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 Instagram. The two samples were independently selected. 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. Compute and interpret a 90% confidence interval for the difference (teen – parent) in proportions who check their online social networking sites.

Answer 1

= 220/1000 = 0.22

= 40/1000 = 0.04

The pooled sample proportion (P) = ( * n1 + * n2)/(n1 + n2) = (0.22 * 1000 + 0.04 * 1000)/(1000 + 1000) = 0.13

SE = sqrt(P(1 - P)(1/n1 + 1/n2))

= sqrt(0.13 * (1 - 0.13) * (1/1000 + 1/1000))

= 0.015

For 90% confidence interval, the critical value is z_0.05 = 1.645

The 90% confidence interval is

( - ) +/- z_0.05 * SE

= (0.22 - 0.04) +/- 1.645 * 0.015

= 0.18 +/- 0.025

= 0.155, 0.205

We are 90% confident that the difference in proportions between teen and parents who check their online social networking sites lies in the above confidence interval.