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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?

Short Answer

Expert verified
The Bay Area theatergoers have a significantly higher mean household income.

Step by step solution

01

Set Up Hypotheses

We need to develop the null and alternative hypotheses. The null hypothesis \( H_0 \) is that the mean income of Bay Area theatergoers is equal to or less than the mean income for all Playbill readers, \( \mu = 119,155 \). The alternative hypothesis \( H_a \) is that the mean income of Bay Area theatergoers is greater than the mean income for all Playbill readers, \( \mu > 119,155 \).
02

Calculate the Test Statistic

We use a z-test for the mean since the population standard deviation is known. The formula for the test statistic \( z \) is: \[z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\] where \( \bar{x} = 126,100 \), \( \mu = 119,155 \), \( \sigma = 20,700 \), and \( n = 60 \). Plugging in these values, we find:\[z = \frac{126,100 - 119,155}{20,700 / \sqrt{60}} = \frac{6,945}{2,672.23} \approx 2.6\]
03

Determine the p-value

The \( p \)-value is the probability that the test statistic is at least as extreme as found under the null hypothesis. For a one-tailed test with \( z = 2.6 \), we find the corresponding \( p \)-value using standard normal distribution tables or a calculator. The \( p \)-value is approximately 0.0047.
04

Compare p-value to Significance Level

The significance level \( \alpha \) is 0.01. Since the \( p \)-value (0.0047) is less than \( \alpha \), we have sufficient evidence to reject the null hypothesis.
05

Conclusion

Because the \( p \)-value is less than the significance level of 0.01, we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that Bay Area theatergoers have a higher mean household income than all Playbill readers.

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Key Concepts

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

Z-Test for the Mean
The z-test for the mean is a statistical method used to determine if there is a significant difference between the sample mean and a known population mean. It requires the population standard deviation to be known, making it applicable in scenarios where this information is available.

The formula for the z-test is: \[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]where:
  • \( \bar{x} \) is the sample mean,
  • \( \mu \) is the population mean,
  • \( \sigma \) is the population standard deviation,
  • \( n \) is the sample size.
In the provided exercise, we used a z-test because we know the population standard deviation \( \sigma = 20,700 \). By using this formula, we calculated the test statistic \( z \approx 2.6 \). This value tells us how many standard deviations the sample mean is away from the population mean. The larger the z-value, the larger the difference between the sample mean and the population mean.
P-Value Calculation
The p-value is a crucial part of hypothesis testing. It represents the probability of obtaining a test statistic at least as extreme as the one calculated if the null hypothesis is true. A smaller p-value indicates that the observed data is unlikely under the null hypothesis. In other words, it helps quantify the evidence against the null hypothesis.

For example, with a one-tailed z-test and a calculated \( z \)-value of \( 2.6 \), we find the p-value using a standard normal distribution table or calculator. In this case, the p-value is approximately 0.0047.

This p-value is quite small, which suggests that the observed sample mean is significantly higher than the expected mean under the null hypothesis. When making decisions, especially in hypothesis testing, understanding the p-value helps determine whether or not to reject the null hypothesis.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold set prior to conducting a hypothesis test. It indicates how willing we are to make a Type I error, which is rejecting a true null hypothesis. Common significance levels used in hypothesis testing include 0.05, 0.01, and 0.10, with smaller values indicating stricter criteria for rejection.

In our exercise, a significance level of \( \alpha = 0.01 \) was chosen, indicating a stringent requirement for rejecting the null hypothesis. A significance level helps determine the critical value or zone of rejection for a test statistic. If the p-value is less than or equal to the significance level, the null hypothesis is rejected.

In our case, the p-value of 0.0047 is less than 0.01, which means we reject the null hypothesis. By setting a low significance level, the analysis aims to minimize errors, requiring strong evidence against the null hypothesis before concluding against it.
Null and Alternative Hypotheses
In hypothesis testing, the null and alternative hypotheses form the foundation of any analysis. The null hypothesis, denoted as \( H_0 \), is a statement that there is no effect or no difference; it serves as a default assumption. Conversely, the alternative hypothesis, denoted as \( H_a \), posits that there is an effect or a difference.

For this exercise, the null hypothesis \( H_0 \) states that the mean income of Bay Area theatergoers is equal to or less than the general mean for all Playbill readers, \( \mu \leq 119,155 \). Meanwhile, the alternative hypothesis \( H_a \) suggests that the mean income is greater than \( 119,155 \).

The process of hypothesis testing allows us to use sample data to decide whether to reject the null hypothesis. It involves comparing the observed data against what is expected under the null hypothesis's framework. Complete understanding of these hypotheses is essential to correctly interpreting the results of hypothesis tests.

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Most popular questions from this chapter

Suppose a new production method will be implemented if a hypothesis test supports the conclusion that the new method reduces the mean operating cost per hour. a. State the appropriate null and alternative hypotheses if the mean cost for the current production method is \(\$ 220\) per hour. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error?

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