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Common Sense Media surveyed 1000 teens and 1000 parents of teens to learn about how teens are using social networking sites such as Facebook and MySpace ("Teens Show, Tell Too Much Online," San Francisco Chronicle, August 10,2009 ). The two samples were independently selected and were chosen in a way that makes it reasonable to regard them as representative of American teens and parents of American teens. a. When asked if they check their online social networking sites more than 10 times a day, 220 of the teens surveyed said yes. When parents of teens were asked if their teen checked his or her site more than 10 times a day, 40 said yes. Use a significance level of .01 to carry out a hypothesis test to determine if there is convincing evidence that the proportion of all parents who think their teen checks a social networking site more than 10 times a day is less than the proportion of all teens who report that they check more than 10 times a day. b. The article also reported that 390 of the teens surveyed said they had posted something on their networking site that they later regretted. Would you use the two-sample \(z\) test of this section to test the hypothesis that more than one-third of all teens have posted something on a social networking site that they later regretted? Explain why or why not. c. Using an appropriate test procedure, carry out a test of the hypothesis given in Part (b). Use \(\alpha=.05\) for this test.

Short Answer

Expert verified
Yes, there is enough evidence to conclude that the proportion of teens who check their social networking sites more than 10 times a day is significantly greater than the proportion of parents who think their teens do the same. Also, there is enough evidence to suggest that more than one-third of all teens have posted something on their social networking sites that they later regretted.

Step by step solution

01

Define the Null and Alternative Hypothesis

Let \(p_1\) be the proportion of teens who check social networking sites more than 10 times a day and \(p_2\) be the proportion of parents who believe their teens do the same. The Null hypothesis is \(H_0: p_1 \leq p_2\), and the alternative hypothesis is \(H_a: p_1 > p_2\).
02

Calculate the Proportions and Perform Z Test

From the survey, \(p_1= 220/1000 = 0.22\) and \(p_2 = 40/1000 = 0.04\). We need to calculate the pooled proportion \(p = (x_1+x_2)/(n_1+n_2) = (220+40)/(1000+1000) = 0.13\) and standard error \(\sqrt{p(1-p)((1/n_1)+(1/n_2))} = \sqrt{0.13(1-0.13)((1/1000)+(1/1000))} = 0.017\). The Z statistic for the difference in proportions is \((p_1-p_2)/SE = (0.22-0.04)/0.017 = 10.59.\)
03

Determine the p-value

The p-value corresponding to z=10.59 is less than 0.01. Hence, we reject the null hypothesis and conclude that there's convincing evidence that the proportion of teens who check their social networking sites more than 10 times a day is significantly greater than the proportion of parents who think their teens do the same.
04

Determine appropriateness of the two-sample Z test for part (b)

Here, it is not appropriate to use a two-sample Z-test, because one-third is a fixed value. Instead, a Z-test for one proportion can be used to test the hypothesis that more than one-third of all teens have posted something on a social networking site that they later regretted.
05

Testing the hypothesis in part (b)

The Null hypothesis for this test is \(H_0: p = 1/3\) and the alternative hypothesis is \(H_a: p > 1/3\). The proportion from the survey is \(p = 390/1000 = 0.39\). The standard error for this case is \(\sqrt{p(1-p)/n} = \sqrt{(1/3)(1-(1/3)/1000)} = 0.0273\). The Z statistic for this test is \((0.39-1/3)/0.0273 = 2.20.\) The p-value corresponding to this Z value is less than our significance level \(\alpha = 0.05\). Therefore, we reject the null hypothesis and there is enough evidence to support the claim that more than one-third of all teens have posted something on their social networking sites that they later regret.

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Key Concepts

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

Z-Test
A Z-Test is a type of statistical test that helps us understand whether there is a significant difference between population proportions or means. In this case, it is used to examine whether there are differences between the proportion of teens and parents regarding online activity.

When we conduct a Z-Test, we calculate a Z-statistic, a measure that helps us figure out how far away our observed data is from the null hypothesis assuming it is true. It requires data that is normally distributed, or larger sample sizes which allows us to assume normality due to the Central Limit Theorem.

In the given exercise, for instance, we want to test the proportion difference: whether more teens claim to check social networking sites frequently than what parents report on behalf of their teens. We compute this by comparing proportions with a pooled standard deviation to understand if the observed difference is significant enough to reject the null hypothesis.
Proportion
Proportion refers to the part of the whole and is expressed as a fraction or percentage. It's one of the core concepts when dealing with hypothesis testing, especially in tests like the Z-Test.

For example, in our exercise, the proportion of teens who say they check social networks more than 10 times a day is calculated as 220 out of 1000, or 22%. Similarly, the proportion of parents who believe their teens exhibit the same behavior is 40 out of 1000, or 4%.

To conduct a hypothesis test in this scenario, we primarily look at the difference between these proportions. Calculating and understanding the proportions correctly is crucial as it gives us the baseline data that we compare in our hypothesis test.
Significance Level
The significance level, often denoted as \(\alpha\), is a threshold set by the researcher which determines whether a result is statistically significant. It essentially represents the probability of rejecting the null hypothesis when it is, in fact, true (Type I error).

In hypothesis testing, we compare the p-value obtained from the Z-Test against this threshold. A common significance level used is 0.05, but in the exercise provided, a stricter level of 0.01 is used, especially for discerning small differences with high confidence.

