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Chapter 8: Q85E (page 452)

Fingerprint expertise.A study published in PsychologicalScience(August 2011) tested the accuracy of experts andnovices in identifying fingerprints. Participants were presentedpairs of fingerprints and asked to judge whetherthe prints in each pair matched. The pairs were presentedunder three different conditions: prints from the same individual (match condition), non-matching but similar prints (similar distracter condition), and nonmatching and very dissimilar prints (non-similar distracter condition). The percentages of correct decisions made by the two groups under each of the three conditions are listed in the table.

Conditions

Fingerprints expert

Novices

Match similar

92.12%

74.55%

Distracter

99.32%

44.82%

Non-similar distracter

100%

77.03%

a.Given a pair of matched prints, what is the probability that an expert failed to identify the match?

b. Given a pair of matched prints, what is the probabilitythat a novice failed to identify the match?

c. Assume the study included 10 participants, 5 experts and 5 novices. Suppose that a pair of matched prints was presented to a randomly selected study participant and the participant failed to identify the match. Is the participant more likely to be an expert or a novice?

Short Answer

Expert verified

(a) The probability is 0.079.

(b) The probability is 0.2545.

(c) The participant more likely to be novice.

Step by step solution

01

Given information

The data is provided in the table,

Let

P-The professional to make the best decision

B-The beginner to make the right decision

C-The condition that is connected or matched

S-The same distracter condition

N-The non-similar distracter.

02

The probability that an expert failed to identify the match.

PPcC=1-PPC=1-0.9212=0.079

So, the probability is 0.079.

03

The probability that a novice failed to identify the match

PScC=1-PSC=1-0.7455=0.2545

Hence, the probability is 0.2545.

04

Is the participant more likely to be an expert or a novice.

From the results of part (a) and (b), its clear that the participant more likely to be a novice.

Therefore, the contestant is more likely to be inexperienced.

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Most popular questions from this chapter

Question: Forecasting daily admission of a water park. To determine whether extra personnel are needed for the day, the owners of a water adventure park would like to find a model that would allow them to predict the day’s attendance each morning before opening based on the day of the week and weather conditions. The model is of the form

where,

y = Daily admission

x1 = 1 if weekend

0 otherwise

X2 = 1 if sunny

0 if overcast

X3 = predicted daily high temperature (°F)

These data were recorded for a random sample of 30 days, and a regression model was fitted to the data.

The least squares analysis produced the following results:

with

  1. Interpret the estimated model coefficients.
  2. Is there sufficient evidence to conclude that this model is useful for predicting daily attendance? Use α = .05.
  3. Is there sufficient evidence to conclude that the mean attendance increases on weekends? Use α = .10.
  4. Use the model to predict the attendance on a sunny weekday with a predicted high temperature of 95°F.
  5. Suppose the 90% prediction interval for part d is (645, 1,245). Interpret this interval.

Homework assistance for accounting students. How much assistance should accounting professors provide students for completing homework? Is too much assistance counterproductive? These were some of the questions of interest in a Journal of Accounting Education (Vol. 25, 2007) article. A total of 75 junior-level accounting majors who were enrolled in Intermediate Financial Accounting participated in an experiment. All students took a pretest on a topic not covered in class; then, each was given a homework problem to solve on the same topic. However, the students were randomly assigned different levels of assistance on the homework. Some (20 students) were given the completed solution, some (25 students) were given check figures at various steps of the solution, and the rest (30 students) were given no help. After finishing the homework, each student was given a posttest on the subject. One of the variables of interest to the researchers was the knowledge gain (or test score improvement), measured as the difference between the posttest and pretest scores. The sample means knowledge gains for the three groups of students are provided in the table.

a. The researchers theorized that as the level of homework assistance increased, the test score improvement from pretest to post test would decrease. Do the sample means reported in the table support this theory?

b. What is the problem with using only the sample means to make inferences about the population mean knowledge gains for the three groups of students?

c. The researchers conducted a statistical test of the Hypothesis to compare the mean knowledge gain of students in the "no solutions" group with the mean knowledge gain of students in the "check figures" group. Based on the theory, part a sets up the null and alternative hypotheses for the test.

d. The observed significance level of the t-test of the partc was reported as8248 Using α=.05, interpret this result.

e. The researchers conducted a statistical test of the hypothesis to compare the mean knowledge gain of students in the "completed solutions" group with the mean knowledge gain of students in the "check figures" group. Based on the theory, part a sets up the null and alternative hypotheses for the test.

f. The observed significance level of the role="math" localid="1652694732458" t-test of part e was reported as 1849. Using α=.05, interpret this result.

g. The researchers conducted a statistical test of the Hypothesis to compare the mean knowledge gain of students in the "no solutions" group with the mean knowledge gain of students in the "completed solutions" group. Based on the theory, part a sets up the null and alternative hypotheses for the test.

h. The observed significance level of the t-test of the part wasg reported as2726. Using role="math" localid="1652694677616" α=.05, interpret this result.

Optimal goal target in soccer. When attempting to score a goal in soccer, where should you aim your shot? Should you aim for a goalpost (as some soccer coaches teach), the middle of the goal, or some other target? To answer these questions, Chance (Fall 2009) utilized the normal probability distribution. Suppose the accuracy x of a professional soccer player’s shots follows a normal distribution with a mean of 0 feet and a standard deviation of 3 feet. (For example, if the player hits his target,x=0; if he misses his target 2 feet to the right, x=2; and if he misses 1 foot to the left,x=-1.) Now, a regulation soccer goal is 24 feet wide. Assume that a goalkeeper will stop (save) all shots within 9 feet of where he is standing; all other shots on goal will score. Consider a goalkeeper who stands in the middle of the goal.

a. If the player aims for the right goalpost, what is the probability that he will score?

b. If the player aims for the center of the goal, what is the probability that he will score?

c. If the player aims for halfway between the right goal post and the outer limit of the goalkeeper’s reach, what is the probability that he will score?

Given that xis a hypergeometric random variable, computep(x)for each of the following cases:

a. N= 8, n= 5, r= 3, x= 2

b. N= 6, n= 2, r= 2, x= 2

c. N= 5, n= 4, r= 4, x= 3

Find a value of the standard normal random variable z, call it z0, such that

a.P(z≤z0)=0.2090b.P(z≤z0)=0.7090c.P(-z0≤z<z0)=0.8472d.P(-z0≤z<z0)=0.1664e.P(z0≤z≤z0)=0.4798f.P(-1<z<z0)
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