91影视

Referring to the State of Cybersecurity (2015) survey of firms from around the world, exercise 1.20, 766 firms that responded to the survey, 628 expect to experience a cyberattack.

The point estimate$\left(\hat{\mathrm{p}}\right)$ of the population proportion p is obtained below,

$\begin{array}{rcl}\hat{\mathrm{p}}& =& \frac{X}{n}\\ & =& \frac{628}{766}\\ & =& 0.8198\\ \hat{p}& 鈮�& 0.82\end{array}$

The mean of the sampling distribution of$\hat{\mathrm{p}}$ is p.

$\hat{\mathrm{p}}$is an unbiased estimator of p.

The mean of the sampling distribution of$\hat{\mathrm{p}}$ is,

$\begin{array}{rcl}{\mathrm{渭}}_{\hat{\mathrm{p}}}& =& \hat{\mathrm{p}}\\ & =& 0.82\end{array}$

The standard deviation of the sampling distribution of$\hat{\mathrm{p}}$ is,

$\begin{array}{rcl}{\mathrm{蟽}}_{\hat{\mathrm{p}}}& =& \sqrt{\frac{\mathrm{p}\left(1-\mathrm{p}\right)}{\mathrm{n}}}\\ & =& \sqrt{\frac{0.82脳0.18}{766}}\\ & =& \sqrt{0.000193}\\ & =& 0.0139\end{array}$

Therefore, the sampling distribution of$\hat{\mathrm{p}}$ follows $\mathrm{N}\left(0.82,0.0139\right)$.

Large-sample confidence interval for p is,

$\begin{array}{rcl}\left(\hat{\mathrm{p}}卤{\mathrm{z}}_{\frac{\mathrm{伪}}{2}}\sqrt{\frac{\hat{\mathrm{p}}\hat{\mathrm{q}}}{\mathrm{n}}}\right)& =& \left(\hat{\mathrm{p}}-{\mathrm{z}}_{\frac{\mathrm{伪}}{2}}\sqrt{\frac{\hat{\mathrm{p}}\hat{\mathrm{q}}}{\mathrm{n}}},\hat{\mathrm{p}}+{\mathrm{z}}_{\frac{\mathrm{伪}}{2}}\sqrt{\frac{\hat{\mathrm{p}}\hat{\mathrm{q}}}{\mathrm{n}}}\right)\\ & =& \left(0.82-\left(1.645脳0.0139\right),0.82+\left(1.645脳0.0139\right)\right)\\ & =& \left(0.82-0.0229,0.82+0.0229\right)\\ & =& \left(0.7971,0.8429\right)\end{array}$

Therefore, the 90% confidence interval for p is $\left(0.7971,0.8429\right)$.

There is 90% chance that the true population proportion for cyberattack belongs to the confidence interval $\left(0.7971,0.8429\right)$.