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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?

Short Answer

Expert verified
There is no significant change in the CNN viewing audience; the null hypothesis is not rejected.

Step by step solution

01

Define the Hypotheses

First, we identify the null and alternative hypotheses. The null hypothesis \( H_0 \) represents the assumption that there is no change in the mean CNN viewing audience, so \( H_0: \mu = 600,000 \). The alternative hypothesis \( H_a \) represents the assumption that there is a change, so \( H_a: \mu eq 600,000 \).
02

Calculate the Test Statistic

The test statistic for the sample mean is calculated using the formula \[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]where \( \bar{x} = 612,000 \) is the sample mean, \( \mu = 600,000 \) is the population mean, \( \sigma = 65,000 \) is the sample standard deviation, and \( n = 40 \) is the sample size.Substituting the values, we get:\[ z = \frac{612,000 - 600,000}{65,000 / \sqrt{40}} = \frac{12,000}{10,271.57} \approx 1.17 \]
03

Determine the p-value

Next, we use the calculated z-value to find the \( p \)-value. Since this is a two-tailed test, we look up the value \( z = 1.17 \) in the standard normal distribution table (or use a calculator) to find the \( p \)-value. The \( p \)-value for \( z = 1.17 \) is approximately 0.242 (for both tails combined).
04

Set the Significance Level

Choose a level of significance \( \alpha \). Common choices are 0.05 or 0.01. Let’s choose \( \alpha = 0.05 \).
05

Make a Conclusion

With a \( p \)-value of 0.242 and \( \alpha = 0.05 \), we compare these values. Since the \( p \)-value is greater than \( \alpha \), we fail to reject the null hypothesis. This suggests there is not enough statistical evidence to indicate a change in the CNN viewing audience.
06

Recommendation to CNN Management

Based on the analysis, it is recommended that CNN management should consider that there is no statistically significant evidence of a change in the viewing audience from 2002 to 2003.

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Key Concepts

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

p-value
The p-value is a crucial concept in hypothesis testing. It helps determine the strength of the evidence against the null hypothesis. In simple terms, the p-value measures how likely it is to observe the sample data, or something more extreme, assuming that the null hypothesis is true.
In this particular example involving CNN viewer numbers, the p-value was found to be approximately 0.242. This means there is a 24.2% chance of seeing the observed result, or something more extreme, if the mean audience size actually hasn't changed from the assumed 600,000 viewers.
A high p-value indicates weak evidence against the null hypothesis, suggesting that the observed data is consistent with the assumption made in the null hypothesis. In contrast, a low p-value suggests strong evidence against the null hypothesis. The primary decision in hypothesis testing revolves around comparing the p-value to the chosen level of significance (often denoted as \( \alpha \)).
  • If the p-value is less than \( \alpha \), we reject the null hypothesis.
  • If the p-value is greater than \( \alpha \), we fail to reject the null hypothesis.
Understanding the p-value helps researchers make informed conclusions about their data.
Level of Significance
The level of significance, often denoted as \( \alpha \), is a threshold set by researchers to determine when to reject the null hypothesis. It represents the probability of making a Type I error, or incorrectly rejecting a true null hypothesis.
In practice, researchers commonly use significance levels such as 0.05 (5%) or 0.01 (1%). For the CNN exercise, a significance level of \( \alpha = 0.05 \) was chosen. This implies that there's a 5% risk of rejecting the null hypothesis if it is indeed true.
Choosing the right level of significance is vital because it balances the risk of errors in decision-making. A smaller \( \alpha \), like 0.01, reduces the chance of a Type I error but increases the risk of a Type II error (failing to reject a false null hypothesis).
  • A higher significance level increases the risk of a Type I error.
  • A lower significance level increases the risk of a Type II error.
When the calculated p-value is compared to \( \alpha \), it helps decide if the null hypothesis should be rejected or not. In this case, since the p-value (0.242) is greater than \( \alpha \) (0.05), the decision was to fail to reject the null hypothesis.
Null and Alternative Hypotheses
In any hypothesis testing, the first step is to establish the null and alternative hypotheses. These hypotheses set the framework for the test.
The null hypothesis, denoted as \( H_0 \), represents the status quo or the statement being tested. It is typically a statement of no effect or no difference. In our CNN example, the null hypothesis was \( H_0: \mu = 600,000 \), indicating there was no change in the mean audience size.
On the other hand, the alternative hypothesis, represented as \( H_a \) or \( H_1 \), suggests a change or effect. This is the hypothesis researchers are typically looking to prove. For CNN, the alternative hypothesis was \( H_a: \mu eq 600,000 \), suggesting there was a change in viewership numbers.
In hypothesis testing:
  • The null hypothesis acts as a default position that indicates no change.
  • The alternative hypothesis indicates the presence of an effect or difference.
The outcome of a statistical test tells us whether there's enough evidence to reject the null hypothesis in favor of the alternative hypothesis. In this analysis, since the p-value was greater than the level of significance, the result led to failing to reject the null hypothesis, implying insufficient evidence of change in CNN's viewership.

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Most popular questions from this chapter

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?

Nielsen reported that young men in the United States watch 56.2 minutes of prime-time \(\mathrm{TV}\) daily (The Wall Street Journal Europe, November 18,2003 ). A researcher believes that young men in Germany spend more time watching prime-time TV, A sample of German young men will be selected by the researcher and the time they spend watching \(\mathrm{TV}\) in one day will be recorded. The sample results will be used to test the following null and alternative hypotheses. \\[ \begin{array}{l} H_{0}: \mu \leq 56.2 \\ H_{\mathrm{a}^{\prime}} \mu>56.2\end{array}\\] a. What is the Type I error in this situation? What are the consequences of making this error? b. What is the Type II error in this situation? What are the consequences of making this error?

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?

According to the federal government, \(24 \%\) of workers covered by their company's health care plan were not required to contribute to the premium (Statistical Abstract of the United States: 2006 . A recent study found that 81 out of 400 workers sampled were not required to contribute to their company's health care plan. a. Develop hypotheses that can be used to test whether the percent of workers not required to contribute to their company's health care plan has declined. b. What is a point estimate of the proportion receiving free company-sponsored health care insurance? c. Has a statistically significant decline occurred in the proportion of workers receiving free company-sponsored health care insurance? Use \(\alpha=.05\).

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: p=.20 \\ H_{\mathrm{a}}: p \neq .20 \end{array} \\] A sample of 400 provided a sample proportion \(\bar{p}=.175\) a. Compute the value of the test statistic. b. What is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

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