Question

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

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

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 \(H_a: p_1 > p_2\).

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 \((p_1-p_2)/SE = (0.22-0.04)/0.017 = 10.59.\)

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.

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.

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.

Key concepts

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.