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PCBs, continued Refer to Exercise \(9.75 .\) a. Calculate \(\beta\) and \(1-\beta\) if the true mean ppm of PCBs is 6 ppm. b. Calculate \(\beta\) and \(1-\beta\) if the true mean ppm of PCBs is 7 ppm. c. Find the power, \(1-\beta,\) when \(\mu=8,9,10,\) and 12\. Use these values to construct a power curve for the test in Exercise \(9.75 .\) d. For what values of \(\mu\) does this test have power greater than or equal to \(90 ?\)

Short Answer

Expert verified
Short Answer: We calculated the probabilities of Type II error (\(\beta\)) and the power of the test for different true mean ppm values of PCBs (6, 7, 8, 9, 10, and 12 ppm). None of these values reached the desired power threshold of 0.90, indicating that our test does not have enough power to detect differences in the true mean ppm larger than the hypothesized value (\(\mu_0 = 5\)).

Step by step solution

01

Find the critical value

To find the critical value for the standard normal distribution, we look up the value where 1 - \(\alpha\) (0.99) of the area is to the left of it. Using a Z-table or calculator, we find that \(z_c \approx 2.33\).
02

Calculate \(\beta\) for \(\mu = 6\) ppm

To find the probability of a Type II error at \(\mu = 6\) ppm, we need to calculate the test statistic at this true mean and use the standard normal distribution to find the corresponding probability. The test statistic for \(\mu = 6\) ppm is: \(z = \frac{(\bar{x} - \mu)}{\frac{s}{\sqrt{n}}} = \frac{(5 - 6)}{\frac{1}{\sqrt{16}}} = -4\) Now, we find the probability under the standard normal curve between the critical value and the test statistic: \(\beta(6) = P(-4 \leq Z \leq 2.33) \approx 0.9896\)
03

Calculate the power of the test for \(\mu = 6\) ppm

The power of the test is the complemetary event, \(1 - \beta(\mu)\). Thus, for \(\mu = 6\) ppm, we have: \(1-\beta(6) = 1 - 0.9896 = 0.0104\)
04

Calculate \(\beta\) for \(\mu = 7\) ppm

Repeat step 2 for the true mean of \(\mu = 7\) ppm. The test statistic for \(\mu = 7\) ppm is: \(z = \frac{(5 - 7)}{\frac{1}{\sqrt{16}}} = -8\) \(\beta(7) = P(-8 \leq Z \leq 2.33) \approx 1\)
05

Calculate the power of the test for \(\mu = 7\) ppm

For \(\mu = 7\) ppm, we have: \(1 -\beta(7) = 1 - 1 = 0\)
06

Calculate the power of the test for other values of \(\mu\)

Repeat steps 2 and 3 for \(\mu = 8, 9, 10,\) and \(12\) ppm, and use these values to construct a power curve for the test. For simplicity, I'll present the calculated values: - For \(\mu = 8\) ppm, \(\beta(8) \approx 1\), \(1 - \beta(8) \approx 0\) - For \(\mu = 9\) ppm, \(\beta(9) \approx 1\), \(1 - \beta(9) \approx 0\) - For \(\mu = 10\) ppm, \(\beta(10) \approx 1\), \(1 - \beta(10) \approx 0\) - For \(\mu = 12\) ppm, \(\beta(12) \approx 1\), \(1 - \beta(12) \approx 0\)
07

Determine values of \(\mu\) for which the test has power greater than or equal to 0.90

We are asked to identify the values of \(\mu\) for which the power of the test is greater than or equal to 0.90, i.e., \(1 - \beta(\mu) \geq 0.90\). Based on our power calculations, it seems that the power of our test does not reach 0.90 for any of the values of \(\mu\) considered. In conclusion, we calculated the \(\beta\) values and the power of the test for different true mean ppm values for PCBs. The power of the test did not reach the desired 0.90 threshold for any of these values.

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Key Concepts

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

Power of a Test
The power of a test is a crucial aspect of hypothesis testing that measures the likelihood that the test will correctly reject a false null hypothesis. Essentially, it tells us how effective the test is. Higher power means a higher probability of detecting a true effect if one exists. The power of a test is calculated as the complement of the Type II error probability, denoted as \(1 - \beta\).
The power depends on a few key factors:
  • The significance level \(\alpha\) - Typically set at 0.05, a lower \(\alpha\) decreases power.
  • The true effect size - Larger effects are easier to detect, increasing power.
  • The sample size - Larger samples yield more accurate results, boosting power.
  • Variability in the data - Less variability provides clearer evidence of an effect, enhancing power.

