Question

Are teens or adults more likely to go online daily? The Pew Internet and American Life Project asked a random sample of $800$ teens and a separate random sample of $2253$ adults how often they use the Internet. In these two surveys, $63\%$ of teens and $68\%$of adults said that they go online every day. Construct and interpret a $90\%$ confidence interval for $p_1-p_2$?

Short answer

The mean and standard deviations are $10$ and $0.205$ respectively.

Step-by-step solution

Given Information

Sample mean $\left({\overline{x}}_1\right)=27$.

Sample mean $\left({\overline{x}}_2\right)=17$.

Standard deviation $\left({\sigma }_1\right)=0.8$.

Standard deviation $\left({\sigma }_2\right)=0.5$.

Explanation

The formulas to compute the mean and standard deviation are:

${\mu }_{1-2}={\mu }_1-{\mu }_2$

${\sigma }_{1-2}=\sqrt{\frac{{\sigma }_1^2+}{n_1}\frac{{\sigma }_2^2}{n_2}}$

The mean and standard deviation could be computed as:

${\mu }_{1-2}={\mu }_1-{\mu }_2$

$=27-17$

$=10$

${\sigma }_{1-2}=\sqrt{\frac{{\sigma }_1^2+}{n_1}\frac{{\sigma }_2^2}{n_2}}$

$=\sqrt{\frac{0.8^2}{25}+\frac{0.5^2}{20}}$

$=0.205$