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91Ó°ÊÓ

The manager of the Danvers-Hilton Resort Hotel stated that the mean guest bill for a weekend is \(\$ 600\) or less. A member of the hotel's accounting staff noticed that the total charges for guest bills have been increasing in recent months. The accountant will use a sample of future weekend guest bills to test the manager's claim. a. Which form of the hypotheses should be used to test the manager's claim? Explain. \\[\begin{array}{lll}H_{0}: \mu \geq 600 & H_{0}: \mu \leq 600 & H_{0}: \mu=600 \\ H_{\mathrm{a}}: \mu<600 & H_{\mathrm{a}}: \mu>600 & H_{\mathrm{a}}: \mu \neq 600\end{array}\\] b. What conclusion is appropriate when \(H_{0}\) cannot be rejected? c. What conclusion is appropriate when \(H_{0}\) can be rejected?

Short Answer

Expert verified
Use \( H_0: \mu \leq 600 \) and \( H_a: \mu > 600 \). If \( H_0 \) cannot be rejected, the claim is plausible. If \( H_0 \) is rejected, the claim is invalid.

Step by step solution

01

Understand the Claim

The manager claims that the mean guest bill for a weekend at the hotel is \( \$600 \) or less. The goal is to test this claim statistically to see if it holds true in light of recent data trends suggesting an increase in bills.
02

Choose the Appropriate Hypotheses

Since the claim suggests that the mean guest bill is \( \\(600 \) or less, we will use the following hypotheses: \( H_{0}: \mu \leq 600 \) (the null hypothesis) and \( H_{a}: \mu > 600 \) (the alternative hypothesis). This sets us up to test if the true mean is actually greater than \( \\)600 \), contradicting the manager's claim.
03

Conclusion When \( H_0 \) Cannot be Rejected

When \( H_0 \) cannot be rejected, it means there is not enough evidence to suggest that the mean guest bill is greater than \( \\(600 \). Therefore, we conclude that the manager's claim of the mean being \( \\)600 \) or less is still plausible based on the given data.
04

Conclusion When \( H_0 \) Can be Rejected

When \( H_0 \) can be rejected, this provides evidence that the mean guest bill is in fact greater than \( \$600 \). This would suggest that the manager's claim is not valid and the average guest bill is higher than originally stated.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is often denoted as \( H_0 \). It essentially represents the starting assumption or the status quo that needs testing. In the context of the Danvers-Hilton Resort Hotel case, we are trying to verify the manager's claim that the mean weekend guest bill is \( \$600 \) or less. This statement forms our null hypothesis: \( H_0: \mu \leq 600 \).

The null hypothesis is typically formulated to represent a statement of "no effect" or "no difference." It sets a benchmark that researchers attempt to challenge with statistical data. By disproving \( H_0 \), we can infer results that are contrary to it.

Here are some key points about the null hypothesis:
  • Serves as the default or original assumption in hypothesis testing.
  • Helps in setting the groundwork for statistical tests.
  • Must be proven false ('rejected') to consider alternative findings.
In this given exercise, the null hypothesis serves as a platform for examining whether the data indicates higher averages than the one claimed by the manager.
Alternative Hypothesis
The alternative hypothesis is a key element of hypothesis testing. It is denoted as \( H_a \) and stands as the contrary statement to the null hypothesis. In the exercise concerning the hotel guest bills, the accountant is interested in whether the mean guest bill is actually more than \( \$600 \). Therefore, the alternative hypothesis is formulated as \( H_a: \mu > 600 \).

The alternative hypothesis indicates the presence of an effect or a significant difference when comparing against the null. It represents the outcome researchers are expecting to support with statistical evidence.

Key aspects of the alternative hypothesis include:
  • Indicates what we aim to demonstrate with data.
  • Suggests deviation from the null hypothesis.
  • Can be 'greater than', 'less than', or 'not equal to' scenarios, depending on the research question.
The alternative hypothesis is what can lead to statistical findings that deviate from the initial claim—prompting a re-evaluation of what is believed to be true.
Statistical Evidence
Statistical evidence is the backbone of hypothesis testing, offering a way to validate or refute claims through data analysis. In the hotel billing case study, statistical evidence would be drawn from a sample of future weekend guest bills to assess if the average charge significantly exceeds \( \$600 \).

In practice:
  • Evidence arises from observing and measuring data.
  • Inferential statistics enable conclusions beyond the immediate data by making predictions or generalizations.
  • Statistical tests, such as t-tests, chi-squared tests, or ANOVA, are instrumental in gauging evidence.
The purpose of gathering this evidence is twofold: to establish whether there is sufficient grounds to reject the null hypothesis (\( H_0 \)), and to determine the validity of the alternative hypothesis (\( H_a \)). If data show a significant increase in the bills, it warrants a conclusion that deviates from the manager's claim.

Conclusively, statistical evidence determines the direction of the conclusion: whether the manager’s assertion holds in light of the new data or needs reconsideration. It plays a central role in backing up the results of hypothesis testing.

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Most popular questions from this chapter

Consider the following hypothesis test: \\[\begin{array}{l}H_{0}: \mu \geq 20 \\ H_{\mathrm{a}}: \mu<20 \end{array}\\] A sample of 50 provided a sample mean of \(19.4 .\) The population standard deviation is \(2 .\) a. Compute the value of the test statistic. b. What is the \(p\) -value? c. \(\quad\) Using \(\alpha=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

A radio station in Myrtle Beach announced that at least \(90 \%\) of the hotels and motels would be full for the Memorial Day weekend. The station advised listeners to make reservations in advance if they planned to be in the resort over the weekend. On Saturday night a sample of 58 hotels and motels showed 49 with a no-vacancy sign and 9 with vacancies. What is your reaction to the radio station's claim after secing the sample evidence? Use \(a=.05\) in making the statistical test. What is the \(p\) -value?

Playbill is a magazine distributed around the country to people attending musicals and other theatrical productions. The mean annual household income for the population of Playbill readers is \(\$ 119,155\) (Playbill, January 2006 ). Assume the standard deviation is \(\sigma=\$ 20,700 .\) A San Francisco civic group has asserted that the mean for theatergoers in the Bay Area is higher. A sample of 60 theater attendees in the Bay Area showed a sample mean household income of \(\$ 126,100\) a. Develop hypotheses that can be used to determine whether the sample data support the conclusion that theater attendees in the Bay Area have a higher mean household income than that for all Playbill readers. b. What is the \(p\) -value based on the sample of 60 theater attendees in the Bay Area? c. Use \(\alpha=.01\) as the level of significance. What is your conclusion?

AOL Time Warner Inc.'s CNN has been the longtime ratings leader of cable television news. Nielsen Media Research indicated that the mean CNN viewing audience was 600,000 viewers per day during 2002 (The Wall Street Journal, March 10,2003 ). Assume that for a sample of 40 days during the first half of \(2003,\) the daily audience was 612,000 viewers with a sample standard deviation of 65,000 viewers. a. What are the hypotheses if CNN management would like information on any change in the CNN viewing audience? b. What is the \(p\) -value? c. Select your own level of significance. What is your conclusion? d. What recommendation would you make to CNN management in this application?

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \leq 12 \\ H_{\mathrm{a}}: \mu>12 \end{array} \\] A sample of 25 provided a sample mean \(\bar{x}=14\) and a sample standard deviation \(s=4.32\) a. Compute the value of the test statistic. b. Use the \(t\) distribution table (Table 2 in Appendix \(B\) ) to compute a range for the \(p\) -value. c. \(\quad\) At \(a=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

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