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Let \(\mu\) denote the true average lifetime (in hours) for a certain type of battery 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 \(\sigma_{\bar{x}}=.1\). 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
\(\alpha\) is the probability of the observing Z-value less than or equal to -1.28. To calculate \(\beta\) for \(\mu=9.8\) and \(\mu=9.5\), subtract the calculated z-scores from -1.28 and find the corresponding probability. As expected, \(\beta\) decreases with a decrease in \(\mu\). The power of the test is \(1 - \beta\), for \(\mu = 9.8\) and \(\mu = 9.5\).

Step by step solution

01

Calculating \(\alpha\)

To calculate \(\alpha\), determine the area to the left of the test statistic value under the standard normal curve. The value where the null hypothesis is rejected is \(z \leq -1.28\). As \(\alpha\) represents the probability of rejecting the null hypothesis when it is true, use the standard normal distribution to find the cumulative probability at \(z = -1.28\). This is given by \(\alpha = P(Z \leq -1.28)\).
02

Calculating \(\beta\) when \(\mu = 9.8\)

To calculate \(\beta\) at \(\mu = 9.8\), first calculate the z-score at \(\mu = 9.8\), then use the standard normal distribution to find the probability of not rejecting the null hypothesis when it is false. The z-score is given by \(z = \frac{(9.8 - 10)}{0.1} = -2\), and so \(\beta = P(Z > -1.28 | \mu = 9.8) = P(Z > -2 + 1.28) = P(Z > -0.72)\).
03

Comparing \(\beta\)s

The value of \(\beta\) decreases with a decrease in the value of \(\mu\) because for smaller values of \(\mu\), it's less likely to not reject \(H_0\) when it is false.
04

Computing \(\beta\) when \(\mu=9.5\)

Calculate the z-score at \(\mu = 9.5\), then use the standard normal distribution to find the probability of not rejecting the null hypothesis when it is false. The z-score is given by \(z = \frac{(9.5 - 10)}{0.1} = -5\), and so \(\beta = P(Z > -1.28 | \mu = 9.5) = P(Z > -5 + 1.28) = P(Z > -3.72)\). As expected, \(\beta\) when \(\mu = 9.5\) is smaller than \(\beta\) when \(\mu = 9.8\).
05

Calculating Power of Test

The power of the test is \(1 - \beta\). So, for \(\mu = 9.8\), the power of the test is \(1 - \beta_{\mu=9.8}\) and for \(\mu = 9.5\), it is \(1 - \beta_{\mu=9.5}\).

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Key Concepts

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

Null Hypothesis
Understanding the concept of the null hypothesis is crucial in hypothesis testing. It is a statement that there is no effect or no difference, and it often represents a position of skepticism or the status quo. For example, in the given exercise, the null hypothesis (\( H_0 \)) is that the true average lifetime (\( \mu \) for a certain type of battery is 10 hours (\( \mu = 10 \) hours).

When conducting a hypothesis test, you aim to determine if there is enough evidence to reject the null hypothesis. The test is set up to control the rate at which we are willing to risk a type I error, which is rejecting the null hypothesis when it is true. This is signified by \( \alpha \) in the exercise, and its value is determined based on the chosen significance level for the test.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to decide whether to reject the null hypothesis. The formula for the test statistic takes into account the sample mean, the hypothesized population mean, and the standard error of the mean.

In our given problem, the test statistic is a z-score calculated by \( z = \frac{\bar{x} - 10}{0.1} \) which measures the number of standard errors the sample mean (\( \bar{x} \) is from the hypothesized population mean of 10 hours. The smaller the test statistic, the less likely the sample mean is drawn from a population where the true mean is 10 hours, providing evidence against the null hypothesis.
Type II Error
A Type II error, represented by \( \beta \), occurs when a hypothesis test fails to reject a false null hypothesis. In other words, it's the error of not detecting an effect that is actually there. The probability of making a Type II error depends on the true population mean. The further the true mean is from the hypothesized mean under the null hypothesis, the lower the probability of making a Type II error.

In the given exercise, we're calculating \( \beta \) when \( \mu = 9.8 \) hours, which indicates the probability of erroneously not rejecting the null hypothesis when the true average lifetime of the battery is actually 9.8 hours. Students are often confused about calculating \( \beta \), so emphasizing that it is the probability of the test statistic being less extreme than the critical value, given that the alternative hypothesis is actually true, can help clarify.
Power of a Test
The power of a test, which is \( 1 - \beta \) , measures the test's ability to correctly reject a false null hypothesis. It is directly related to Type II error (\( \beta \) in that increasing power decreases the probability of a Type II error. It is affected by several factors including the significance level (\( \alpha \), sample size, effect size, and population variance.

In the context of the exercise, students are asked to calculate the power of the test for two different true means of the battery life (\( \mu \) . As one moves further away from the null hypothesis value, the power of the test increases because the likelihood of correctly rejecting the false null hypothesis rises. Hence, when the true mean battery life (\( \mu \) decreases from 9.8 to 9.5 hours, the power of the test increases, indicating a higher chance of detecting the actual difference.

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Most popular questions from this chapter

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal probability plot of the resulting data supports the use of a one-sample \(t\) test. The relevant hypotheses are \(H_{0}: \mu=10\) versus \(H_{a}: \mu<10 .\) a. If \(t=-2.3\) and \(\alpha=.05\) is selected, what conclusion is appropriate? b. If \(t=-1.83\) and \(\alpha=.01\) is selected, what conclusion is appropriate? c. If \(t=0.47,\) what conclusion is appropriate?

Use the definition of the \(P\) -value to explain the following: a. Why \(H_{0}\) would be rejected if \(P\) -value \(=.0003\) b. Why \(H_{0}\) would not be rejected if \(P\) -value \(=.350\)

According to a Washington Post-ABC News poll, 331 of 502 randomly selected U.S. adults interviewed said they would not be bothered if the National Security Agency collected records of personal telephone calls they had made. Is there sufficient evidence to conclude that a majority of U.S. adults feel this way? Test the appropriate hypotheses using a .01 significance level.

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above \(150^{\circ} \mathrm{F}\), researchers will take 50 water samples at randomly selected times and record the temperature of each sample. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ} \mathrm{F}\) versus \(H_{d}: \mu>\) \(150^{\circ} \mathrm{F}\). In the context of this example, describe Type I and Type II errors. Which type of error would you consider more serious? Explain.

10.52 - Medical research has shown that repeated wrist extension beyond 20 degrees increases the risk of wrist and hand injuries. Each of 24 students at Cornell University used a proposed new computer mouse design, and while using the mouse, each student's wrist extension was recorded. Data consistent with summary values given in the paper "Comparative Study of Two Computer Mouse Designs" (Cornell Human Factors Laboratory Technical Report \(\mathrm{RP} 7992\) ) are given. Use these data to test the hypothesis that the mean wrist extension for people using this new mouse design is greater than 20 degrees. Are any assumptions required in order for it to be appropriate to generalize the results of your test to the population of Cornell students? To the population of all university students? \(\begin{array}{llllllllllll}27 & 28 & 24 & 26 & 27 & 25 & 25 & 24 & 24 & 24 & 25 & 28 \\ 22 & 25 & 24 & 28 & 27 & 26 & 31 & 25 & 28 & 27 & 27 & 25\end{array}\)

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