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Chapter 10: Problem 59

Refer to the instructions given prior to Exercise \(10.57 .\) The paper referenced in the previous exercise also reported that when each of the 1178 students who participated in the study was asked if he or she played video games at least once a day, 271 responded "yes." The researchers were interested in using this information to decide if there is convincing evidence that more than \(20 \%\) of students age 8 to 18 play video games at least once a day.

Short Answer

Expert verified
In this hypothesis test, we have \(H_0: p = 0.20\) and \(H_a: p > 0.20\). We calculated the sample proportion \(\hat{p} \approx 0.230\), the standard error \(SE \approx 0.0126\), and the test statistic \(Z \approx 2.38\). The p-value is approximately \(0.0087\), which is less than the significance level \(\alpha = 0.05\). Therefore, we reject the null hypothesis and conclude that there is convincing evidence that more than 20% of students age 8 to 18 play video games at least once a day.

Step by step solution

01

State the null and alternative hypotheses.

Let \(p\) be the proportion of students aged 8 to 18 who play video games at least once a day. Null Hypothesis (\(H_0\)): \(p = 0.20\) Alternative Hypothesis (\(H_a\)): \(p > 0.20\) We're interested in testing if there is convincing evidence that more than 20% of students (aged 8 to 18) play video games at least once a day.
02

Calculate the sample proportion and determine the standard error.

Sample proportion (\(\hat{p}\)) is equal to the number of students who played video games daily divided by the total number of students in the sample. \(\hat{p} = \frac{271}{1178} \approx 0.230\) Now we need to find the standard error. In this case, since we're dealing with proportions, the standard error is given by the formula: \(SE = \sqrt{\frac{p_0(1-p_0)}{n}}\) Here, \(p_0 = 0.20\) (from hypothesis test) and \(n = 1178.\) \(SE \approx \sqrt{\frac{0.20(1-0.20)}{1178}} \approx 0.0126\)
03

Calculate the test statistic (Z-score).

The Z-score for the test can be calculated using the formula: \(Z = \frac{\hat{p} - p_0}{SE}\) \(Z \approx \frac{0.230 - 0.20}{0.0126} \approx 2.38\)
04

Calculate the p-value from the Z-score.

Now, we look up the Z-score in a standard normal table or use a calculator to find the one-sided p-value. Using a calculator or a standard normal table, we find that the area to the right of \(Z = 2.38\) is approximately \(0.0087\). This is our p-value.
05

Make a conclusion based on the p-value.

To make a conclusion, we need to compare the p-value to a chosen significance level, usually denoted as \(\alpha\). A common choice is \(\alpha = 0.05\). Our p-value (\(0.0087\)) is less than \(\alpha = 0.05\). Therefore, we reject the null hypothesis in favor of the alternative hypothesis. Conclusion: There is convincing evidence that more than 20% of students age 8 to 18 play video games at least once a day.

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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 fundamental concept in hypothesis testing. It is a statement that there is no effect or no difference, and it serves as a baseline or default position. In the context of the exercise, the null hypothesis (\(H_0\)) suggests that "20% of students aged 8 to 18 play video games at least once a day." In mathematical terms, we express this as \(p = 0.20\), where \(p\) represents the proportion of students.
  • Essentially, the null hypothesis allows for a point of comparison.
  • We test whether observed data significantly deviates from this hypothesis.
  • If evidence suggests otherwise, we may reject the null hypothesis.
This is crucial since it forms the baseline; all calculations and interpretations start here.
Alternative Hypothesis
The alternative hypothesis is what you would accept if the null hypothesis is rejected. It indicates that there is an effect or a difference worth considering. For this exercise, the alternative hypothesis (\(H_a\)) states that "more than 20% of students aged 8 to 18 play video games daily." We mathematically express this as \(p > 0.20\).
  • The alternative hypothesis represents the researcher's query or theory in question.
  • In hypothesis testing, the aim is to gather enough evidence to support this hypothesis over the null.
  • It is directional in this case, specifically testing if the proportion is greater than \(0.20\).
Evaluating the likelihood of this hypothesis helps determine the direction of statistical conclusions.
P-value
In hypothesis testing, the p-value helps determine the statistical significance of the results. It quantifies the probability of obtaining an observed effect, assuming the null hypothesis is true.For example, in this exercise, a p-value of approximately \(0.0087\) is discovered.
  • A lower p-value indicates stronger evidence against the null hypothesis.
  • This means our observed data is less predictable under the null hypothesis assumption.
  • If the p-value is below the predetermined significance level (\(\alpha\) usually set at \(0.05\)), it leads us to reject the null hypothesis.
The p-value is crucial in deciding whether to support or reject the null hypothesis, making it a pivotal part of hypothesis testing.
Z-score
The Z-score is a measure indicating how many standard deviations an element is from the mean. In hypothesis testing, it is used as a test statistic to understand how far the sample proportion is from the population proportion under the null hypothesis.In this exercise, the calculated Z-score is approximately \(2.38\).
  • A higher absolute value of the Z-score suggests a more significant deviation from the null hypothesis.
  • The formula used here is \(Z = \frac{\hat{p} - p_0}{SE}\), where \(\hat{p}\) is the sample proportion, \(p_0\) is the null hypothesis proportion, and \(SE\) is the standard error.
  • In this context, the Z-score helps identify how extreme the sample observation is, within the assumption of the null hypothesis.
The Z-score hence plays an instrumental role in understanding the comparative scale of our data against typical expectations.

