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The paper "Teens and Distracted Driving" (Pew Internet \& American Life Project, 2009) reported that in a representative sample of 283 American teens ages 16 to \(17,\) there were 74 who indicated that they had sent a text message while driving. For purposes of this exercise, assume that this sample is a random sample of 16 - to 17 -year-old Americans. Do these data provide convincing evidence that more than a quarter of Americans ages 16 to 17 have sent a text message while driving? Test the appropriate hypotheses using a significance level of 0.01 . (Hint: See Example 10.11 )

Short Answer

Expert verified
No, the data do not provide convincing evidence to support the claim that more than a quarter of American teens aged 16 to 17 have sent a text message while driving.

Step by step solution

01

Setup the Hypotheses

Firstly, set up the null hypothesis (H0) and the alternative hypothesis (H1). The null hypothesis is H0: p = 0.25 and the alternative hypothesis is H1: p > 0.25, where p represents the proportion of teens aged 16 to 17 who text while driving.
02

Calculate Test Statistic

Next, calculate the test statistic using the formula: z = (p̂ - p0) / sqrt[(p0(1 - p0) / n], where p̂ is the sample proportion, p0 is the hypothesized population proportion under H0, and n is the sample size. In this case p̂ = 74 / 283 = 0.2615, p0 = 0.25, and n = 283. Substituting these in the formula gives z = (0.2615 - 0.25) / sqrt[(0.25(1 - 0.25) / 283] = 0.5115.
03

Calculate P-value

Now, calculate the p-value. The p-value is the probability that we would observe a test statistic as extreme as our calculated z-score, assuming the null hypothesis is true. In this case, we use a z-table or calculator to find the area to the right of our test statistic (0.5115) in a standard normal distribution, which gives a p-value of approximately 0.3042.
04

Make Conclusion

Finally, compare the p-value with the given significance level (α=0.01). In this case, the p-value is much larger than the significance level; therefore, we do not reject the null hypothesis. There isn't enough evidence to support the claim that more than 25% of American teenagers aged 16 to 17 text while driving.

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Key Concepts

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

Null Hypothesis
In the realm of hypothesis testing in statistics, a null hypothesis is a statement that there is no effect or no difference, and it serves as a starting point for testing. It is denoted as H0. For example, in the context of the exercise provided, the null hypothesis states that the proportion (p) of 16 to 17 year old Americans who text while driving is 25% (H0: p = 0.25).

Establishing the null hypothesis is a fundamental step because it creates a benchmark against which the alternative hypothesis is compared. It assumes the status quo and puts the onus on the researcher to provide evidence to support any claim of departure from this norm.
Alternative Hypothesis
Conversely, the alternative hypothesis suggests that there is a statistically significant effect or a difference. It is the statement researchers are trying to provide evidence for and it is denoted as H1 or Ha. In the provided exercise, the alternative hypothesis posits that more than a quarter (25%) of the same age group texts while driving (H1: p > 0.25).

The alternative hypothesis is the researcher's assertion which is considered only after it is concluded that the null hypothesis does not hold based on the data. This hypothesis is the main subject of verification in any statistical test and provides the direction for the kind of test to be performed, whether one-tailed or two-tailed.
Test Statistic
A test statistic is a calculated number from the sample data that, under the null hypothesis, follows a known probability distribution. It measures how far off the observed data is from the null hypothesis. The calculation of the test statistic takes into account the sample proportion, the hypothesized proportion under the null hypothesis, and the sample size.

As outlined in the given exercise, the test statistic is calculated using the formula:
\( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \)
where:\( \hat{p} \) is the sample proportion, \( p_0 \) is the hypothesized proportion under H0, and n is the sample size. This statistic helps to determine how extreme the sample data is, given the null hypothesis is true.
P-value
The p-value is a key component of hypothesis testing and represents the probability of obtaining test results at least as extreme as the observed results, assuming that the null hypothesis is true. It is a decimal between 0 and 1 and is used as a benchmark against the significance level to decide whether to reject the null hypothesis.

