \( n = 9 \)).
  • When the mean shifts to values like 103 or 105, the distribution of \( \bar{x} \) also shifts, which helps to understand how \( \beta \) changes.
Utilizing the properties of the normal distribution is crucial for performing accurate hypothesis testing.
Acceptance Region
The acceptance region is central to hypothesis testing because it specifies the range of values for which we fail to reject the null hypothesis. In the exercise, it is defined as \(98.5 \leq \bar{x} \leq 101.5\). When the sample mean lies within this region, we conclude that there's no sufficient evidence to say the mean is not 100. Conversely, if it falls outside, the null hypothesis gets rejected.
  • This region is used to calculate both Type I and Type II errors as it determines when an error occurs if the true population mean does not align with our hypothesis.
  • The width and placement of the acceptance region are influenced by the desired level of significance (\( \alpha \)) and the standard deviation of the sample mean. They directly affect the probability of making Type I and Type II errors.
Understanding the acceptance region is fundamental to deciding whether the sample provides enough evidence against the null hypothesis.

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