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Let \(X_{1}, \ldots, X_{n}\) denote a random sample from a normal population distribution with a known value of \(\sigma\). a. For testing the hypotheses \(H_{0}: \mu=\mu_{0}\) versus \(H_{\mathrm{a}}: \mu>\mu_{0}\) (where \(\mu_{0}\) is a fixed number), show that the test with test statistic \(\bar{X}\) and rejection region \(\bar{x} \geq \mu_{0}+2.33 \sigma / \sqrt{n}\) has significance level \(.01\). b. Suppose the procedure of part (a) is used to test \(H_{0}: \mu \leq \mu_{0}\) versus \(H_{\mathrm{a}}: \mu>\mu_{0}\). If \(\mu_{0}=100, n=25\), and \(\sigma=5\), what is the probability of committing a type I error when \(\mu=99\) ? When \(\mu=98\) ? In general, what can be said about the probability of a type I error when the actual value of \(\mu\) is less than \(\mu_{0}\) ? Verify your assertion.

Short Answer

Expert verified
The test has \( \alpha = 0.01 \). Type I error probability is zero when \( \mu < \mu_0 \).

Step by step solution

01

Understanding the Hypotheses

We have two hypotheses to test: the null hypothesis \( H_0: \mu = \mu_0 \) and the alternative hypothesis \( H_a: \mu > \mu_0 \). Essentially, we are testing whether the true population mean \( \mu \) is greater than a specific value \( \mu_0 \).
02

Formulating the Rejection Region

The rejection region is the range of values for which we reject the null hypothesis. For a significance level \( \alpha = 0.01 \), we use the test statistic \( \bar{X} \) and reject \( H_0 \) if \( \bar{X} \geq \mu_0 + 2.33\frac{\sigma}{\sqrt{n}} \). The constant 2.33 comes from the z-table and corresponds to the 99th percentile, reflecting a very high cutoff for significance at \( \alpha = 0.01 \).
03

Calculating Type I Error for \( \mu = 99 \)

To find the probability of a Type I error for \( \mu = 99 \), observe the rejection criteria \( \bar{X} \geq 100 + 2.33\frac{5}{5} = 102.33 \). Type I error occurs when we reject \( H_0 \) when it is true, which only happens under \( H_0: \mu \leq 100 \): but since \( \mu=99 < 100 \), the z-score \( \zeta = \frac{99 - 100}{1} = -1 \), giving a probability lower than the critical threshold, so Type I error isn't relevant since we're focusing on finding its significance at \( \mu = \mu_0 \).
04

Calculating Type I Error for \( \mu = 98 \)

Similarly, for \( \mu=98 \), using \( \bar{X} \geq 102.33 \), a Type I error for the scenario \( H_0: \mu=100 \) implies rejection when true. Since \( \mu=98 < 100 \), \( \zeta = \frac{98 - 100}{1} = -2 \) is used to find standard z-tables, resulting in an overwhelmingly small p-value and confirming no variance from \( H_0 \).
05

General Assertion about Type I Error Probability

If the actual \( \mu < \mu_0 \), the calculations involving z-scores from less than 0 confirm no rejection probabilities because \( \mu < \mu_0 \). In statistical practice, these cases aren't significant for Type I error under these test conditions. The general assertion is that when \( \mu < \mu_0 \), the Type I error probability is zero since \( \bar{X} \) being greater than the specified cutoff means it's not typically errors, but the true state won't reject \( H_0 \).

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Key Concepts

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

normal distribution
In statistics and probability theory, the **normal distribution** is a fundamental concept that is often used to describe real-valued random variables with a symmetric distribution. It is characterized by its bell-shaped curve, known as the Gaussian curve, and is determined by its mean (µ) and standard deviation (σ). Here are some key properties of the normal distribution:

  • **Symmetry:** The curve is symmetric around the mean, meaning that the left and right sides of the curve are mirror images of each other.

  • **Mean, Median, and Mode:** In a perfectly normal distribution, the mean, median, and mode of the dataset are identical and located at the center of the distribution.

  • **68-95-99.7 Rule:** This rule helps to quantify the spread of data in a normal distribution:
    • About 68% of data falls within one standard deviation of the mean.
    • About 95% falls within two standard deviations.
    • About 99.7% is within three standard deviations.

  • **Standard Normal Distribution:** When a normal distribution has a mean of 0 and a standard deviation of 1, it is referred to as the standard normal distribution or Z-distribution.

The normal distribution is widely utilized in hypothesis testing, including the exercise provided, where the sampling distribution of the sample mean is considered normal. This scenario allows researchers to infer population parameters and form conclusions based on sample data.
significance level
The **significance level** in hypothesis testing is a critical threshold that defines how much evidence we require to reject the null hypothesis. Denoted by the symbol \( \alpha \), it represents the probability of rejecting the null hypothesis when it is actually true, also known as the risk of a Type I error.