In the context of the exercise, the p-value is obtained by finding the area to the right of the test statistic (in this case a z-score) in the standard normal distribution. If this p-value is less than the significance level, the null hypothesis is rejected; otherwise, we fail to reject the null hypothesis.
Significance Level
The significance level (denoted as α) represents the threshold against which the p-value is compared to determine the strength of the evidence against the null hypothesis. Common levels of significance are 0.05, 0.01, and 0.10.

In the exercise, a significance level of 0.01 indicates that there is a 1% risk of concluding that a difference exists when there is no actual difference. It is chosen before examining the data to decide the critical region. Since the p-value in the exercise was greater than 0.01, we do not have sufficient evidence to reject the null hypothesis, indicating that we do not have strong enough evidence to support the claim that more than 25% of teenagers text while driving.

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Most popular questions from this chapter

A Newsweek article titled "America the Ignorant" (www.newsweek.com) described a Gallup poll that asked adult Americans if they believe that there are real witches and warlocks. Suppose that the poll used a random sample of 800 adult Americans and that you want to use the poll data to decide if there is evidence that more than \(10 \%\) of adult Americans believe in witches and warlocks. Let \(p\) be the proportion of all adult Americans who believe in witches and warlocks. a. Describe the shape, center, and spread of the sampling distribution of \(\hat{p}\) for random samples of size 800 if the null hypothesis \(H_{0}: p=0.10\) is true. b. Would you be surprised to observe a sample proportion as large as \(\hat{p}=0.16\) for a sample of size 800 if the null hypothesis \(H_{0}: p=0.10\) were true? Explain why or why not. c. Would you be surprised to observe a sample proportion as large as \(\hat{p}=0.12\) for a sample of size 800 if the null hypothesis \(H_{0}: p=0.10\) were true? Explain why or why not. d. The actual sample proportion observed in the study was \(\hat{p}=0.21 .\) Based on this sample proportion, is there convincing evidence that more than \(10 \%\) of adult Americans believe in witches and warlocks, or is the sample proportion consistent with what you would expect to see when the null hypothesis is true? Support your answer with a probability calculation.

The paper referenced in the previous exercise also reported that when each of the 1,178 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 ages 8 to 18 play video games at least once a day.

The article "Irritated by Spam? Get Ready for Spit" (USA Today, November 10,2004 ) predicts that "spit," spam that is delivered via Internet phone lines and cell phones, will be a growing problem as more people turn to web- based phone services. In a poll of 5,500 cell phone users, \(20 \%\) indicated that they had received commercial messages and ads on their cell phones. These data were used to test \(H_{o}: p=0.13\) versus \(H_{a}: p>0.13\) where 0.13 was the proportion reported for the previous year. The null hypothesis was rejected. a. Based on the hypothesis test, what can you conclude about the proportion of cell phone users who received commercial messages and ads on their cell phones in the year the poll was conducted? 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?

The National Cancer Institute conducted a 2-year study to determine whether cancer death rates for areas near nuclear power plants are higher than for areas without nuclear facilities (San Luis Obispo Telegram-Tribune, September 17,1990 ). A spokesperson for the Cancer Institute said, "From the data at hand, there was no convincing evidence of any increased risk of death from any of the cancers surveyed due to living near nuclear facilities. However, no study can prove the absence of an effect." a. Suppose \(p\) denotes the true proportion of the population in areas near nuclear power plants who die of cancer during a given year. The researchers at the Cancer Institute might have considered two rival hypotheses of the form \(H_{0}: p\) is equal to the corresponding value for areas without nuclear facilities \(H_{a}: p\) is greater than the corresponding value for areas without nuclear facilities Did the researchers reject \(H_{0}\) or fail to reject \(H_{0} ?\) b. If the Cancer Institute researchers are incorrect in their conclusion that there is no evidence of increased risk of death from cancer associated with living near a nuclear power plant, are they making a Type I or a Type II error? Explain. c. Comment on the spokesperson's last statement that no study can prove the absence of an effect. Do you agree with this statement?

In a hypothesis test, what does it mean to say that the null hypothesis was not rejected?

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