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

Expert verified

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

01

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.

P¯c=xcnc=2040=0.5

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

P¯s=xsns=4154=0.759

The weighted average P of Pc and Ps is,

P¯=xc+xsnc+ns=20+4140+54=6194=0.649

02

State the test hypotheses

Null hypothesis:

H0: PS– Pc= 0

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

Alternative hypothesis:

Ha= Ps- Pc< 0

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

03

Test statistics

z=p¯1q¯1p¯q¯[1n1+1n2]=0.50.759(0.649)(10.649)(140+154)=0.25*0.0996=2.60

Thus, the test statistic is -2.60.

04

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

α2= 0.025

From appendix D, table 2, the critical value for the two-tailed test with α = 0.05 isza2(-0.025)= 1.96.

05

Conclusion

Here, the test statistic falls in the rejection region.

So, the null hypothesis was rejected.

06

 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.

=(P¯sP¯c)±za5p¯s(1P¯s)n5+P¯c(1P¯c)nc=(0.759-.05)±1.960.759(10.759)54+0.5(10.05)40=0.259±1.96(0.09817)=0.259±0.1924=(0.067,0.451)

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

07

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

4.111 Personnel dexterity tests. Personnel tests are designed to test a job applicant’s cognitive and/or physical abilities. The Wonderlic IQ test is an example of the former; the Purdue Pegboard speed test involving the arrangement of pegs on a peg board is an example of the latter. A particular

dexterity test is administered nationwide by a private testing service. It is known that for all tests administered last year, the distribution of scores was approximately normal with mean 75 and standard deviation 7.5.

a. A particular employer requires job candidates to score at least 80 on the dexterity test. Approximately what percentage of the test scores during the past year exceeded 80?

b. The testing service reported to a particular employer that one of its job candidate’s scores fell at the 98th percentile of the distribution (i.e., approximately 98% of the scores were lower than the candidate’s, and only 2%were higher). What was the candidate’s score?

Question: Find the following probabilities for the standard normal random variable z:

a.P(z>1.46)b.P(z<-1.56)c.P(.67z<2.41)d.P(-1.96z-.33)e.P(Z0)f.P(-2.33<z<1.50)

Enough money has been budgeted to collect independent random samples of size n1=n2=100from populations 1 and 2 to estimate localid="1664867109106" μ1-μ2. Prior information indicates that σ1=σ2=10. Have sufficient funds been allocated to construct a 90% confidence interval forμ1-μ2of width 5 or less? Justify your answer.

Assume that σ12222. Calculate the pooled estimator σ2 for each of the following cases:

a.s12=120,s22=100,n1=n2=25

b.s12=12,s22=20,n1=20,n2=10

c.s12=.15,s22=.20,n1=6,n2=10

d.s12=3000,s22=2500,n1=16,n2=17

Note that the pooled estimate is a weighted average of the sample variances. To which of the variances does the pooled estimate fall nearer in each of the above cases?

Question: Performance ratings of government agencies. The U.S. Office of Management and Budget (OMB) requires government agencies to produce annual performance and accounting reports (PARS) each year. A research team at George Mason University evaluated the quality of the PARS for 24 government agencies (The Public Manager, Summer 2008), where evaluation scores ranged from 12 (lowest) to 60 (highest). The accompanying file contains evaluation scores for all 24 agencies for two consecutive years. (See Exercise 2.131, p. 132.) Data for a random sample of five of these agencies are shown in the accompanying table. Suppose you want to conduct a paired difference test to determine whether the true mean evaluation score of government agencies in year 2 exceeds the true mean evaluation score in year 1.

Source: J. Ellig and H. Wray, “Measuring Performance Reporting Quality,” The Public Manager, Vol. 37, No. 2, Summer 2008 (p. 66). Copyright © 2008 by Jerry Ellig. Used by permission of Jerry Ellig.

a. Explain why the data should be analyzedusing a paired difference test.

b. Compute the difference between the year 2 score and the year 1 score for each sampled agency.

c. Find the mean and standard deviation of the differences, part

b. Use the summary statistics, part c, to find the test statistic.

e. Give the rejection region for the test using a = .10.

f. Make the appropriate conclusion in the words of the problem.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free