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Chapter 10: Problem 64

Optical fibers are used in telecommunications to transmit light. Suppose current technology allows production of fibers that transmit light about \(50 \mathrm{~km} .\) Researchers are trying to develop a new type of glass fiber that will increase this distance. In evaluating a new fiber, it is of interest to test \(H_{0}: \mu=50\) versus \(H_{a}: \mu>50\), with \(\mu\) denoting the mean transmission distance for the new optical fiber. a. Assuming \(\sigma=10\) and \(n=10,\) use Appendix Table 5 to find \(\beta,\) the probability of a Type II error, for each of the given alternative values of \(\mu\) when a test with significance level .05 is employed: i. 52 ii. 55 \(\begin{array}{ll}\text { iii. } 60 & \text { iv. } 70\end{array}\) b. What happens to \(\beta\) in each of the cases in Part (a) if \(\sigma\) is actually larger than 10? Explain your reasoning.

Short Answer

Expert verified
In Part a, using the formula for z-score and the standard normal distribution table, the beta values for each \( \mu \) can be calculated. In Part b, as the standard deviation increases, the data points become more spread out which could make it harder to reject the null hypothesis when it is false, thus increasing the probability of a Type II error .

Step by step solution

01

Calculate z-score

For each given \( \mu \), calculate the z-score using the formula: \( z = (\mu - \mu_0) / (\sigma / \sqrt{n}) \) where \( \mu_0 \) = 50 km (the hypothesized mean), \( \sigma \) = 10 km (standard deviation), and n = 10 (sample size). For \( \mu \) = 52, 55, 60, and 70 km respectively, the z-scores are: \( z_{52} \), \( z_{55} \), \( z_{60} \), and \( z_{70} \).
02

Calculate beta

Use Appendix Table 5 to find the probability corresponding to each z-score obtained in Step 1. Since the test is one-tailed and considering the test is at .05 level of significance, one should look up the probability of the z-score and subtract from 1 to find \( \beta \). Repeat for all z-scores.
03

Explain variance effect on beta

In Part b, provide a general explanation for how increasing the standard deviation (increasing spread or dispersion of the data) affects the probability of Type II error \( \beta \). This does not involve any calculations.

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Key Concepts

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

Type II Error
In hypothesis testing, a Type II error occurs when we fail to reject the null hypothesis, even though the alternative hypothesis is true. In simpler terms, it's like saying the new optical fiber is not better than the current one, even though it actually is.

- **Understanding Beta (\( \beta \))**: The probability of making a Type II error is denoted by \( \beta \). Calculating \( \beta \) helps us understand the likelihood of missing an improvement when evaluating the new fiber.- **Importance in Testing**: Knowing the probability of a Type II error helps researchers decide the value of continuing or stopping research efforts on the new fiber technology.

- **Connected Factors**: The probability \( \beta \) is influenced by factors such as significance level, sample size, and the effect size (how much better or different the new fiber is compared to the old one). Enhance your intuition by imagining a net trying to catch improvements 鈥� a larger net (more samples, higher significance) catches more true improvements.
Z-Score Calculation
The z-score is a statistical measurement that indicates how many standard deviations an element is from the mean. In this exercise, it helps compare the observed transmission distances of the optical fibers to the hypothesized mean.

- **Calculation Formula**: To find the z-score, use the formula: \[ z = \frac{\mu - \mu_0}{\sigma / \sqrt{n}} \]where \( \mu \) is the observed mean, \( \mu_0 \) is the hypothesized mean (50 km), \( \sigma \) is the standard deviation, and \( n \) is the sample size (10).

- **Why Z-Scores Matter**: They allow us to determine how far off the sampled mean is from the hypothesized average. A higher z-score implies a greater degree of separation from the hypothesized mean, suggesting significant innovation in the fiber's distance transmission.

- **Using Z-Values**: Once computed, these z-scores help locate the probability of the research data (or more extreme) under the null hypothesis using standard normal distribution tables.
Standard Deviation Effect
Standard deviation reflects the variation or spread of data points around the mean. It plays a crucial role in hypothesis testing and significantly influences the Type II error.

- **Effect on Data Spread**: A smaller standard deviation means the data points are closer to the mean, suggesting more consistent results. Conversely, a larger standard deviation means more diverse outcomes, making it challenging to detect true changes.

- **Relation with \( \beta \)**: When the standard deviation increases, the value of \( \beta \) (probability of Type II error) increases as well, making it harder to detect improvements in the new optical fiber's transmission distance, even if they exist.
- **Reasoning for Variance Effect**: High variability implies samples might overlap with the hypothesized mean, reducing the distinguishability between the current and new fiber technologies.

Understanding how the standard deviation impacts \( \beta \) provides insights into the reliability and precision needed for testing new fibers.
Significance Level
The significance level, typically denoted by alpha (\( \alpha \)), is the threshold for rejecting the null hypothesis. In this exercise, a significance level of 0.05 is used, meaning there is a 5% risk of rejecting the null hypothesis when it is actually true.

