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Chapter 9: Problem 37

A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times, and determines that 14 of the plates have blistered. a. Does this provide compelling evidence for concluding that more than \(10 \%\) of all plates blister under such circumstances? State and test the appropriate hypotheses using a significance level of .05. In reaching your conclusion, what type of error might you have committed? b. If it is really the case that \(15 \%\) of all plates blister under these circumstances and a sample size of 100 is used, how likely is it that the null hypothesis of part (a) will not be rejected by the level 05 test? Answer this question for a sample size of 200 . c. How many plates would have to be tested to have \(\beta(.15)=.10\) for the test of part (a)?

Short Answer

Expert verified
a. Do not reject null; potential Type II error. b. Calculate \( \beta \) for \( n=100 \) or \( n=200 \); increases sample detection. c. Use calculations for \( n \) to achieve \( \beta =0.10 \).

Step by step solution

01

Understanding the Problem

We are given a sample of 100 nickel plates and found that 14 of them have blistered. We need to determine if there is evidence that more than 10% of all plates blister.First, identify the sample proportion as \( \hat{p} = \frac{14}{100} = 0.14 \). Our null hypothesis \( H_0 \) is that the true proportion \( p = 0.10 \) and the alternative \( H_a \) is that \( p > 0.10 \). We will use a significance level (\( \alpha \)) of 0.05.
02

Calculate the Test Statistic

Use the formula for the test statistic for proportions:\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]Where \( \hat{p} = 0.14 \), \( p_0 = 0.10 \), and \( n = 100 \).Substitute these values, we get:\[ Z = \frac{0.14 - 0.10}{\sqrt{\frac{0.10 \times 0.90}{100}}} = \frac{0.04}{0.03} \approx 1.33 \]
03

Determine the Critical Value and Decision

For a right-tailed test at a significance level of 0.05, the critical value of \( Z \) is approximately 1.645.Since our test statistic \( Z = 1.33 \) is less than the critical value, we do not reject the null hypothesis \( H_0 \). Thus, we do not have enough evidence to conclude that more than 10% of all plates blister.
04

Consider the Type of Error

Since we failed to reject the null hypothesis, the error we could potentially commit is a Type II error, where we incorrectly retain \( H_0 \) even though it is false.
05

Determine Probability for Part b with n=100

The probability that we fail to reject \( H_0 \) even though it is false is the Type II error probability, \( \beta \). Given \( p = 0.15 \), use:\[ Z = \frac{\hat{p} - 0.10}{\sqrt{\frac{0.10 \times 0.90}{100}}} \]Setting \( \hat{p} \) such that \( Z < 1.645 \), solve for additional probability calculations (requires a detailed application of power calculation formulas or a normal distribution table for precise \( \beta \)).
06

Extend Sample Size to n=200 for Part b

Repeat the calculations with \( n = 200 \). This effectively changes the standard error in the formula, potentially reducing \( \beta \) (the probability of a Type II error) due to the increased sample size, enhancing the power of the test to detect \( p > 0.10 \).
07

Calculate Required Sample Size for Part c

To achieve \( \beta = 0.10 \) when \( p = 0.15 \), use the formula for determining sample size in hypothesis testing. This involves setting the desired \( \beta \) into the power function, and solving for \( n \) such that both \( \alpha \) and \( \beta \) criteria are met. This might require iterative calculations or simulations utilizing statistical software for an exact solution.

