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Ch. 11 - Goodness-of-Fit and Contingency Tables
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
Chapter 11, Problem 11.2.8

Accuracy of Fingerprint Identifications An experiment was conducted to compare the accuracy of fingerprint experts to the accuracy of novices (based on data from “Identifying Fingerprint Expertise,” by Tangen, Thompson, and McCarthy, Psychological Science, Vol. 22, No. 8). The data in the table are based on trials in which the evaluators were given matching fingerprints. Use a 0.05 significance level to determine whether correct identification is independent of whether the evaluator is an expert or a novice.


Table comparing fingerprint identification accuracy: Novices 331 correct, 113 wrong; Experts 409 correct, 35 wrong.

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Step 1: Define the null hypothesis (H₀) and the alternative hypothesis (H₁). H₀: Correct identification is independent of whether the evaluator is an expert or a novice. H₁: Correct identification is not independent of whether the evaluator is an expert or a novice.
Step 2: Organize the data into a contingency table. The table is already provided: Rows represent evaluator type (Novice and Expert), and columns represent identification outcome (Correct and Wrong).
Step 3: Calculate the expected frequencies for each cell in the contingency table using the formula: E = (row total × column total) / grand total. For example, calculate the expected frequency for 'Novice-Correct' as: E = ((331 + 113) × (331 + 409)) / (331 + 113 + 409 + 35). Repeat for all cells.
Step 4: Compute the chi-square test statistic using the formula: χ² = Σ((O - E)² / E), where O is the observed frequency and E is the expected frequency. Perform this calculation for each cell in the table and sum the results.
Step 5: Compare the calculated χ² value to the critical value from the chi-square distribution table at a significance level of 0.05 with degrees of freedom df = (number of rows - 1) × (number of columns - 1). If χ² > critical value, reject H₀; otherwise, fail to reject H₀.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Chi-Square Test of Independence

The Chi-Square Test of Independence is a statistical method used to determine if there is a significant association between two categorical variables. In this context, it helps assess whether the accuracy of fingerprint identification (correct or wrong) is independent of the evaluator's expertise (novice or expert). The test compares observed frequencies in each category to expected frequencies under the assumption of independence.
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Significance Level (α)

The significance level, denoted as α, is the threshold for determining whether a statistical result is significant. In this case, a significance level of 0.05 indicates that there is a 5% risk of concluding that a difference exists when there is none. If the p-value obtained from the Chi-Square test is less than 0.05, we reject the null hypothesis, suggesting that the accuracy of identification is not independent of the evaluator's expertise.
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Step 4: State Conclusion Example 4

Contingency Table

A contingency table is a matrix that displays the frequency distribution of variables, allowing for the analysis of the relationship between them. In this scenario, the table shows the number of correct and wrong identifications made by novices and experts. This format is essential for conducting the Chi-Square Test, as it organizes the data needed to evaluate the independence of the two categorical variables.
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Related Practice
Textbook Question

Equivalent Tests A x^2 test involving a 2 x 2 table is equivalent to the test for the difference between two proportions, as described in Section 9-1. Using Table 11-1 from the Chapter Problem, verify that the x^2 test statistic and the z test statistic (found from the test of equality of two proportions) are related as follows: z^2 = x^2 Also show that the critical values have that same relationship.

Textbook Question

Clinical Trial of Echinacea In a clinical trial of the effectiveness of echinacea for preventing colds, the results in the table below were obtained (based on data from “An Evaluation of Echinacea Angustifolia in Experimental Rhinovirus Infections,” by Turner et al., New England Journal of Medicine, Vol. 353, No. 4). Use a 0.05 significance level to test the claim that getting a cold is independent of the treatment group. What do the results suggest about the effectiveness of echinacea as a prevention against colds?

Textbook Question

In Exercises 5–20, conduct the hypothesis test and provide the test statistic and the P-value and/or critical value, and state the conclusion.


Flat Tire and Missed Class A classic story involves four carpooling students who missed a test and gave as an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn’t have a flat tire, would they be able to identify the same tire? The author asked 41 other students to identify the tire they would select. The results are listed in the following table (except for one student who selected the spare). Use a 0.05 significance level to test the author’s claim that the results fit a uniform distribution. What does the result suggest about the likelihood of four students identifying the same tire when they really didn’t have a flat?


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Textbook Question

Gender and Eye Color The following table describes the distribution of eye colors reported by male and female statistics students (based on data from “Does Eye Color Depend on Gender? It Might Depend on Who or How You Ask,” by Froelich and Stephenson, Journal of Statistics Education, Vol. 21, No. 2). Is there sufficient evidence to warrant rejection of the belief that gender and eye color are independent traits? Use a 0.01 significance level.


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Textbook Question

Ghosts The following table summarizes results from a Pew Research Center survey in which subjects were asked whether they had seen or been in the presence of a ghost. Use a 0.01 significance level to test the claim that gender is independent of response. Does the conclusion change if the significance level is changed to 0.05?


Textbook Question

In Exercises 5–20, conduct the hypothesis test and provide the test statistic and the P-value and/or critical value, and state the conclusion.


Heights Measured or Reported? A random sample of the last digits of heights (in.) of males from Data Set 4 “Measured and Reported” is summarized in the table below. Use these last digits to determine whether they occur with about the same frequency. Use a 0.05 significance level. Do the corresponding heights appear to be measured or reported?


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