Here are some important points about significance level:

  • **Common Levels:** Typically, significance levels are set at 0.05, 0.01, or 0.10, where \( \alpha = 0.05 \) is the most common in many fields.

  • **Choice of \( \alpha \):** The choice of \( \alpha \) depends on the acceptable balance of risk for making a Type I error versus the power of the test to detect true effects. A smaller \( \alpha \) (like 0.01) means stricter evidence is required to reject the null hypothesis.

  • **Connection with Confidence Level:** The confidence level of a test is complementary to its significance level, with a confidence level of \( 1 - \alpha \). For example, a significance level of 0.01 corresponds to a 99% confidence level.

In the provided exercise, the significance level is set at 0.01, indicating that we are willing to accept a 1% chance of wrongfully rejecting the null hypothesis. This decision allows us to confidently interpret results from the statistical test within this error margin.
type I error
A **Type I error** occurs when the null hypothesis \( H_0 \) is rejected even though it is true. It is a false positive result. Understanding its implications is crucial in hypothesis testing as it affects the overall validity of research findings.

Here are key aspects of Type I error:

  • **Probability:** The probability of making a Type I error is equivalent to the significance level \( \alpha \) of the test. If \( \alpha \) is set to 0.01, there is a 1% risk of committing a Type I error.

  • **Consequences:** The consequence of a Type I error can be significant, leading to incorrect conclusions. For instance, concluding that a treatment works when it actually does not.

  • **Controlling Type I Error:** Setting a lower \( \alpha \) reduces the risk of committing a Type I error but may increase the risk of a Type II error (failing to reject a false null hypothesis).

  • **Application in the Exercise:** In the given exercise, when the null hypothesis \( H_0: \mu \leq \mu_0 \) is tested with an actual \( \mu < \mu_0 \), the probability of a Type I error is illustrated to be zero. This is because under such circumstances, the test statistic does not exceed the critical value threshold to prompt a rejection of \( H_0 \).

It is important for researchers to carefully choose their \( \alpha\) level to balance the probability of Type I and Type II errors, ensuring that their study's conclusions are both robust and reliable.

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Most popular questions from this chapter

A random sample of 150 recent donations at a certain blood bank reveals that 82 were type A blood. Does this suggest that the actual percentage of type A donations differs from \(40 \%\), the percentage of the population having type A blood? Carry out a test of the appropriate hypotheses using a significance level of \(.01\). Would your conclusion have been different if a significance level of \(.05\) had been used?

Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery with minimal aftereffects so that horses can be left unattended. The article "A Field Trial of Ketamine Anesthesia in the Horse" (Equine Vet. J., 1984: 176-179) reports that for a sample of \(n=73\) horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was \(18.86 \mathrm{~min}\) and the standard deviation was \(8.6 \mathrm{~min}\). Does this data suggest that true average lateral recumbency time under these conditions is less than \(20 \mathrm{~min}\) ? Test the appropriate hypotheses at level of significance . 10 .

To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected, and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose the specifications state that mean strength of welds should exceed \(100 \mathrm{lb} / \mathrm{in}^{2}\); the inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100\). Explain why it might be preferable to use this \(H_{\mathrm{a}}\) rather than \(\mu<100\).

One method for straightening wire before coiling it to make a spring is called "roller straightening." The article "The Effect of Roller and Spinner Wire Straightening on Coiling Performance and Wire Properties" (Springs, 1987: 27-28) reports on the tensile properties of wire. Suppose a sample of 16 wires is selected and each is tested to determine tensile strength \(\left(\mathrm{N} / \mathrm{mm}^{2}\right)\). The resulting sample mean and standard deviation are 2160 and 30, respectively. a. The mean tensile strength for springs made using spinner straightening is \(2150 \mathrm{~N} / \mathrm{mm}^{2}\). What hypotheses should be tested to determine whether the mean tensile strength for the roller method exceeds 2150 ? b. Assuming that the tensile strength distribution is approximately normal, what test statistic would you use to test the hypotheses in part (a)? c. What is the value of the test statistic for this data? d. What is the \(P\)-value for the value of the test statistic computed in part (c)? e. For a level \(.05\) test, what conclusion would you reach?

The article referenced in Example \(8.11\) also reported that in a sample of 106 wine consumers, \(22(20.8 \%)\) thought that screw tops were an acceptable substitute for natural corks. Suppose a particular winery decided to use screw tops for one of its wines unless there was strong evidence to suggest that fewer than \(25 \%\) of wine consumers found this acceptable. a. Using a significance level of \(.10\), what would you recommend to the winery? b. For the hypotheses tested in (a), describe in context what the type I and II errors would be, and say which type of error might have been committed.

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