- **Setting the Threshold**: Choosing the right significance level depends on the balance between Type I errors (false positives) and Type II errors (false negatives).

- **Impact on Testing Outcomes**: A lower significance level means stronger evidence is needed to reject the null hypothesis, decreasing the probability of a Type I error. However, this can inadvertently raise \( \beta \) if care isn't taken.- **Importance in Fiber Testing**: Since researchers want to confirm the superiority of the new fiber with strong evidence, a 0.05 level balances sensitivity with fairness. If the distance exceeds 50 km consistently, the fiber's potential for better transmission becomes credible.

Selecting a proper significance level helps ensure that the improvements observed in the testing phase are not due to random chance.

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Most popular questions from this chapter

For the following pairs, indicate which do not comply with the rules for setting up hypotheses, and explain why: a. \(H_{0}: \mu=15, H_{a}: \mu=15\) b. \(H_{0}: p=.4, H_{a}: p>.6\) c. \(H_{0}: \mu=123, H_{a}: \mu<123\) d. \(H_{0}: \mu=123, H_{d}: \mu=125\) e. \(\quad H_{0}: \hat{p}=.1, H_{a}: \hat{p} \neq .1\)

The article "Poll Finds Most Oppose Return to Draft, Wouldn't Encourage Children to Enlist" (Associated Press, December 18,2005 ) reports that in a random sample of 1000 American adults, 700 indicated that they oppose the reinstatement of a military draft. Is there convincing evidence that the proportion of American adults who oppose reinstatement of the draft is greater than two-thirds? Use a significance level of .05 .

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 \(150^{\circ} \mathrm{F}\), a scientist will take 50 water samples at randomly selected times and will record the water temperature of each sample. She will then use a \(z\) statistic $$ z=\frac{\bar{x}-150}{\frac{\sigma}{\sqrt{n}}} $$ to decide between the hypotheses \(H_{0}: \mu=150\) and \(H_{a}: \mu>150,\) where \(\mu\) is the mean temperature of discharged water. Assume that \(\sigma\) is known to be 10 . a. Explain why use of the \(z\) statistic is appropriate in this setting. b. Describe Type I and Type II errors in this context. \(c\). The rejection of \(H_{0}\) when \(z \geq 1.8\) corresponds to what value of \(\alpha\) ? (That is, what is the area under the \(z\) curve to the right of \(1.8 ?\) ) d. Suppose that the actual value for \(\mu\) is 153 and that \(H_{0}\) is to be rejected if \(z \geq 1.8 .\) Draw a sketch (similar to that of Figure 10.5 ) of the sampling distribution of \(\bar{x},\) and shade the region that would represent \(\beta\), the probability of making a Type II error. e. For the hypotheses and test procedure described, compute the value of \(\beta\) when \(\mu=153\). f. For the hypotheses and test procedure described, what is the value of \(\beta\) if \(\mu=160\) ? g. What would be the conclusion of the test if \(H_{0}\) is rejected when \(z \geq 1.8\) and \(\bar{x}=152.4\) ? What type of error might have been made in reaching this conclusion?

Researchers at the University of Washington and Harvard University analyzed records of breast cancer screening and diagnostic evaluations ("Mammogram Cancer Scares More Frequent than Thought," USA Today, April 16, 1998). Discussing the benefits and downsides of the screening process, the article states that, although the rate of false-positives is higher than previously thought, if radiologists were less aggressive in following up on suspicious tests, the rate of false-positives would fall but the rate of missed cancers would rise. Suppose that such a screening test is used to decide between a null hypothesis of \(H_{0}:\) no cancer is present and an alternative hypothesis of \(H_{a}:\) cancer is present. (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) a. Would a false-positive (thinking that cancer is present when in fact it is not) be a Type I error or a Type II error? b. Describe a Type I error in the context of this problem, and discuss the consequences of making a Type I error. c. Describe a Type II error in the context of this problem, and discuss the consequences of making a Type II error. d. What aspect of the relationship between the probability of Type I and Type II errors is being described by the statement in the article that if radiologists were less aggressive in following up on suspicious tests, the rate of false-positives would fall but the rate of missed cancers would rise?

In a survey of 1005 adult Americans, \(46 \%\) indicated that they were somewhat interested or very interested in having web access in their cars (USA Today, May I. 2009 ). Suppose that the marketing manager of a car manufacturer claims that the \(46 \%\) is based only on a sample and that \(46 \%\) is close to half, so there is no reason to believe that the proportion of all adult Americans who want car web access is less than \(.50 .\) Is the marketing manager correct in his claim? Provide statistical evidence to support your answer. For purposes of this exercise, assume that the sample can be considered as representative of adult Americans.
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