Respuesta :
Answer:
[tex]0.491 - 1.64\sqrt{\frac{0.491(1-0.491)}{900}}=0.464[/tex]
[tex]0.491 + 1.64\sqrt{\frac{0.491(1-0.491)}{900}}=0.518[/tex]
We are 90% confident that the true proportion of teenagers who have used at least one informal element in school writing assignments is between 0.464 and 0.518
Step-by-step explanation:
We can begin finding the estimator for the proportion of interest:
[tex]\hat p =\frac{442}{900}= 0.491[/tex]
The confidence level is 90% , and the significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/tex]
The confidence interval for the mean is given by the following formula: Â
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
Replacing the info given we got:
[tex]0.491 - 1.64\sqrt{\frac{0.491(1-0.491)}{900}}=0.464[/tex]
[tex]0.491 + 1.64\sqrt{\frac{0.491(1-0.491)}{900}}=0.518[/tex]
We are 90% confident that the true proportion of teenagers who have used at least one informal element in school writing assignments is between 0.464 and 0.518