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Chapter 7: Q16E (page 398)

Authorizing computer users with palm prints. Access to computers, email, and Facebook accounts is achieved via a password鈥攁 collection of symbols (usually letters and numbers) selected by the user. One problem with passwords is that persistent hackers can create programs that enter millions of combinations of symbols into a target system until the correct password is found. An article in IEEE Pervasive Computing (October-December 2007) investigated the effectiveness of using palm prints to identify authorized users. For example, a system developed by Palmguard, Inc. tests the hypothesis

\({H_0}\): The proposed user is authorized

\({H_a}\): The proposed user is unauthorized

by checking characteristics of the proposed user鈥檚 palm print against those stored in the authorized users鈥� data bank.

a. Define a Type I error and Type II error for this test. Which is the more serious error? Why?

Short Answer

Expert verified
  1. Type II error is the more serious

Step by step solution

01

Given Information

The null and alternative hypothesis are given by

\({H_0}\): The proposed user is authorized

\({H_a}\): The proposed user is unauthorized

02

Type I error

In this case, Type 1 error happens when the user is labeled as unauthorized in spite of authorized.

03

Type II error

In this case, Type II error indicates that an unauthorized user would be considered authorized.

  • An unauthorized may commit significant damage to the data that the security system is designed to defend whereas an unauthorized user presumably has alternative, safe methods of obtaining access.
    Hence, Type II error is the more serious.

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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 伪 and 尾 ; 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伪 increase or decrease? Explain.

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

Performance of stock screeners. Recall, from Exercise 6.36 (p. 350), that stock screeners are automated tools used by investment companies to help clients select a portfolio of stocks to invest in. The data on the annualized percentage return on investment (as compared to the Standard & Poor鈥檚 500 Index) for 13 randomly selected stock screeners provided by the American Association of Individual Investors (AAII) are repeated in the accompanying table. You want to determine whether \(\mu \) , the average annualized return for all AAII stock screeners, is positive (which implies that the stock screeners perform better, on average, than the S&P 500). An XLSTAT printout of the analysis is shown on the top of page 418.

9.0 -.1 -1.6 14.6 16.0 7.7 19.9 9.8 3.2 24.8 17.6 10.7 9.1

  1. State \({H_0}\,and\,{H_a}\) for this test

A sample of five measurements, randomly selected from a normally distributed population, resulted in the following summary statistics: \(\bar x = 4.8\), \(s = 1.3\) \(\) .

a. Test the null hypothesis that the mean of the population is 6 against the alternative hypothesis, 碌<6. Use\(\alpha = .05.\)

b. Test the null hypothesis that the mean of the population is 6 against the alternative hypothesis, 碌\( \ne 6\). Use\(\alpha = .05.\)

c. Find the observed significance level for each test.

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 伪, while the probability of a missed detection (i.e., no warning given when an intrusion occurs) is defined by the symbol 尾. 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 伪and 尾.

Customers who participate in a store鈥檚 free loyalty card program save money on their purchases but allow the store to keep track of the customer鈥檚 shopping habits and potentially sell these data to third parties. A Pew Internet & American Life Project Survey (January 2016) revealed that 225 of a random sample of 250 U.S. adults would agree to participate in a store loyalty card program, despite the potential for information sharing. Letp represent the true proportion of all customers who would participate in a store loyalty card program.

a. Compute a point estimate ofp

b. Consider a store owner who claims that more than 80% of all customers would participate in a loyalty card program. Set up the null and alternative hypotheses for testing whether the true proportion of all customers who would participate in a store loyalty card program exceeds .8

c. Compute the test statistic for part b.

d. Find the rejection region for the test if 伪=.01.

e. Find the p-value for the test.

f. Make the appropriate conclusion using the rejection region.

g. Make the appropriate conclusion using the p-value.
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