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Chapter 9: Problem 65

The manager of a restaurant in a large city claims that waiters working in all restaurants in his city earn an average of \(\$ 150\) or more in tips per week. A random sample of 25 waiters selected from restaurants of this city yielded a mean of \(\$ 139\) in tips per week with a standard deviation of \(\$ 28\). Assume that the weekly tips for all waiters in this city have a normal distribution. a. Using the \(1 \%\) significance level, can you conclude that the manager's claim is true? Use both approaches. b. What is the Type I error in this exercise? Explain. What is the probability of making such an error?

Short Answer

Expert verified
The claim of the manager that the waiters earn \$150 or more in tips per week cannot be rejected at 1% level of significance. The probability of making a Type I error, i.e., wrongly rejecting the manager's claim when it is true, is 1%.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) is that the average tips waiters earn is $150 or more per week. The alternative hypothesis (\(H_1\)) is that the average tips are less than $150. Mathematically, it can be represented as: \(H_0: \mu \geq \$150\) and \(H_1: \mu < \$150\).
02

Compute Test Statistic

The test statistic for a one-sample t-test is calculated as: \(t = \frac{\bar{x} - \mu}{s / \sqrt{n}}\). Substituting the provided values: \(t = \frac{\$139 - \$150}{\$28 / \sqrt{25}} = -1.96\).
03

Determine the Critical Value

The critical value for a one-tailed test at the 1% significance level with degrees of freedom (df) = n - 1 = 24 can be obtained from the t-distribution table, and it is approximately -2.5.
04

Make a Decision

Since the computed test statistic (-1.96) is not less than the critical value (-2.5), do not reject the null hypothesis. \(-1.96 > -2.5\)
05

Interpret the Result

The claim of the manager that waiters earn an average of \$150 or more in tips per week cannot be rejected at 1% level of significance using the critical value approach.
06

Type I Error

A Type I error would occur if one wrongly rejected the manager's claim when it was true. The probability of making a Type I error is determined by the significance level of the test. In this case, the probability of making a Type I error is 1%.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a foundational concept in hypothesis testing. It is essentially a default assumption that there is no effect or no difference. In this problem, the null hypothesis (\(H_0\)) claims that waiters earn an average of \(\$150\) or more in tips per week in the large city.
The purpose of setting up a null hypothesis is to provide a statement that can be tested statistically. In research, we typically assume the null hypothesis is true until evidence suggests otherwise.
Key points about the null hypothesis:
  • Standard Assumption: It's like the baseline or control condition.
  • Testable: Researchers gather data to test its validity.
  • Rejection: It can be rejected or not rejected based on the statistical evidence.
In summary, considering the null hypothesis is crucial in determining whether observed data can challenge existing assumptions.
T-test
A t-test is a type of statistical test used to determine if there is a significant difference between the means of two groups. In this exercise, a one-sample t-test compares the sample mean of waiters' tips (\(\\(139\)) with the claimed population mean (\(\\)150\)).
Here's how the t-test works in this scenario:
  • The test statistic is computed using the formula: \[t = \frac{\bar{x} - \mu}{s / \sqrt{n}}\]where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the standard deviation, and \(n\) is the sample size.
  • The test statistic of \(-1.96\) is compared with the critical value to determine if the null hypothesis should be rejected.
The t-test is essential for making inferences about population means based on sample data, especially when sample sizes are small.
Type I Error
A Type I error occurs when the null hypothesis is incorrectly rejected when it is true. In simpler terms, it's like sounding a false alarm. In the restaurant manager's case, a Type I error would mean concluding that the average tips are less than \(\$150\) when they are actually not.
Understanding Type I error is critical because it reflects the risk of making a wrong decision:
  • False Positive: It indicates a finding when there isn't one.
  • Measured by Significance Level: The probability of a Type I error is equal to the significance level, in this case, \(1\%\).
By keeping the significance level low, researchers aim to reduce the chances of committing a Type I error.
Significance Level
The significance level is a threshold that determines whether the null hypothesis should be rejected. It reflects how certain we need to be before concluding that an observed effect is real and not due to random chance.
In the example problem, the significance level is set at \(1\%\) (or \(0.01\) in decimal form). This setting indicates a high standard for evidence before rejecting the null hypothesis:
  • Rigorous Testing: A low level means stronger evidence is needed to reject \(H_0\)
  • Type I Error Probability: It represents the maximum risk for making a Type I error.
  • Judgment Call: The choice of significance level can affect research conclusions.
Selecting an appropriate significance level is crucial to balancing the risk of Type I errors and the need for credible results.

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Most popular questions from this chapter

A food company is planning to market a new type of frozen yogurt. However, before marketing this yogurt, the company wants to find what percentage of the people like it. The company's management has decided that it will market this yogurt only if at least \(35 \%\) of the people like it. The company's research department selected a random sample of 400 persons and asked them to taste this yogurt. Of these 400 persons, 112 said they liked it. a. Testing at the \(2.5 \%\) significance level, can you conclude that the company should market this yogurt? b. What will your decision be in part a if the probability of making a Type I error is zero? Explain. c. Make the test of part a using the \(p\) -value approach and \(\alpha=.025\).

A 2008 AARP survey reported that \(85 \%\) of U.S. workers aged 50 years and older with at least one 4-year college degree had taken employer-based training within the previous 2 years, compared to only \(50 \%\) of workers aged 50 years and older with a high school degree or less. In a current survey of \(640 \mathrm{U.S}\). workers aged 50 years and older with a high school degree or less, 341 had taken employer-based training within the previous 2 years. a. Using the critical-value approach and \(\alpha=.05\), test whether the current percentage of all U.S. workers aged 50 years and older with a high school degree or less who have taken employerbased training within the previous 2 years is different from \(50 \%\). b. How do you explain the Type I error in part a? What is the probability of making this error in part a? c. Calculate the \(p\) -value for the test of part a. What is your conclusion if \(\alpha=.05 ?\)

According to a 2008 survey by the Royal Society of Chemistry, \(30 \%\) of adults in Great Britain stated that they typically run the water for a period of 6 to 10 minutes while taking the shower (http://www.rsc.org/ AboutUs/News/PressReleases/2008/EuropeanShowerHabits.asp). Suppose that in a recent survey of 400 adults in Great Britain, 104 stated that they typically run the water for a period of 6 to 10 minutes when they take a shower. At the \(5 \%\) significance level, can you conclude that the proportion of all adults in Great Britain who typically run the water for a period of 6 to 10 minutes when they take a shower is less than 30 ? Use both the \(p\) -value and the critical value approaches.

According to the Magazine Publishers of America (www.magazine.org), the average visit at the magazines' Web sites during the fourth quarter of 2007 lasted \(4.145\) minutes. Forty-six randomly selected visits to magazine's Web sites during the fourth quarter of 2009 produced a sample mean visit of \(4.458\) minutes, with a standard deviation of \(1.14\) minutes. Using the \(10 \%\) significance level and the critical value approach, can you conclude that the length of an average visit to these Web sites during the fourth quarter of 2009 was longer than \(4.145\) minutes? Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value range and \(\alpha=.10\) ?

A business school claims that students who complete a 3 -month typing course can type, on average, at least 1200 words an hour. A random sample of 25 students who completed this course typed, on average, 1125 words an hour with a standard deviation of 85 words. Assume that the typing speeds for all students who complete this course have an approximately normal distribution. a. Suppose the probability of making a Type I error is selected to be zero. Can you conclude that the claim of the business school is true? Answer without performing the five steps of a test of hypothesis. b. Using the \(5 \%\) significance level, can you conclude that the claim of the business school is true? Use both approaches.
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