If the p-value is less than \(\alpha\), we reject the null hypothesis. In Part (a), since the p-value is less than 0.01, it means there's a considerable discrepancy between what teens report and what parents believe, far beyond what could be attributed to random chance.
Null and Alternative Hypothesis
Every hypothesis test begins with establishing the null and alternative hypotheses. These hypotheses provide a structure for what needs to be tested.

The null hypothesis, \(H_0\), states that there is no effect or difference. For our scenario, \(H_0: p_1 \leq p_2\), suggesting the proportion of teens actively checking social networking sites is at most equal to what parents think.

Meanwhile, the alternative hypothesis, \(H_a\), suggests the opposite of \(H_0\). In this exercise, \(H_a: p_1 > p_2\) posits that a greater proportion of teens claim frequent checks compared to parental beliefs.

These hypotheses then drive the direction of testing. The outcome of the test will indicate which hypothesis could be supported or rejected based on the data and p-values derived from the statistical test.

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Most popular questions from this chapter

The director of the Kaiser Family Foundation's Program for the Study of Entertainment Media and Health said, "It's not just teenagers who are wired up and tuned in, its babies in diapers as well." A study by Kaiser Foundation provided one of the first looks at media use among the very youngest children \(-\) those from 6 months to 6 years of age (Kaiser Family Foundation, \(2003,\) www .kff.org). Because previous research indicated that children who have a TV in their bedroom spend less time reading than other children, the authors of the Foundation study were interested in learning about the proportion of kids who have a TV in their bedroom. They collected data from two independent random samples of parents. One sample consisted of parents of children age 6 months to 3 years old. The second sample consisted of parents of children age 3 to 6 years old. They found that the proportion of children who had a TV in their bedroom was . 30 for the sample of children age 6 months to 3 years and .43 for the sample of children age 3 to 6 years old. Suppose that the two sample sizes were each 100 . a. Construct and interpret a \(95 \%\) confidence interval for the proportion of children age 6 months to 3 years who have a TV in their bedroom. Hint: This is a one-sample confidence interval. b. Construct and interpret a \(95 \%\) confidence interval for the proportion of children age 3 to 6 years who have a TV in their bedroom. c. Do the confidence intervals from Parts (a) and (b) overlap? What does this suggest about the two population proportions? d. Construct and interpret a \(95 \%\) confidence interval for the difference in the proportion that have TVs in the bedroom for children age 6 months to 3 years and for children age 3 to 6 years. e. Is the interval in Part (d) consistent with your answer in Part (c)? Explain.

"Doctors Praise Device That Aids Ailing Hearts" (Associated Press, November 9,2004 ) is the headline of an article that describes the results of a study of the effectiveness of a fabric device that acts like a support stocking for a weak or damaged heart. In the study, 107 people who consented to treatment were assigned at random to either a standard treatment consisting of drugs or the experimental treatment that consisted of drugs plus surgery to install the stocking. After two years, \(38 \%\) of the 57 patients receiving the stocking had improved and \(27 \%\) of the patients receiving the standard treatment had improved. Do these data provide convincing evidence that the proportion of patients who improve is higher for the experimental treatment than for the standard treatment? Test the relevant hypotheses using a significance level of \(.05 .\)

"Smartest People Often Dumbest About Sunburns" is the headline of an article that appeared in the San Luis Obispo Tribune (July 19,2006 ). The article states that "those with a college degree reported a higher incidence of sunburn that those without a high school degree- \(43 \%\) versus \(25 \%\)." For purposes of this exercise, suppose that these percentages were based on random samples of size 200 from each of the two groups of interest (college graduates and those without a high school degree). Is there convincing evidence that the proportion experiencing a sunburn is higher for college graduates than it is for those without a high school degree? Answer based on a test with a .05 significance level.

Suppose that you were interested in investigating the effect of a drug that is to be used in the treatment of patients who have glaucoma in both eyes. A comparison between the mean reduction in eye pressure for this drug and for a standard treatment is desired. Both treatments are applied directly to the eye. a. Describe how you would go about collecting data for your investigation. b. Does your method result in paired data? c. Can you think of a reasonable method of collecting data that would not result in paired samples? Would such an experiment be as informative as a paired experiment? Comment.

The paper "Ready or Not? Criteria for Marriage Readiness among Emerging Adults" (Journal of \(\underline{\text { Ado- }}\) lescent Research [2009]: 349-375) surveyed emerging adults (defined as age 18 to 25 ) from five different colleges in the United States. Several questions on the survey were used to construct a scale designed to measure endorsement of cohabitation. The paper states that "on average, emerging adult men \((\mathrm{M}=3.75, \mathrm{SD}=1.21)\) reported higher levels of cohabitation endorsement than emerging adult women \((\mathrm{M}=3.39, \mathrm{SD}=1.17) . "\) The sample sizes were 481 for women and 307 for men. a. Carry out a hypothesis test to determine if the reported difference in sample means provides convincing evidence that the mean cohabitation endorsement for emerging adult women is significantly less than the mean for emerging adult men for students at these five colleges. b. What additional information would you want in order to determine whether it is reasonable to generalize the conclusion of the hypothesis test from Part (a) to all college students?

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