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You read that a statistical test at the \(\alpha=0.01\) level has probability 0.14 of making a Type II error when a specific alternative is true. What is the power of the test against this alternative?

Short Answer

Expert verified
The power of the test is 0.86.

Step by step solution

01

Understand the Definitions

In statistical hypothesis testing, a Type II error occurs when we fail to reject a false null hypothesis. The probability of making a Type II error is denoted by \(\beta\). Power is defined as the probability of correctly rejecting a false null hypothesis and is calculated as \(1 - \beta\).
02

Identify Known Values

From the problem, we are given that the probability of making a Type II error (\(\beta\)) is 0.14. We need to find the power of the test against this alternative.
03

Calculate the Power

The power of the test is calculated using the formula \(\text{Power} = 1 - \beta\). Plug in the given value of \(\beta\): \(\text{Power} = 1 - 0.14 = 0.86\).

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Key Concepts

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

Type I Error
A Type I error, often denoted by the Greek letter \( \alpha \), is a fundamental concept in hypothesis testing. It occurs when we reject a true null hypothesis. Essentially, it's a false positive result, where you conclude that there is an effect or a difference when, in reality, there isn’t one. This can lead to incorrect conclusions and can be influenced by the significance level you choose during testing.

The significance level \( \alpha \) is pre-determined by the researcher and typically set at values like 0.05 or 0.01. It represents the probability of making a Type I error. A lower \( \alpha \) means you are less likely to incorrectly reject a true null hypothesis, but it can also reduce the sensitivity of the test. Choosing the appropriate value for \( \alpha \) is important to balance the risk of Type I errors against other factors in your study.
Type II Error
Type II error occurs when we fail to reject a false null hypothesis. It’s a 'missed opportunity' where a true effect exists, but our test does not capture it. The probability of making a Type II error is denoted by \( \beta \). Smaller \( \beta \) values indicate a more sensitive test.

Several factors can increase the likelihood of Type II errors:
  • Small sample size - larger samples give more reliable results.
  • Low effect size - when the difference we are trying to detect is very small.
  • High variability in data - when data points are very spread out or inconsistent.
By understanding and minimizing the factors that contribute to Type II errors, researchers can increase the power of their tests, which is the ability to correctly identify true effects.
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions or inferences about populations based on sample data. This process typically starts with constructing a null hypothesis \( H_0 \), which states that there is no effect or difference in the population, and an alternative hypothesis \( H_a \), which suggests the opposite.

The testing process involves:
  • Choosing a significance level \( \alpha \), which affects Type I error probability.
  • Calculating a test statistic from the data that needs to be compared to a threshold (critical value) determined by \( \alpha \).
  • Deciding based on this comparison whether to reject or fail to reject \( H_0 \).
Hypothesis testing is crucial in scientific research, helping to validate findings. It helps in distinguishing real differences from those occurring by chance. However, it’s also important to consider both Type I and Type II errors in this process, which can affect the conclusions drawn from the hypothesis test.

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Most popular questions from this chapter

For the study of Jordanian children in Exercise 2 , the sample mean hemoglobin level was \(11.3 \mathrm{~g} / \mathrm{dl}\) and the sample standard deviation was \(1.6 \mathrm{~g} / \mathrm{dl} .\) A significance test yields a \(P\) -value of 0.0016 . (a) Explain what it would mean for the null hypothesis to be true in this setting. (b) Interpret the \(P\) -value in context.

Which of the following \(95 \%\) confidence intervals would lead us to reject \(H_{0}: p=0.30\) in favor of \(H_{a}: p \neq 0.30\) at the \(5 \%\) significance level? (a) (0.19,0.27) (c) (0.27,0.31) (e) None of these (b) (0.24,0.30) (d) (0.29,0.38)

In early 2012 , the Pew Internet and American Life Project asked a random sample of U.S. adults, "Do you ever ... use Twitter or another service to share updates about yourself or to see updates about others?" According to Pew, the resulting \(95 \%\) confidence interval is \((0.123\), 0.177)\(.^{13}\) Does this interval provide convincing evidence that the actual proportion of U.S. adults who would say they use Twitter differs from \(0.16 ?\) Justify your answer.

Does Friday the 13 th have an effect on people's behavior? Researchers collected data on the number of shoppers at a sample of 45 nearby grocery stores on Friday the 6 th and Friday the 1 3th in the same month. The dotplot and computer output below summarize the data on the difference in the number of shoppers at each store on these two days (subtracting in the order 6 th minus 13 th \() .^{25}\) Researchers would like to carry out a test of \(H_{0}: \mu_{d}=0\) versus \(H_{a}: \mu_{d} \neq 0,\) where \(\mu_{d}\) is the true mean difference in the number of grocery shoppers on these two days. Which of the following conditions for performing a paired \(t\) test are clearly satisfied? I. Random II. \(10 \%\) III. Normal/Large Sample (a) I only (b) II only (c) III only (d) I and II only (e) I, II, and III

You manufacture and sell a liquid product whose electrical conductivity is supposed to be \(5 .\) You plan to make six measurements of the conductivity of each lot of product. If the product meets specifications, the mean of many measurements will be \(5 .\) You will therefore test $$ \begin{array}{l} H_{0}: \mu=5 \\ H_{a}: \mu \neq 5 \end{array} $$ If the true conductivity is \(5.1,\) the liquid is not suitable for its intended use. You learn that the power of your test at the \(5 \%\) significance level against the alternative \(\mu=5.1\) is 0.23. (a) Explain in simple language what "power \(=0.23 "\) means in this setting. (b) You could get higher power against the same alternative with the same \(\alpha\) by changing the number of measurements you make. Should you make more measurements or fewer to increase power? (c) If you decide to use \(\alpha=0.10\) in place of \(\alpha=0.05\), with no other changes in the test, will the power increase or decrease? Justify your answer. (d) If you shift your interest to the alternative \(\mu=5.2\), with no other changes, will the power increase or decrease? Justify your answer.

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