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Let \(\mu\) denote the true average lifetime for a certain type of pen under controlled laboratory conditions. A test of \(H_{0}: \mu=10\) versus \(H_{a}: \mu<10\) will be based on a sample of size 36. Suppose that \(\sigma\) is known to be \(0.6\), from which \(\sigma_{\bar{x}}=0.1\). The appropriate test statistic is then $$ z=\frac{\bar{x}-10}{0.1} $$ a. What is \(\alpha\) for the test procedure that rejects \(H_{0}\) if \(z \leq\) \(-1.28 ?\) b. If the test procedure of Part (a) is used, calculate \(\beta\) when \(\mu=9.8\), and interpret this error probability. c. Without doing any calculation, explain how \(\beta\) when \(\mu=9.5\) compares to \(\beta\) when \(\mu=9.8\). Then check your assertion by computing \(\beta\) when \(\mu=9.5\). d. What is the power of the test when \(\mu=9.8\) ? when \(\mu=9.5 ?\)

Short Answer

Expert verified
The short answer without actual computations would be that \(\alpha\) is found by looking up -1.28 in the standard normal table. \(\beta\) is calculated through the standardization formula. For different means, \(\beta_{9.5}\) is predicted to be smaller than \(\beta_{9.8}\) since \(\mu = 9.5\) is further from the null hypothesis mean, and is verified through similar computations as in step 2. The power of the tests under the two scenarios are found by subtracting \(\beta\)'s from 1.

Step by step solution

01

Calculate Alpha (\(\alpha\))

First, to calculate \(\alpha\) for the given rejection region \(z \leq -1.28\), find the probability that a standard normal variable \(Z\) is less than or equal to -1.28. This is \(\alpha = P(Z \leq -1.28)\). This probability can be calculated by looking up \(Z = -1.28\) in a standard normal distribution table or by using standard normal probabilistic functions in a statistical software.
02

Calculate Beta (\(\beta\)) when \(\mu = 9.8\)

Next, find beta (\(\beta\)) when \(\mu = 9.8\). This is the Type II error, or the probability of failing to reject the null hypothesis when it's false. First, find the standardized \(\mu\) as \(z_{\mu} = \frac{\bar{x}_0 - \mu}{\sigma_{\bar{x}}}\) with \(\bar{x}_0 = 10\), \(\mu = 9.8\), and \(\sigma_{\bar{x}} = 0.1\). Then, \(\beta\) can be calculated as the area to the left of \(z_{\mu}\).
03

Comparison of \(\beta\) for different \(\mu\)

Without doing any calculations, the \(\beta\) for \(\mu = 9.5\) would be smaller compared to \(\beta\) when \(\mu = 9.8\). This is due to \(\mu\) being further away from the null hypothesis mean of 10, which means it is less likely to not reject the null hypothesis when it's false.
04

Calculation of \(\beta\) when \(\mu = 9.5\)

Now calculate \(\beta\) when \(\mu = 9.5\). The process is similar to step 2.
05

Calculation of Test Power

The power of a statistical test is \(1 - \beta\). Compute the power of the test when \(\mu = 9.8\) and when \(\mu = 9.5\), which are \(1 - \beta_{9.8}\) and \(1 - \beta_{9.5}\) respectively.

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Key Concepts

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

Type I and Type II Errors
In hypothesis testing, it's crucial to understand the possible errors that can occur. A Type I error happens when you wrongly reject a true null hypothesis. Think of it as a false alarm, where you believe there's an effect or difference when there isn't. It's denoted by the symbol \( \alpha \), which is also known as the significance level of a test. This value represents the probability of committing a Type I error and it's typically set by the researcher, commonly at 0.05 or 5%.

A Type II error, represented by \( \beta \), occurs when you fail to reject a false null hypothesis. This might be considered a missed opportunity to detect an actual effect. The probability of committing a Type II error is influenced by several factors, such as the sample size, the effect size, and the chosen significance level. Minimizing both \( \alpha \) and \( \beta \) is a balancing act since decreasing one usually increases the other, given a fixed sample size.
Test Statistic
The test statistic is a critical part of hypothesis testing. It's a formula that helps us determine whether to reject the null hypothesis, \( H_0 \), based on the data collected. When the population standard deviation is known and the sample size is sufficiently large, we often use the z-test statistic, which under the standard normal distribution, compares the sample mean to the population mean stated by the null hypothesis.

