Question

Common Sense Media surveyed 1,000 teens and 1,000 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 to be representative of American teens and parents of American teens. When asked if they check 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 networking sites more than 10 times a day, 40 said yes. Use a significance level of 0.01 to determine if there is convincing evidence that the proportion of all parents who think their teen checks social networking sites more than 10 times a day is less than the proportion of all teens who check more than 10 times a day.

Short answer

Yes, there is convincing evidence at the 0.01 level of significance to suggest that the proportion of all parents who think their teen checks social networking sites more than 10 times a day is less than the proportion of all teens who check these sites more than 10 times a day.

Step-by-step solution

State the Hypotheses

The null hypothesis is that the proportions are equal, and the alternative hypothesis is that the proportion of parents who think their teen checks social networking sites more than 10 times a day is less than the proportion of teens who check more than 10 times a day. So, \n\nNull Hypothesis \(H_0: p1 = p2\), \nAlternative Hypothesis \(H_a: p1 < p2\)

Calculate the Proportions and Test Statistic

First, calculate the proportions. For the teens, the proportion \(p1\) is 220 out of 1000, or 0.22. For the parents, the proportion \(p2\) is 40 out of 1000, or 0.04. Now, calculate the pooled proportion \(p\) which is equal to \((x1 + x2) / (n1 + n2)\) = (220 + 40) / (1000 + 1000) = 0.13. Then, calculate the test statistic using the formula for the z score in testing of two proportions: \(z = (p1 - p2) / sqrt [p(1-p)(1/n1 + 1/n2)]\). Substituting the given numbers, our test statistic becomes \(z = (0.22 - 0.04) / sqrt [0.13(1 - 0.13)(1/1000 + 1/1000)] = 15.94.

Determine the P-Value and Decision

Use a standard normal distribution (z-distribution) to find the p-value associated with your calculated z-test statistic. In this case, our z-value is so large that the p-value will be effectively 0. Comparing the p-value with our significance level of 0.01, since p-value (0) < \(\alpha\) (0.01), we reject the null hypothesis.