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What is meant by a type I error? A type II error? How are they related?

Short Answer

Expert verified
Type I error is rejecting a true null hypothesis; Type II is failing to reject a false null. They are inversely related; reducing one increases the other.

Step by step solution

01

Understanding Type I Error

In hypothesis testing, a Type I error occurs when we reject the null hypothesis when it is actually true. This is also known as a 'false positive'. It's like finding a difference when there isn't one. The probability of making a Type I error is denoted by \( \alpha \), and is called the significance level of the test, typically set at 0.05 or 5%.
02

Understanding Type II Error

A Type II error happens when we fail to reject the null hypothesis when it is actually false. This is also known as a 'false negative'. It occurs when we don't detect a difference when there is one. The probability of making a Type II error is denoted by \( \beta \).
03

Relationship Between Type I and Type II Errors

Type I and Type II errors are inversely related. Reducing the risk of one generally increases the risk of the other. By choosing a lower significance level (\( \alpha \)) to reduce Type I errors, you might increase the risk of Type II errors (\( \beta \)), assuming other factors constant. They are a trade-off to consider when designing experiments.

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Key Concepts

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

Type I Error
When conducting hypothesis testing, one common mistake is known as a Type I error. This occurs when we incorrectly reject the null hypothesis despite it being true. Imagine you're a scientist testing if a new drug has an effect, and you conclude that it does, but in reality, it doesn't. This is known as a 'false positive'. In simple terms, you are finding a difference or effect when none actually exists.

The probability of making a Type I error is labeled as \( \alpha \). This is known as the **significance level** of the test, and it reflects the risk you are willing to take of making this mistake. Typically, scientists set \( \alpha \) at 0.05, or 5%, meaning there's a 5% risk of concluding that there is an effect when there isn't one.

  • Key Takeaway: Type I error is a 'false positive'.
  • Probability: Denoted by \( \alpha \), often set at 0.05.
  • Significance Level: Represents how confidently we are rejecting the null hypothesis.
Type II Error
A Type II error, in contrast, happens when we fail to reject the null hypothesis even though it's false. This is often referred to as a 'false negative'. Continuing with the same drug test example, this would mean not detecting the drug's effect when it indeed has one. It’s like failing to recognize a true difference or effect when it’s there.

The probability of committing a Type II error is symbolized as \( \beta \). Unlike the significance level, \( \beta \) does not have a standard value, but it is important to minimize it, as failing to identify a true effect can have serious implications, particularly in fields like medicine.

  • Key Takeaway: Type II error is a 'false negative'.
  • Probability: Denoted by \( \beta \), should be minimized.
  • Outcome: Overlooked true effect or difference.
Significance Level
The significance level, denoted by \( \alpha \), represents the probability of making a Type I error in hypothesis testing. It's a threshold set by researchers to determine if the results of a test are statistically significant. By default, many researchers use a significance level of 0.05.

Choosing a significance level is a balance act. If it's too low, you may miss genuine effects (increasing the chance of a Type II error). If it's too high, you might identify effects that aren’t there (increasing the chance of a Type I error). By adjusting the significance level according to the context and consequences of the errors, researchers can better control the risk of making wrongful conclusions.

  • Importance: Balances between minimizing Type I and Type II errors.
  • Balance: Key in designing experiments and interpreting results.
  • Common Value: Often set at 5% or 0.05.

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Most popular questions from this chapter

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. A machine fills 12 -ounce bottles with soda. For the machine to function properly, the standard deviation of the population must be less than or equal to 0.03 ounce. A random sample of 8 bottles is selected, and the number of ounces of soda in each bottle is given. At \(\alpha=0.05,\) can we reject the claim that the machine is functioning properly? Use the \(P\) -value method. \(\begin{array}{llll}12.03 & 12.10 & 12.02 & 11.98 \\ 12.00 & 12.05 & 11.97 & 11.99\end{array}\)

Suppose a statistician chose to test a hypothesis at \(\alpha=0.01\). The critical value for a right-tailed test is \(+2.33 .\) If the test value were \(1.97,\) what would the decision be? What would happen if, after seeing the test value, she decided to choose \(\alpha=0.05 ?\) What would the decision be? Explain the contradiction, if there is one.

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. A random sample of second-round golf scores from a major tournament is listed below. At \(\alpha=0.10\), is there sufficient evidence to conclude that the population variance exceeds \(9 ?\) \(\begin{array}{lllll}75 & 67 & 69 & 72 & 70 \\ 66 & 74 & 69 & 74 & 71\end{array}\)

Find the critical value (or values) for the \(t\) test for each. a. \(n=12, \alpha=0.01,\) left-tailed b. \(n=16, \alpha=0.05,\) right-tailed c. \(n=7, \alpha=0.10,\) two-tailed d. \(n=11, \alpha=0.025,\) right-tailed e. \(n=10, \alpha=0.05,\) two-tailed

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. The average cost for teeth straightening with metal braces is approximately \(\$ 5400\). A nationwide franchise thinks that its cost is below that figure. A random sample of 28 patients across the country had an average cost of \(\$ 5250\) with a standard deviation of \(\$ 629 .\) At \(\alpha=0.025,\) can it be concluded that the mean is less than \(\$ 5400 ?\)

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