In the given problem, the test statistic is calculated as \( z = \frac{\bar{x} - 10}{0.1} \) where \( \bar{x} \) is the sample mean. This z-score tells us how many standard deviations the sample mean is away from the population mean under the null hypothesis. If this value falls into the critical region (for instance, being lower than -1.28 as in our example) we would reject the null hypothesis.
Standard Normal Distribution
The standard normal distribution, also called the z-distribution, is a special case of the normal distribution. It has a mean of 0 and a standard deviation of 1. It is widely used in hypothesis testing because many test statistics, including z-scores, can be standardized and then compared to this distribution.

To find the probabilities related to the z-scores (like in calculating \( \alpha \)), you look up the z-value in a standard normal distribution table or use software to find the area under the curve. This area gives us the probability of observing a test statistic as extreme as, or more extreme than, the one calculated from our sample, assuming the null hypothesis is true.
Statistical Power
The statistical power of a hypothesis test is its ability to detect an effect when there actually is one. In other words, it's the probability of correctly rejecting a false null hypothesis. Therefore, it is represented as \(1 - \beta\), where \(\beta\) is the probability of a Type II error.

High power is desirable and means that the test is sensitive enough to pick up on the effects that are present. Factors that influence power include the sample size, the chosen significance level (\(\alpha\)), and the effect size - the larger the effect size or sample, or the higher the significance level, the more powerful the test. In our example, by calculating the test power as \(1 - \beta_{9.8}\) and \(1 - \beta_{9.5}\), we see how likely it is that the test will find a true difference when the actual average lifetime of the pens differs from the hypothesized mean of 10 hours.

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Most popular questions from this chapter

Suppose that you are an inspector for the Fish and Game Department and that you are given the task of determining whether to prohibit fishing along part of the Oregon coast. You will close an area to fishing if it is determined that fish in that region have an unacceptably high mercury content. a. Assuming that a mercury concentration of \(5 \mathrm{ppm}\) is considered the maximum safe concentration, which of the following pairs of hypotheses would you test: $$ H_{0}: \mu=5 \text { versus } H_{a}: \mu>5 $$ or $$ H_{0}: \mu=5 \text { versus } H_{a}: \mu<5 $$ Give the reasons for your choice. b. Would you prefer a significance level of \(.1\) or \(.01\) for your test? Explain.

When a published article reports the results of many hypothesis tests, the \(P\) -values are not usually given. Instead, the following type of coding scheme is frequently used: \(^{*} p<.05,^{* *} p<.01,^{*} *^{*} p<.001, *\) Which of the symbols would be used to code for each of the following \(P\) -values? a. 037 c. \(.072\) b. \(.0026\) d. \(.0003\)

A survey of teenagers and parents in Canada conducted by the polling organization Ipsos ("Untangling the Web: The Facts About Kids and the Internet," January 25 . 2006 ) included questions about Internet use. It was reported that for a sample of 534 randomly selected teens, the mean number of hours per week spent online was \(14.6\) and the standard deviation was \(11.6\). a. What does the large standard deviation, \(11.6\) hours, tell you about the distribution of online times for this sample of teens? b. Do the sample data provide convincing evidence that the mean number of hours that teens spend online is greater than 10 hours per week?

Newly purchased automobile tires of a certain type are supposed to be filled to a pressure of 30 psi. Let \(\mu\) denote the true average pressure. Find the \(P\) -value associated with each of the following given \(z\) statistic values for testing \(H_{0}: \mu=30\) versus \(H_{a}: \mu \neq 30\) when \(\sigma\) is known: a. \(2.10\) d. \(1.44\) b. \(-1.75\) e. \(-5.00\) c. \(0.58\)

Many people have misconceptions about how profitable small, consistent investments can be. In a survey of 1010 randomly selected U.S. adults (Associated Press, October 29,1999 ), only 374 responded that they thought that an investment of \(\$ 25\) per week over 40 years with a \(7 \%\) annual return would result in a sum of over \(\$ 100,000\) (the correct amount is \(\$ 286,640\) ). Is there sufficient evidence to conclude that less than \(40 \%\) of U.S. adults are aware that such an investment would result in a sum of over \(\$ 100,000 ?\) Test the relevant hypotheses using \(\alpha=.05\).

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