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Chapter 9: Problem 8

Consider \(H_{0}=\mu=55\) versus \(H_{1}: \mu \neq 55\). a. What type of error would you make if the null hypothesis is actually false and you fail to reject it? b. What type of error would you make if the null hypothesis is actually true and you reject it?

Short Answer

Expert verified
a) Thinking the mean is \(55\) when it's not is a Type II error. b) Rejecting the idea that the mean is \(55\) when indeed it is a Type I error.

Step by step solution

01

Identifying Type II Error

Type II error would occur if the null hypothesis \((H_{0})\) is actually false and it is not rejected. This means the true mean is not equal to \(55\) but it is concluded that it is. So in the context of the exercise, a type II error is when the mean is not actually \(55\) (it could be less than or greater than \(55\)), but the tests fail to show this and we continue to believe that the mean is \(55\). That would be a mistake.
02

Identifying Type I Error

Type I error would occur if the null hypothesis \((H_{0})\) is actually true and it is rejected. This means the true mean is \(55\) but it is concluded that it is not. So in the context of this exercise, a type I error is when the mean is actually \(55\), but due to the sample data, it is incorrectly concluded that it is not equal to \(55\) (could have been mistakenly determined as lesser or greater). That would be an error.

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Key Concepts

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

Type I Error
In hypothesis testing, a Type I error occurs when we mistakenly reject the null hypothesis \((H_{0})\) even though it is true. Think of it as a false alarm or a false positive. In other words, it suggests that there is an effect or a difference when, in fact, there is none. It confuses the absence of effect with an actual effect.

For example, consider the hypothesis test where the null hypothesis is that the population mean \(\mu = 55\). A Type I error would occur if we mistakenly conclude that \(\mu eq 55\) based on our sample data, even though in reality it is exactly \(55\).
  • Occurs when \(H_{0}\) is true but rejected
  • Leads to the incorrect belief of a false truth
  • Threshold is often represented by \(\alpha\) (alpha), commonly set at 0.05
Type II Error
A Type II error happens when we fail to reject a false null hypothesis. It's often referred to as a missed discovery or a false negative. This essentially means that we conclude there is no effect or difference, when there actually is one.

In our exercise, if the true mean is not \(55\) and our test fails to identify this, then a Type II error has occurred. It means our statistical test did not have enough power to detect the deviation from the hypothesized mean.
  • Occurs when \(H_{0}\) is false but not rejected
  • Leads to missing a true effect or difference
  • Related to the concept of \(\beta\) (beta), which is the probability of making a Type II error
Null Hypothesis
The null hypothesis, denoted as \(H_{0}\), is a statement or hypothesis we wish to test. It is often a statement of 'no effect' or 'no difference'. The null hypothesis acts as a starting point for statistical testing and is assumed true until proven otherwise through evidence.

In our context, \(H_{0}: \mu = 55\) indicates that the population mean is thought to be \(55\). The aim of our hypothesis testing is to determine whether the evidence (sample data) robustly supports this claim or not.
  • Acts as the default assumption
  • Assumed to be true unless evidence suggests otherwise
  • It's what we seek to test and potentially reject
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_{1}\) or \(H_{a}\), is the statement we wish to provide evidence for if the null hypothesis is rejected. It is a statement that contradicts the null hypothesis and is typically what the researcher believes might be true.

For the exercise, \(H_{1}: \mu eq 55\) suggests that the population mean indeed differs from \(55\). Proving the alternative hypothesis involves showing that the null hypothesis does not hold, which means that there is a statistically significant difference in the results;
  • Contradicts the null hypothesis
  • Becomes the accepted theory if \(H_{0}\) is rejected
  • Represents the possibility of an effect or difference

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Most popular questions from this chapter

A past study claimed that adults in America spent an average of 18 hours a week on leisure activi ties. A researcher wanted to test this claim. She took a sample of 12 adults and asked them about the time they spend per week on leisure activities. Their responses (in hours) are as follows \(\begin{array}{lllll}13.6 & 14.0 & 24.5 & 24.6 & 22.9\end{array}\) \(\begin{array}{llllllll}37.7 & 14.6 & 14.5 & 21.5 & 21.0 & 17.8 & 21.2\end{array}\) Assume that the times spent on leisure activities by all American adults are normally distributed. Us ng a \(10 \%\) significance level, can you conclude that the average amount of time spent by American adults on leisure activities has changed? (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections 3.1.1 and 3.2.2 of Chapter 3. Then make the test of hypothesis about \(\mu .\)

A consumer advocacy group suspects that a local supermarket's 10 -ounce packages of cheddar cheese actually weigh less than 10 ounces. The group took a random sample of 20 such packages and found that the mean weight for the sample was \(9.955\) ounces. The population follows a normal distribution with the population standard deviation of \(.15\) ounce. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the mean weight of all such packages is less than 10 ounces. Will you reject the null hypothesis at \(\alpha=.01\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01\).

For each of the following examples of tests of hypothesis about \(\mu\), show the rejection and nonrejection regions on the \(t\) distribution curve. a. A two-tailed test with \(\alpha=.02\) and \(n=20\) b. A left-tailed test with \(\alpha=.01\) and \(n=16\) c. A right-tailed test with \(\alpha=.05\) and \(n=18\)

According to the U.S. Postal Service, the average weight of mail received by Americans in 2011 through the Postal Service was \(57.2\) pounds (The New York Times, December 4,2011 ). One hundred randomly selected Americans were asked to keep all their mail for last year. It was found that they received an average of \(55.3\) pounds of mail last year. Suppose that the population standard deviation is \(8.4\) pounds. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the average weight of mail received by all Americans last year was less than \(57.2\) pounds. Will you reject the null hypothesis at \(\alpha=.01 ?\) Explain. What if \(\alpha=.025\) ? b. Test the hypothesis of part a using the critical-value approach. Will you reject the null hypothesis at \(\alpha=.01\) ? What if \(\alpha=.025\) ?

According to a New York Times/CBS News poll conducted during June \(24-28,2011,55 \%\) of the American adults polled said that owning a home is a very important part of the American Dream (The New York Times, June 30,2011 ). Suppose this result was true for the population of all American adults in \(2011 .\) In a recent poll of 1800 American adults, \(61 \%\) said that owning a home is a very important part of the American Dream. Perform a hypothesis test to determine whether it is reasonable to conclude that the percentage of all American adults who currently hold this opinion is higher than \(55 \%\). Use a \(2 \%\) significance level, and use both the \(p\) -value and the critical-value approaches.
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