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A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean thread elongation of 12 kilograms with a standard deviation of 0.5 kilograms. The company wishes to test the hypothesis \(H_{0}: \mu=12\) against \(H_{1}: \mu<12,\) using a random sample of four specimens. (a) What is the type I error probability if the critical region is defined as \(\bar{x}<11.5\) kilograms? (b) Find \(\beta\) for the case in which the true mean elongation is 11.25 kilograms. (c) Find \(\beta\) for the case in which the true mean is 11.5 kilograms.

Short Answer

Expert verified
(a) 0.0228 (b) 0.8413 (c) 0.5

Step by step solution

01

Understanding Type I Error

Type I error occurs when we reject the null hypothesis, \(H_0: \mu = 12\), when it is actually true. The critical region for this test is defined as \(\bar{x} < 11.5\) kg. This means that a Type I error would occur if the sample mean \(\bar{x}\) falls below 11.5 kg while the true population mean is, in fact, 12 kg.
02

Calculating Critical Value for Type I Error

Calculate the critical value using the formula for the standard normal distribution. Since the sample size \(n = 4\), the standard error \(SE = \frac{\sigma}{\sqrt{n}} = \frac{0.5}{\sqrt{4}} = 0.25\). The z-score formula, when \(\bar{x} < 11.5\), is \(z = \frac{11.5 - 12}{0.25} = -2\).
03

Finding Type I Error Probability

The probability of a Type I error, often denoted as \(\alpha\), is the probability of observing a test statistic as extreme as the critical value or more extreme under the null hypothesis. Using the standard normal distribution, find \(P(Z < -2)\). This results in \(\alpha = 0.0228\), meaning there is about a 2.28% chance of a Type I error.
04

Understanding Beta (\(\beta\) Error)

\(\beta\) error is the probability of failing to reject the null hypothesis when it is false. Here, we are to find \(\beta\) for two cases: \(\mu = 11.25\) kg and \(\mu = 11.5\) kg.
05

Calculating \(\beta\) for \(\mu = 11.25\) kg

Using the alternate mean, calculate the z-score for the critical region defined by \(\bar{x} < 11.5\). The z-score is now \(z = \frac{11.5 - 11.25}{0.25} = 1\). Find \(P(Z < 1)\), which is \(0.8413\). Thus, \(\beta = 0.8413\).
06

Calculating \(\beta\) for \(\mu = 11.5\) kg

Similarly, for \(\mu = 11.5\), the z-score becomes \(z = \frac{11.5 - 11.5}{0.25} = 0\). Find \(P(Z < 0)\), which is \(0.5\). Thus, \(\beta = 0.5\).

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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 happens when we mistakenly reject the null hypothesis, even though it is true. In simpler terms, it's like saying there's a difference when there isn’t one in reality. Imagine this as a false alarm, where we believe something changed when it didn't. This type of error is often denoted by the symbol \(\alpha\).
In our exercise related to the textile fiber, the company sets the null hypothesis \(H_0: \mu=12\), which claims the mean thread elongation is 12 kilograms. The critical region, which dictates when we'll reject the null hypothesis, is when the sample mean \(\bar{x} < 11.5\) kilograms. If we conclude that the mean is less than 12 when it truly is 12, we're making a Type I error. Here, \(\alpha = 0.0228\), meaning there's about a 2.28% risk of this false positive.
Being aware of Type I errors is crucial because it reflects how often we might falsely believe our hypothesis test is showing something noteworthy.
Type II Error
A Type II error occurs when we fail to reject the null hypothesis when it is actually false. Simply put, it's a missed detection or a false negative, where we overlook a real difference. This error is represented by \(\beta\).
In the exercise context, when determining the actual mean elongation for the yarn, we check if it is below 12 kilograms. If the true mean is different, such as 11.25 kilograms or 11.5 kilograms, not rejecting the null hypothesis \(H_0\) when in fact it should be rejected constitutes a Type II error.
For a true mean of 11.25 kg, \(\beta = 0.8413\), indicating an 84.13% chance that we won't notice the yarn doesn’t meet the claimed standard. In the case of a true mean of 11.5 kg, \(\beta = 0.5\), which means there’s a 50% likelihood of failing to detect the actual decrease in elongation. Understanding these errors is vital to grasp the reliability and power of a hypothesis test.
Standard Normal Distribution
The standard normal distribution is a special version of the normal distribution that is pivotal in hypothesis testing. It is characterized by a mean (\(\mueq\)) of 0 and a standard deviation (\(\sigma\)) of 1. This distribution helps in determining probabilities and critical values during hypothesis testing.
In the given problem, we use the standard normal distribution to find our z-scores. Z-scores are standardized values that show how many standard deviations away a data point is from the mean. For instance, when the sample mean \(\bar{x} < 11.5\) and the population mean \(\mu = 12\), the z-score calculated is \(-2\), which tells us how much the sample mean deviates from the known mean.
Using the z-score, we find the probability under the curve which translates to the chance of observing that sample mean or more extreme ones, assuming the null hypothesis is true. Using this, we established the Type I error rate by looking up \(P(Z < -2)\) in standard normal distribution tables. This method is essential in making statistical inferences about population parameters.
Null Hypothesis
The null hypothesis is a fundamental part of hypothesis testing, representing the default position that there is no effect or no difference. Symbolized by \(H_0\), it is what we attempt to reject or disprove in many experiments.
In our yarn elongation problem, the null hypothesis posits that the mean elongation is exactly 12 kilograms. The alternative hypothesis \(H_1\) suggests that the mean is actually less than 12. Testing starts with the assumption that \(H_0\) is true. Only with sufficient statistical evidence do we reject it in favor of \(H_1\).
The beauty of the null hypothesis is its role as a benchmark for the status quo. It helps gauge whether the new observations or results differ significantly from what was expected. Applying this concept systematically aids researchers in maintaining objectivity and reduces the risks of being swayed by extraordinary claims without sufficient scientific evidence.

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Most popular questions from this chapter

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