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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
When conducting hypothesis tests, two types of errors can occur. Understanding these is crucial in the context of our battery testing scenario. A **Type I error** occurs when you wrongly reject the null hypothesis \(H_0\) even though it is true. This is akin to a false alarm. For example, in our scenario, concluding that more than 10% of all plates blister when they actually do not, due to sampling error.
  • Type I errors are controlled by the significance level \(\alpha\), which in this case is set to 0.05. This means there is a 5% risk of making a Type I error.
A **Type II error** happens when the null hypothesis is not rejected even though it is false. Essentially, it's a missed detection. In our exercise:
  • A Type II error would mean failing to recognize that more than 10% of plates are indeed problematic, leading to inadequate corrective measures in manufacturing.
  • Type II errors are denoted by \(\beta\), the risk of missing the truth.
Understanding these errors helps prioritize whether we should emphasize accuracy (lower Type I risk) or certainty (lower Type II risk).
Sample Size Calculation
Calculating an adequate sample size is vital for reliable hypothesis testing results. In the context of the battery test, it allows us to efficiently detect whether more than 10% of plates blister. As seen in the solution, the sample size directly affects the power of the test, which is the probability of correctly rejecting \(H_0\) when it is false.
  • Increasing the sample size from 100 to 200 in Part b means a smaller standard error. This enhancement means we're more likely to correctly detect when more than 10% of plates blister, thereby reducing the Type II error probability \(\beta\).
  • For improving test power to achieve \(\beta = 0.10\), part c involves calculating the exact necessary sample size. This requires balancing the significance level while keeping the desired power.
Ultimately, careful sample size determination is crucial for validating conclusions drawn from hypothesis testing with confidence.
Significance Level
The significance level \(\alpha\) is a threshold for determining how extreme the data must be before rejecting the null hypothesis \(H_0\). In hypothesis testing, it quantifies the acceptable Type I error rate.
In our battery example, \(\alpha\) is set to 0.05, a common choice, meaning a 5% risk of rejecting \(H_0\) when it is actually true.
  • Practically, choosing 0.05 as the significance level indicates a willingness to accept a 5% chance of mistakenly concluding that more than 10% of plates blister.
  • It helps in establishing a critical value that our test statistic is compared against, determining whether to reject \(H_0\) or not.
By using this benchmark, the manufacturer balances between cautious error avoidance and detecting meaningful deviations in the plate's performance. This understanding helps make informed decisions in quality control scenarios.

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Most popular questions from this chapter

Suppose the population distribution is normal with known \(\sigma\). Let \(\gamma\) be such that \(0<\gamma<\alpha\). For testing \(H_{0}: \mu=\mu_{0}\) versus \(H_{\mathrm{a}}: \mu \neq \mu_{0}\), consider the test that rejects \(H_{0}\) if either \(z \geq z_{\gamma}\) or \(z \leq-z_{\alpha-\gamma}\), where the test statistic is \(Z=\left(\bar{X}-\mu_{0}\right) /(\sigma / \sqrt{n})\). a. Show that \(P\) (type I error) \(=\alpha\). b. Derive an expression for \(\beta\left(\mu^{\prime}\right)\). [Hint: Express the test in the form "reject \(H_{0}\) if either \(\bar{x} \geq c_{1}\) or \(\left.\leq \mathrm{c}_{2} . "\right]\) c. Let \(\Delta>0\). For what values of \(\gamma\) (relative to \(\alpha\) ) will \(\beta\left(\mu_{0}+\Delta\right)<\beta\left(\mu_{0}-\Delta\right)\) ?

For each of the following assertions, state whether it is a legitimate statistical hypothesis and why: a. \(H: \sigma>100\) b. \(H: \tilde{x}=45\) c. \(H: s \leq .20\) d. \(H: \sigma_{1} / \sigma_{2}<1\) e. \(H: \bar{X}-\bar{Y}=5\) f. \(H: \lambda \leq .01\), where \(\lambda\) is the parameter of an exponential distribution used to model component lifetime

For a random sample of \(n\) individuals taking a licensing exam, let \(X_{i}=1\) if the \(i\) th individual in the sample passes the exam and \(X_{i}=0\) otherwise \((i=1, \ldots, n)\). a. With \(p\) denoting the proportion of all examtakers who pass, show that the most powerful test of \(H_{0}: p=.5\) versus \(H_{\mathrm{a}}: p=.75\) rejects \(H_{0}\) when \(\Sigma x_{i} \geq c\). b. If \(n=20\) and you want \(\alpha \leq .05\) for the test of (a), would you reject \(H_{0}\) if 15 of the 20 individuals in the sample pass the exam? c. What is the power of the test you used in (b) when \(p=.75[\) i.e., what is \(\pi(.75)]\) ? d. Is the test derived in (a) UMP for testing the hypotheses \(H_{0}: p=.5\) versus \(H_{\mathrm{a}}: p>.5\) ? Explain your reasoning. e. Graph the power function \(\pi(p)\) of the test for the hypotheses of (d) when \(n=20\) and \(\alpha \leq .05\). f. Return to the scenario of (a), and suppose the test is based on a sample size of 50 . If the probability of a type II error is approximately \(.025\), what is the approximate significance level of the test (use a normal approximation)?

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 \(150^{\circ}, 50\) water samples will be taken at randomly selected times, and the temperature of each sample recorded. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ}\) versus \(H_{\mathrm{a}}: \mu>150^{\circ}\). In the context of this situation, describe type I and type II errors. Which type of error would you consider more serious? Explain.

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)?
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