Answer: [tex](13.22\%,\ 19.34\%)[/tex]
Step-by-step explanation:
Given : Sample space : n= 559
Sample proportion : [tex]\hat{p}=\dfrac{91}{559}=0.162790697674\approx0.1628[/tex]
[tex]=16.28\%[/tex]
Significance level : [tex]\alpha= 1-0.95=0.05[/tex]
Critical value : [tex]z_{\alpha/2}=1.96[/tex]
Confidence level for population proportion:-
[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.1628\pm (1.96)\sqrt{\dfrac{0.1628(1-0.1628)}{559}}\\\\=0.1628\pm0.030604971367\\\\\approx 0.1628\pm0.0306=(0.1322,\ 0.1934)=(13.22\%,\ 19.34\%)[/tex]
Hence, 95% confidence interval for the percentage of all auto accidents that involve teenage drivers.= [tex](13.22\%,\ 19.34\%)[/tex]
