Cybersecurity survey. Refer to the State of Cybersecurity (2015) survey of firms from around the world, Exercise 1.20 (p. 50). Recall that of the 766 firms that responded to the survey, 628 (or 82%) expect to experience a cyberattack (e.g., a Malware, hacking, or phishing attack) during the year. Estimate the probability of an expected cyberattack at a firm during the year with a 90% confidence interval. Explain how 90% is used as a measure of reliability for the interval.

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,

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

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,

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

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

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

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

Large-sample confidence interval for p is,

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

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 $\left(0.7971,0.8429\right)$.