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Most popular questions from this chapter

The article "Public Acceptability in the UK and the USA of Nudging to Reduce Obesity: The Example of Reducing Sugar-Sweetened Beverages" (PLOS One, June 8,2016 ) describes a survey in which each person in a representative sample of 1082 adult Americans was asked about whether they would find different types of interventions acceptable in an effort to reduce consumption of sugary beverages. When asked about a tax on sugary beverages, 459 of the people in the sample said they thought that this would be an acceptable intervention. These data were used to test \(H_{0}: p=0.5\) versus \(H_{a^{*}}: p<0.5\) and the null hypothesis was rejected. a. Based on the hypothesis test, what can you conclude about the proportion of adult Americans who think that taxing sugary beverages is an acceptable intervention in an effort to reduce consumption of sugary beverages? b. Is it reasonable to say that the data provide strong support for the alternative hypothesis? c. Is it reasonable to say that the data provide strong evidence against the null hypothesis?

Let \(p\) denote the proportion of students living on campus at a large university who plan to move off campus in the next academic year. For a large sample \(z\) test of \(H_{0}: p=0.70\) versus \(H_{a}: p>0.70,\) find the \(P\) -value associated with each of the following values of the \(z\) test statistic. a. 1.40 b. 0.92 c. 1.85 d. 2.18 e. -1.40

Assuming a random sample from a large population, for which of the following null hypotheses and sample sizes is the large-sample \(z\) test appropriate? a. \(H_{0}: p=0.2, n=25\) b. \(H_{0}: p=0.6, n=200\) c. \(H_{0}: p=0.9, n=100\) d. \(H_{0}: p=0.05, n=75\)

The paper "Living Near Nuclear Power Plants and Thyroid Cancer Risks" (Environmental International [2016]: \(42-48\) ) investigated whether living near a nuclear power plant increases the risk of thyroid cancer. The authors of this paper concluded that there was no evidence of increased risk of thyroid cancer in areas that were near a nuclear power plant. a. Suppose \(p\) denotes the true proportion of the population in areas near nuclear power plants who are diagnosed with thyroid cancer during a given year. The researchers who wrote this paper might have considered two rival hypotheses of the form \(H_{0}: p\) is equal to the corresponding value for areas without nuclear power plants \(H_{a^{*}} p\) is greater than the corresponding value for areas without nuclear power plants Did the researchers reject \(H_{0}\) or fail to reject \(H_{0} ?\) b. If the researchers are incorrect in their conclusion that there is no evidence of increased risk of thyroid cancer associated with living near a nuclear power plant, are they making a Type I or a Type II error? Explain. c. Can the result of this hypothesis test be interpreted as meaning that there is strong evidence that the risk of thyroid cancer is not higher for people living near nuclear power plants? Explain.

In a survey conducted by CareerBuilder.com, employers were asked if they had ever fired an employee for holiday shopping online while at work ("Cyber Monday Shopping at Work? You're Not Alone," November \(22,2016,\) retrieved November 30,2016 ). Of the 2379 employers responding to the survey, 262 said they had fired an employee for shopping online while at work. Suppose that it is reasonable to assume that the sample is representative of employers in the United States. Do the sample data provide convincing evidence that more than \(10 \%\) of employers have fired an employee for shopping online at work? Test the relevant hypotheses using \(\alpha=0.01\)
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