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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

The accompanying observations on residual flame time (sec) for strips of treated children's nightwear were given in the article "An Introduction to Some Precision and Accuracy of Measurement Problems" ( \(J\). of Testing and Eval., 1982: 132-140). Suppose a true average flame time of at most \(9.75\) had been mandated. Does the data suggest that this condition has not been met? Carry out an appropriate test after first investigating the plausibility of assumptions that underlie your method of inference. $$ \begin{array}{lllllll} 9.85 & 9.93 & 9.75 & 9.77 & 9.67 & 9.87 & 9.67 \\ 9.94 & 9.85 & 9.75 & 9.83 & 9.92 & 9.74 & 9.99 \\ 9.88 & 9.95 & 9.95 & 9.93 & 9.92 & 9.89 & \end{array} $$

The article "Orchard Floor Management Utilizing SoilApplied Coal Dust for Frost Protection" (Agri. and Forest Meteorology, 1988: 71-82) reports the following values for soil heat flux of eight plots covered with coal dust. \(\begin{array}{llllllll}34.7 & 35.4 & 34.7 & 37.7 & 32.5 & 28.0 & 18.4 & 24.9\end{array}\) The mean soil heat flux for plots covered only with grass is 29.0. Assuming that the heat-flux distribution is approximately normal, does the data suggest that the coal dust is effective in increasing the mean heat flux over that for grass? Test the appropriate hypotheses using \(\alpha=.05\).

A sample of 12 radon detectors of a certain type was selected, and each was exposed to \(100 \mathrm{pCi} / \mathrm{L}\) of radon. The resulting readings were as follows: \(\begin{array}{rrrrrr}105.6 & 90.9 & 91.2 & 96.9 & 96.5 & 91.3 \\ 100.1 & 105.0 & 99.6 & 107.7 & 103.3 & 92.4\end{array}\) a. Does this data suggest that the population mean reading under these conditions differs from 100 ? State and test the appropriate hypotheses using \(\alpha=.05\). b. Suppose that prior to the experiment a value of \(\sigma=7.5\) had been assumed. How many determinations would then have been appropriate to obtain \(\beta=.10\) for the alternative \(\mu=95\) ?

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.

A new design for the braking system on a certain type of car has been proposed. For the current system, the true average braking distance at \(40 \mathrm{mph}\) under specified conditions is known to be \(120 \mathrm{ft}\). It is proposed that the new design be implemented only if sample data strongly indicates a reduction in true average braking distance for the new design. a. Define the parameter of interest and state the relevant hypotheses. b. Suppose braking distance for the new system is normally distributed with \(\sigma=10\). Let \(\bar{X}\) denote the sample average braking distance for a random sample of 36 observations. Which of the following three rejection regions is appropriate: \(R_{1}=\\{\bar{x}: \bar{x} \geq 124.80\\}, \quad R_{2}=\\{\bar{x}: \bar{x} \leq 115.20\\}\), \(R_{3}=\\{\bar{x}\) : either \(\bar{x} \geq 125.13\) or \(\bar{x} \leq 114.87\\}\) ? c. What is the significance level for the appropriate region of part (b)? How would you change the region to obtain a test with \(\alpha=.001\) ? d. What is the probability that the new design is not implemented when its true average braking distance is actually \(115 \mathrm{ft}\) and the appropriate region from part (b) is used? e. Let \(Z=(\bar{X}-120) /(\sigma / \sqrt{n})\). What is the significance level for the rejection region \(\\{z: z \leq-2.33\\}\) ? For the region \(\\{z: z \leq-2.88\\}\) ?

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