Question

Vulnerability of counting party Web spots. When you subscribe to your Facebook account, you're granted access to further. Then 1 million counting parties (RP) Web spots. Vulnerabilities in this sign-on system may permit a bushwhacker to gain unauthorized access to your profile, allowing the bushwhacker to impersonate you on the RP Web point. Computer and systems Masterminds delved into the vulnerability of counting party Web spots and presented their results at the Proceedings of the 5th AMC Factory on Computers & Communication Security (October 2012). RP Web spots were distributed as Garçon- inflow or customer- inflow Web spots. Of the 40 garçon- inflow spots studied, 20 were planted to be vulnerable to impersonation attacks. Of the 54 customer-inflow spots examined, 41 were. Plant to be vulnerable to impersonation attacks. Give your opinion on whether a customer- inflow Web point is more likely to be vulnerable to an impersonation attack than a garçon- inflow Website. However, how much more likely? If so.

Short answer

There is 95% confidence that a client-flow website is more likely to be vulnerable to a special attack than a server-flow website is, from 0.067 to 0.451.

Step-by-step solution

Step-by-Step Solution Step 1: Check whether a client-flow website is more likely to an impressive attack than a server-flow site

The sample proportion for the client-flow website is obtained below.

$\begin{array}{l}{\overline{P}}_c=\frac{x_c}{n_c}\\ =\frac{20}{40}\\ =0.5\end{array}$

The sample proportion for the serve-flow website is obtained below.

$\begin{array}{l}{\overline{P}}_s=\frac{x_s}{n_s}\\ =\frac{41}{54}\\ =0.759\end{array}$

The weighted average $\left(\overline{P}\right)$ of P_c and P_s is,

$\begin{array}{l}\overline{P}=\frac{x_c+x_s}{n_c+n_s}\\ =\frac{20+41}{40+54}\\ =\frac{61}{94}\\ =0.649\end{array}$

State the test hypotheses

Null hypothesis:

H_0: P_S– P_c= 0

The client-flow website is not more likely to be vulnerable to a special attack than a serve-flow website.

Alternative hypothesis:

H_a= P_s- P_c< 0

The client-flow website is more likely to be vulnerable to a special attack than a serve-flow website.

Test statistics

$\begin{array}{l}z=\frac{{\overline{p}}_1{\overline{q}}_1}{\overline{p}\overline{q}[\frac{1}{n_1}+\frac{1}{n_2}]}\\ =\frac{0.5-0.759}{\left(0.649\left)\right(1-0.649\right)(\frac{1}{40}+\frac{1}{54})}\\ =\frac{-0.25*}{0.0996}\\ =-2.60\end{array}$

Thus, the test statistic is -2.60.

Critical values

The critical value is obtained below.

Here, the test is two-tailed, and the significance level is α = 0.05.

The confidence coefficient is 0.95.

So,

(1-α)=0.95

Α=0.05

$\frac{\alpha }{2}$= 0.025

From appendix D, table 2, the critical value for the two-tailed test with α = 0.05 is$z_{\frac{a}{2}}$_(-0.025)= 1.96.

Conclusion

Here, the test statistic falls in the rejection region.

So, the null hypothesis was rejected.

 Get a 95% confidence interval

Determine how much a client-flow website is more likely to be vulnerable to a special attack than a server-flow website.

$$\begin{gathered} =\left[\begin{array}{c}({\overline{P}}_s-{\overline{P}}_c)\pm z_{\frac{a}{5}}\\ \sqrt{\frac{{\overline{p}}_s(1-{\overline{P}}_s)}{n_5}+\frac{{\overline{P}}_c(1-{\overline{P}}_c)}{n_c}}\end{array}\right] \\ =(0.759-.05)\pm 1.96\sqrt{\frac{0.759(1-0.759)}{54}+\frac{0.5(1-0.05)}{40}} \\ =0.259\pm 1.96(0.09817) \\ =0.259\pm 0.1924 \\ =(0.067,0.451) \end{gathered}$$

So, 95% confidence interval is (0.067,0.451).

Final answer

There is 95% confidence that a client-flow website is more likely to be vulnerable to a special attack than a server-flow website is, from 0.067 to 0.451.