Respuesta :
Answer:
The 95% confidence interval would be given (0.307;0.373).
Step-by-step explanation:
Data given and notation
n=800 represent the random sample taken
X=272 represent the teens who admit texting while driving
[tex]\hat p=\frac{272}{800}=0.34[/tex] proportion estimated for teens who admit texting while driving
[tex]\alpha=0.05[/tex] represent the significance level (no given, but is assumed)
Confidence =0.95 or 95%
p= population proportion of teens who admit texting while driving
The confidence interval for the population proportion would be given by this formula:
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.34 - 1.96 \sqrt{\frac{0.34(1-0.34)}{800}}=0.307[/tex]
[tex]0.34 + 1.96 \sqrt{\frac{0.34(1-0.34)}{800}}=0.373[/tex]
And the 95% confidence interval would be given (0.307;0.373).
We are confident (95%) that about 30.7% to 37.3% of the teens are texting while driving
