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Chapter 9: Problem 5

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?

Short Answer

Expert verified
Type I error is incorrectly concluding that German young men watch more TV; it leads to false assumptions. Type II error is failing to recognize they actually watch more; it results in missed insights.

Step by step solution

01

Understanding Type I Error

A Type I error occurs when the null hypothesis \( H_0 \) is true, but we incorrectly reject it. In this case, the null hypothesis states that the average time young men in Germany watch prime-time TV is 56.2 minutes or less. A Type I error would mean concluding that they spend more time watching TV when they actually do not.
02

Consequence of Type I Error

The consequence of making a Type I error in this context would be incorrectly asserting that German young men watch more TV than those in the United States, which may lead to misinformed decisions or perceptions about viewing habits.
03

Understanding Type II Error

A Type II error occurs when the null hypothesis \( H_0 \) is false, but we fail to reject it. Here, it would mean the true average time is greater than 56.2 minutes, yet we conclude it is not.
04

Consequence of Type II Error

The consequence of a Type II error would be failing to recognize that young men in Germany actually do spend more time watching TV. This could result in missed opportunities to understand differences in media consumption or misallocation of resources aimed at targeting the correct audience.

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Key Concepts

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

Type I Error
A Type I error, also known as a "false positive," happens when the null hypothesis is actually true, but our test incorrectly rejects it. In simpler terms, we assert something is happening, when in fact, it is not.
In the context of the Nielsen report exercise, the null hypothesis is that young men in Germany watch 56.2 minutes or less of TV daily. If a Type I error occurs, we would incorrectly conclude that they watch more than 56.2 minutes, despite the fact they do not.
The consequences of such an error revolve around making false claims. This may lead to erroneous conclusions about viewing habits, potentially influencing policy or media strategy based on incorrect assumptions.
Type II Error
A Type II error, sometimes called a "false negative," arises when the null hypothesis is false, but we fail to reject it. Essentially, this means missing out on detecting a real effect.
In our scenario, this would mean the average time spent watching TV by German young men is greater than 56.2 minutes, but our test indicates it is not. Failing to identify this could mean overlooking genuine differences in media consumption patterns between the U.S. and Germany.
  • In practice, this may lead to missed opportunities for broadcasters and advertisers to tailor their strategies effectively.
  • Moreover, it can prevent understanding cultural viewing behaviors and preferences accurately.
Null Hypothesis
The null hypothesis is a foundational concept in statistical hypothesis testing. It represents the default or initial claim that is assumed to be true unless there is strong evidence to suggest otherwise.
Typically denoted as \(H_0\), in the Nielsen report context, the null hypothesis states that young men in Germany watch no more than 56.2 minutes of TV per day.
This hypothesis serves as a baseline for comparison. Before any sample data is collected, it asserts that there is no difference or effect present. Rejecting the null hypothesis usually means you have found sufficient evidence supporting an alternative cause or relationship. However, if the null hypothesis is not rejected, it suggests any observed differences could be due to random variation.
  • Understanding the null hypothesis is crucial because it guides the decision-making process throughout the test.
  • Furthermore, the risk of making a Type I or Type II error is inherently linked to the assumptions made by the null hypothesis.
Alternative Hypothesis
The alternative hypothesis offers a contrasting proposition to the null hypothesis. It's what researchers aim to support with evidence; it reflects the outcome of interest or the effect they suspect.
Denoted as \(H_a\), in the study of TV watching habits, the alternative hypothesis would be that young men in Germany watch more than 56.2 minutes daily.
The alternative hypothesis captures what you hope to prove with your research. It suggests there is a statistically significant effect or difference, thereby contradicting the null hypothesis. This hypothesis becomes the focus once the null hypothesis is found lacking substantial support based on the data.
  • When formulating hypotheses, the alternative hypothesis is often an exact opposite to the null, clearly defining the expected result.
  • Successfully supporting the alternative hypothesis provides justification for claims outside of the base assumption.
Understanding both hypotheses helps in designing experiments and interpreting the resulting data carefully.

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Most popular questions from this chapter

The manager of an automobile dealership is considering a new bonus plan designed to increase sales volume. Currently, the mean sales volume is 14 automobiles per month. The manager wants to conduct a research study to see whether the new bonus plan increases sales volume. To collect data on the plan, a sample of sales personnel will be allowed to sell under the new bonus plan for a one-month period. a. Develop the null and alternative hypotheses most appropriate for this situation. b. Comment on the conclusion when \(H_{0}\) cannot be rejected. c. Comment on the conclusion when \(H_{0}\) can be rejected.

A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious problem and when detected requires the operator to shut down the production line to readjust the filling mechanism. From past data, a population standard deviation \(\sigma=.8\) ounces is assumed. A quality control inspector selects a sample of 30 items every hour and at that time makes the decision of whether to shut down the line for readjustment. The level of significance is \(\alpha=.05\) a. State the hypothesis test for this quality control application. b. If a sample mean of \(\bar{x}=16.32\) ounces were found, what is the \(p\) -value? What action would you recommend? c. If a sample mean of \(\bar{x}=15.82\) ounces were found, what is the \(p\) -value? What action would you recommend? d. Use the critical value approach. What is the rejection rule for the preceding hypothesis testing procedure? Repeat parts (b) and (c). Do you reach the same conclusion?

A study by Consumer Reports showed that \(64 \%\) of supermarket shoppers believe supermarket brands to be as good as national name brands. To investigate whether this result applies to its own product, the manufacturer of a national name-brand ketchup asked a sample of shoppers whether they believed that supermarket ketchup was as good as the national brand ketchup. a. Formulate the hypotheses that could be used to determine whether the percentage of supermarket shoppers who believe that the supermarket ketchup was as good as the national brand ketchup differed from \(64 \%\) b. If a sample of 100 shoppers showed 52 stating that the supermarket brand was as good as the national brand, what is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. Should the national brand ketchup manufacturer be pleased with this conclusion? Explain.

On December \(25,2009,\) an airline passenger was subdued while attempting to blow up a Northwest Airlines flight headed for Detroit, Michigan. The passenger had smuggled explosives hidden in his underwear past a metal detector at an airport screening facility. As a result, the Transportation Security Administration (TSA) proposed installing full-body scanners to replace the metal detectors at the nation's largest airports. This proposal resulted in strong objections from privacy advocates who considered the scanners an invasion of privacy. On January \(5-6,2010,\) USA Today conducted a poll of 542 adults to learn what proportion of airline travelers approved of using full-body scanners (USA Today, January 11 2010)\(.\) The poll results showed that 455 of the respondents felt that full- body scanners would improve airline security and 423 indicated that they approved of using the devices. a. Conduct a hypothesis test to determine if the results of the poll justify concluding that over \(80 \%\) of airline travelers feel that the use of full- body scanners will improve airline security. Use \(\alpha=.05\) b. Suppose the TSA will go forward with the installation and mandatory use of full-body scanners if over \(75 \%\) of airline travelers approve of using the devices. You have been told to conduct a statistical analysis using the poll results to determine if the TSA should require mandatory use of the full-body scanners, Because this is viewed as a very sensitive decision, use \(\alpha=.01 .\) What is your recommendation?

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \leq 50 \\ H_{\mathrm{a}}: \mu>50 \end{array} \\] A sample of 60 is used and the population standard deviation is \(8 .\) Use the critical value approach to state your conclusion for each of the following sample results, Use \\[\alpha=.05\\] a. \(\quad \bar{x}=52.5\) b. \(\bar{x}=51\) c. \(\quad \bar{x}=51.8\)
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