To increase the power of a test, researchers can adjust these factors. For instance, they might increase the sample size, or opt for a higher \(\alpha\) if it's appropriate for the context. The power curve visually depicts how power changes with different true mean values (\(\mu\)) and helps identify situations where the test is or is not effective.
Type II Error
A Type II error occurs when a test fails to reject a false null hypothesis. In simpler terms, it's when we mistakenly conclude there's no effect or difference when, in reality, there is one. This is represented by the symbol \(\beta\), which signifies the probability of making this mistake.
Understanding Type II error is essential because:
  • It helps us gauge the reliability of our test results.
  • Lowering \(\beta\) (and hence increasing the power) means reducing the risk of this oversight.
  • It's particularly important in contexts where missing a true effect has significant consequences, like drug testing.

Factors that influence \(\beta\) include the same ones affecting power: sample size, effect size, significance level, and variability. By optimizing these elements, we can minimize the chance of making a Type II error. An awareness of this type of error aids researchers in improving study designs to better capture true effects.
Standard Normal Distribution
The standard normal distribution is a vital concept in statistics, especially in hypothesis testing. It's a special case of the normal distribution with a mean of 0 and a standard deviation of 1. Often represented by the letter \(Z\), it serves as the foundation for Z-scores and Z-tables, which are crucial for determining probabilities in hypothesis tests.
The standard normal distribution is used to calculate:
  • Probabilities related to test statistics, allowing us to assess the likelihood of obtaining specific sample outcomes.
  • Critical values, which tell us the threshold at which we reject or fail to reject the null hypothesis.

Z-scores transform raw data into a standard form, making it easier to compare results across different contexts and conditions. By doing this, researchers can determine how far a sample statistic is from the population mean, in terms of standard deviation. This transformation is essential for calculating Type I and Type II errors and defining a test's power, providing a unified framework for understanding statistical significance and decision-making in hypothesis testing.

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Most popular questions from this chapter

Independent random samples of \(n_{1}=140\) and \(n_{2}=140\) observations were randomly selected from binomial populations 1 and \(2,\) respectively. Sample 1 had 74 successes, and sample 2 had 81 successes. a. Suppose you have no preconceived idea as to which parameter, \(p_{1}\) or \(p_{2}\), is the larger, but you want to detect only a difference between the two parameters if one exists. What should you choose as the alternative hypothesis for a statistical test? The null hypothesis? b. Calculate the standard error of the difference in the two sample proportions, \(\left(\hat{p}_{1}-\hat{p}_{2}\right) .\) Make sure to use the pooled estimate for the common value of \(p\). c. Calculate the test statistic that you would use for the test in part a. Based on your knowledge of the standard normal distribution, is this a likely or unlikely observation, assuming that \(H_{0}\) is true and the two population proportions are the same? d. \(p\) -value approach: Find the \(p\) -value for the test. Test for a significant difference in the population proportions at the \(1 \%\) significance level. e. Critical value approach: Find the rejection region when \(\alpha=.01 .\) Do the data provide sufficient evidence to indicate a difference in the population proportions?

A random sample of 100 observations from a quantitative population produced a sample mean of 26.8 and a sample standard deviation of \(6.5 .\) Use the \(p\) -value approach to determine whether the population mean is different from \(28 .\) Explain your conclusions.

Find the \(p\) -value for the following large-sample \(z\) tests: a. A right-tailed test with observed \(z=1.15\) b. A two-tailed test with observed \(z=-2.78\) c. A left-tailed test with observed \(z=-1.81\)

Independent random samples of 80 measurements were drawn from two quantitative populations, 1 and \(2 .\) Here is a summary of the sample data: $$\begin{array}{lcc} & \text { Sample 1 } & \text { Sample 2 } \\\\\hline \text { Sample Size } & 80 & 80 \\\\\text { Sample Mean } & 11.6 & 9.7 \\\\\text { Sample Variance } & 27.9 & 38.4\end{array}$$ a. If your research objective is to show that \(\mu_{1}\) is larger than \(\mu_{2}\), state the alternative and the null hypotheses that you would choose for a statistical test. b. Is the test in part a one- or two-tailed? c. Calculate the test statistic that you would use for the test in part a. Based on your knowledge of the standard normal distribution, is this a likely or unlikely observation, assuming that \(H_{0}\) is true and the two population means are the same? d. \(p\) -value approach: Find the \(p\) -value for the test. Test for a significant difference in the population means at the \(1 \%\) significance level. e. Critical value approach: Find the rejection region when \(\alpha=.01 .\) Do the data provide sufficient evidence to indicate a difference in the population anc?

Treatment versus Control An experiment was conducted to test the effect of a new drug on a viral infection. The infection was induced in 100 mice, and the mice were randomly split into two groups of 50\. The first group, the control group, received no treatment for the infection. The second group received the drug. After a 30 -day period, the proportions of survivors, \(\hat{p}_{1}\) and \(\hat{p}_{2}\), in the two groups were found to be .36 and \(.60,\) respectively. a. Is there sufficient evidence to indicate that the drug is effective in treating the viral infection? Use \(\alpha=.05 .\) b. Use a \(95 \%\) confidence interval to estimate the actual difference in the cure rates for the treated versus the control groups.

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