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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 age 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 age 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
In conclusion, based on the sample data of 283 American teens aged 16-17, there is insufficient evidence to claim that more than a quarter of Americans age 16 to 17 have sent a text message while driving, as the p-value (0.121) is greater than the given significance level (0.01).

Step by step solution

01

Calculate sample proportion and test statistic

The sample proportion, \(\bar{p}\), is the number of teens who sent text messages while driving divided by the total number of teens in the sample: \(\bar{p} = \frac{74}{283} \approx 0.262\) To find the test statistic (z), we can use the formula: \(z = \frac{\bar{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\) where \(p_0\) is the proportion in the null hypothesis, 0.25, and n is the sample size, 283.
02

Find the test statistic value

Substituting the values, we can find the z value: \(z = \frac{0.262 - 0.25}{\sqrt{\frac{0.25(1-0.25)}{283}}}\) Calculating the values inside the radical: \(z = \frac{0.262 - 0.25}{\sqrt{\frac{0.1875}{283}}}\) Now, solving for z: \(z \approx 1.171\)
03

Find the p-value

Since this is a one-tailed test, and we are looking for evidence that more than a quarter of teens have sent a text message while driving, we will find the area to the right of the z value: p-value = P(Z > 1.171) Using a z-table or calculator, we find: p-value ≈ 0.121
04

Compare p-value with the significance level

We now compare the p-value to the significance level (α) given to us, which is 0.01: 0.121 > 0.01 The p-value is greater than the significance level.
05

Make a conclusion

Since the p-value is greater than the significance level, we fail to reject the null hypothesis. This means that there is insufficient evidence to claim that more than a quarter of Americans age 16 to 17 have sent a text message while driving, based on this sample data.

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Key Concepts

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

Sample Proportion
The sample proportion is a crucial part of hypothesis testing because it provides an estimate about the population parameter based on sample data. In this exercise, the sample proportion, denoted as \(\bar{p}\), is the ratio of individuals in the sample that meet a certain criterion—in this case, American teens who have texted while driving. To compute it, we divide the number of teens who reported texting while driving (74) by the total number in the sample (283).
Thus, the sample proportion is \(\bar{p} = \frac{74}{283} \approx 0.262\).
This means about 26.2% of the surveyed teens admitted to texting while driving. This proportion will be used to compare against what is expected, according to the null hypothesis.
Significance Level
The significance level, often denoted by \(\alpha\), is an essential component in hypothesis testing used to determine the threshold for statistical significance. It represents the probability of rejecting the null hypothesis when it is, in fact, true.
In this exercise, the significance level is set at 0.01. This signifies that there is a 1% risk of concluding that more than a quarter of American teens text while driving, assuming that this is not actually the case.
Choosing a significance level is a balance between type I errors (false positives) and type II errors (false negatives). A lower significance level, like 0.01, indicates a stringent test to reduce the likelihood of a false positive.
P-value
The p-value helps us understand the strength of the evidence against the null hypothesis. It is the probability, under the null hypothesis, of obtaining a result at least as extreme as the one observed.
To calculate the p-value, we find the probability that a sample proportion is as extreme as, or more extreme than, what we observed, given that the null hypothesis is true.
From our exercise, with a test statistic of approximately 1.171, the p-value was calculated as 0.121. This is done by assessing the area to the right of the z-value on a standard normal distribution.
A p-value greater than the chosen significance level (0.01) suggests that the evidence is insufficient to reject the null hypothesis.
Null Hypothesis
The null hypothesis is a statement used in hypothesis testing that suggests there is no effect or no difference. It serves as the starting point, or a default position, which indicates no change from the current understanding.
In this particular case, the null hypothesis \(H_0\) can be stated as the proportion of American teens texting while driving being equal to or less than 25% (\(p_0 = 0.25\)).
Testing aims to provide evidence to either reject or fail to reject the null hypothesis. If the evidence from the sample data is strong enough (based on the significance level and p-value calculations), the null hypothesis can be rejected. Otherwise, as in this exercise, the data does not provide sufficient evidence to reject it.
Test Statistic
The test statistic is a standardized value used in hypothesis testing to decide whether to reject the null hypothesis. It represents the number of standard deviations the sample proportion is from the hypothesized population proportion.
For our exercise, the z-test statistic is calculated using the formula:\[z = \frac{\bar{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]where \(\bar{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and \(n\) is the sample size.
Substituting the given values, the z-test statistic comes out to be approximately 1.171. This statistic is then used to find the p-value to assess the strength of evidence against the null hypothesis.

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Most popular questions from this chapter

A television station has been providing live coverage of a sensational criminal trial. The station's program director wants to know if more than half of potential viewers prefer a return to regular daytime programming. A survey of randomly selected viewers is conducted. With \(p\) representing the proportion of all viewers who prefer regular daytime programming, what hypotheses should the program director test?

A representative sample of 1000 likely voters in the United States included 440 who indicated that they think that women should not be required to register for the military draft ("Most Women Oppose Having to Register for the Draft," www .rasmessenreports.com, February 10, 2016, retrieved November 30,2016 ). Using the five-step process for hypothesis testing \(\left(\mathrm{HMC}^{3}\right)\) and a 0.05 significance level, determine if there is convincing evidence that less than half of likely voters in the United States think that women should not be required to register for the military draft.

Refer to the instructions given prior to this exercise. The paper "College Students' Social Networking Experiences on Facebook" (Journal of Applied Developmental Psychology [2009]: \(227-238\) ) summarized a study in which 92 students at a private university were asked how much time they spent on Facebook on a typical weekday. You would like to estimate the average time spent on Facebook by students at this university.

Refer to the instructions given prior to this exercise. The paper "College Students' Social Networking Experiences on Facebook" (Journal of Applied Developmental Psychology [2009]: \(227-238\) ) summarized a study in which 92 students at a private university were asked whether they used Facebook just to pass the time. Twenty-three responded "yes" to this question. The researchers were interested in estimating the proportion of students at this college who use Facebook just to pass the time.

The article "Poll Finds Most Oppose Return to Draft, Wouldn't Encourage Children to Enlist" (Associated Press, December 18,2005 ) reports that in a random sample of 1000 American adults, 430 answered "yes" to the following question: "If the military draft were reinstated, would you favor drafting women as well as men?" The data were used to test \(H_{0}: p=0.5\) versus \(H_{a}: p<0.5,\) and the null hypothesis was rejected. (Hint: See discussion following Example \(10.5 .\) ) a. Based on the result of the hypothesis test, what can you conclude about the proportion of American adults who fayor drafting women if a military draft were reinstated? 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?

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