/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q18E Intrusion detection systems. The... [FREE SOLUTION] | 91影视

91影视

Intrusion detection systems. The Journal of Research of the National Institute of Standards and Technology (November鈥 December 2003) published a study of a computer intrusion detection system (IDS). The IDS is designed to provide an alarm whenever unauthorized access (e.g., an intrusion) to a computer system occurs. The probability of the system giving a false alarm (i.e., providing a warning when no intrusion occurs) is defined by the symbol , while the probability of a missed detection (i.e., no warning given when an intrusion occurs) is defined by the symbol . These symbols are used to represent Type I and Type II error rates, respectively, in a hypothesis-testing scenario

a. What is the null hypothesis, H0?

b. What is the alternative hypothesis,Ha?

c. According to actual data collected by the Massachusetts Institute of Technology Lincoln Laboratory, only 1 in 1,000 computer sessions with no intrusions resulted in a false alarm. For the same system, the laboratory found that only 500 of 1,000 intrusions were actually detected. Use this information to estimate the values of and .

Short Answer

Expert verified
  1. H0:No warning given when no intrusion occurs
  2. Ha:Warning given when an intrusion occurs
  3. The value of andare 0.001 and 0.5.

Step by step solution

01

Null hypothesis

a. The null hypothesis is given by:

H0:No warning given when no intrusion occurs

02

Alternative hypothesis

b. The alternative hypothesis is given by:

Ha:Warning given when an intrusion occurs

03

Computing the value of α and β

c. The value of is computed as,

=Number of computer sessions with no intrusionsTotal number of computer sessions=11000=0.001

The value of is computed as,

=Number of intrusions were actually detectedTotal number of intrusions=5001000=0.5

Therefore, the value of andare 0.001 and 0.5.

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 91影视!

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

Jury trial outcomes. Sometimes, the outcome of a jury trial defies the 鈥渃ommon sense鈥 expectations of the general public (e.g., the 1995 O. J. Simpson verdict and the 2011 Casey Anthony verdict). Such a verdict is more acceptable if we understand that the jury trial of an accused murderer is analogous to the statistical hypothesis-testing process. The null hypothesis in a jury trial is that the accused is innocent. (The status-quo hypothesis in the U.S. system of justice is innocence, which is assumed to be true until proven beyond a reasonable doubt.) The alternative hypothesis is guilt, which is accepted only when sufficient evidence exists to establish its truth. If the vote of the jury is unanimous in favor of guilt, the null hypothesis of innocence is rejected, and the court concludes that the accused murderer is guilty. Any vote other than a unanimous one for guilt results in a 鈥渘ot guilty鈥 verdict. The court never accepts the null hypothesis; that is, the court never declares the accused 鈥渋nnocent.鈥 A 鈥渘ot guilty鈥 verdict (as in the Casey Anthony case) implies that the court could not find the defendant guilty beyond a reasonable doubt

a. Define Type I and Type II errors in a murder trial.

b. Which of the two errors is the more serious? Explain.

c. The court does not, in general, know the values of and ; but ideally, both should be small. One of these probabilities is assumed to be smaller than the other in a jury trial. Which one, and why?

d. The court system relies on the belief that the value of is made very small by requiring a unanimous vote before guilt is concluded. Explain why this is so.

e. For a jury prejudiced against a guilty verdict as the trial begins, will the value of increase or decrease? Explain.

f. For a jury prejudiced against a guilty verdict as the trial begins, will the value of increase or decrease? Explain

Stability of compounds in new drugs. Refer to the ACS Medicinal Chemistry Letters (Vol. 1, 2010) study of the metabolic stability of drugs, Exercise 2.22 (p. 83). Recall that two important values computed from the testing phase are the fraction of compound unbound to plasma (fup) and the fraction of compound unbound to microsomes (fumic). A key formula for assessing stability assumes that the fup/fumic ratio is 1:1. Pharmacologists at Pfizer Global Research and Development tested 416 drugs and reported the fup/fumic ratio for each. These data are saved in the FUP file, and summary statistics are provided in the accompanying Minitab printout. Suppose the pharmacologists want to determine if the true mean ratio, 渭, differs from 1.

a. Specify the null and alternative hypotheses for this test.

b. Descriptive statistics for the sample ratios are provided in the Minitab printout on page 410. Note that the sample mean,\(\overline x = .327\)is less than 1. Consequently, a pharmacologist wants to reject the null hypothesis. What are the problems with using such a decision rule?

c. Locate values of the test statistic and corresponding p-value on the printout.

d. Select a value of\(\alpha \)the probability of a Type I error. Interpret this value in the words of the problem.

e. Give the appropriate conclusion based on the results of parts c and d.

f. What conditions must be satisfied for the test results to be valid?

Cooling method for gas turbines. During periods of high electricity demand, especially during the hot summer months, the power output from a gas turbine engine can drop dramatically. One way to counter this drop in power is by cooling the inlet air to the gas turbine. An increasingly popular cooling method uses high-pressure inlet fogging. The performance of a sample of 67 gas turbines augmented with high-pressure inlet fogging was investigated in the Journal of Engineering for Gas Turbines and Power (January 2005). One performance measure is heat rate (kilojoules per kilowatt per hour). Heat rates for the 67 gas turbines are listed in the table below. Suppose that standard gas turbines have heat rates with a standard deviation of 1,500 kJ/kWh. Is there sufficient evidence to indicate that the heat rates of the augmented gas turbine engine are more variable than the heat rates of the standard gas turbine engine? Test using a = .05.

Feminized faces in TV commercials. Television commercials most often employ females or 鈥渇eminized鈥 males to pitch a company鈥檚 product. Research published in Nature (August 27 1998) revealed that people are, in fact, more attracted to 鈥渇eminized鈥 faces, regardless of gender. In one experiment, 50 human subjects viewed both a Japanese female face and a Caucasian male face on a computer. Using special computer graphics, each subject could morph the faces (by making them more feminine or more masculine) until they attained the 鈥渕ost attractive鈥 face. The level of feminization x (measured as a percentage) was measured.

a. For the Japanese female face, x = 10.2% and s = 31.3%. The researchers used this sample information to test the null hypothesis of a mean level of feminization equal to 0%. Verify that the test statistic is equal to 2.3.

b. Refer to part a. The researchers reported the p-value of the test as p = .021. Verify and interpret this result.

c. For the Caucasian male face, x = 15.0% and s = 25.1%. The researchers reported the test statistic (for the test of the null hypothesis stated in part a) as 4.23 with an associated p-value of approximately 0. Verify and interpret these results.

Play Golf America program. The Professional Golf Association (PGA) and Golf Digest have developed the Play Golf America program, in which teaching professionals at participating golf clubs provide a free 10-minute lesson to new customers. According to Golf Digest, golf facilities that participate in the program gain, on average, \(2,400 in greens fees, lessons, or equipment expenditures. A teaching professional at a golf club believes that the average gain in greens fees, lessons, or equipment expenditures for participating golf facilities exceeds \)2,400.

a. In order to support the claim made by the teaching professional, what null and alternative hypotheses should you test?

b. Suppose you select = 0.05. Interpret this value in the words of the problem.

c. For = 0.05, specify the rejection region of a large sample test.

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.