91影视

If such amount of views in a spontaneous dataset consisting $\widehat{p}$than adhere to a classi铿乪r is $n$, an estimate $100(1鈭�伪)\mathrm{\%}$significance level upon that share $p$ of the sample that falls to all or any this kind type is $n$

$\widehat{p}鈭�z_{\frac{伪}{2}}\sqrt{\frac{\widehat{p}(1鈭�\widehat{p})}{n}}鈮�p鈮�\widehat{p}+z_{\frac{伪}{2}}\sqrt{\frac{\widehat{p}(1鈭�\widehat{p})}{n}}$

Where $Z\frac{a}{2}$ signifies probability the conventional Tarbell top $\frac{a}{2}$ number of percent.

Nine $x=9$of $n=48$ respondents during a previous Zogby World Research ranked the likelihood of a act of terrorism within the region as "likely" or "extremely likely."

whether or not they utilize the bonus procedure,$n^{\mathrm{鈥�}}=n+4=52$and $x^{\mathrm{鈥�}}=x+2=11$are the outputs. As an answer, the info point of both the population share $p$is

$\widehat{p}=\frac{x^{\mathrm{鈥�}}}{n^{\mathrm{鈥�}}}=\frac{11}{52}=0.21$

$\frac{伪}{2}=\frac{1鈭�0.97}{2}$

$=0.015$

$Z\frac{伪}{2}=2.17$

$0.21鈭�2.17\sqrt{\frac{0.21(1鈭�0.21)}{52}}鈮�p鈮�0.21+2.17\sqrt{\frac{0.21(1鈭�0.21)}{52}}$

That's shortened to

$0.089鈮�p鈮�0.334$