Ch. 11 - Goodness-of-Fit and Contingency Tables

Triola - Elementary Statistics 14th Edition

Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook

Triola 14th Edition

Ch. 11 - Goodness-of-Fit and Contingency Tables

Problem 11.2.21

Chapter 11, Problem 11.2.21

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.

Verified step by step guidance

Understand the relationship: The problem states that the chi-square test statistic (χ²) and the z-test statistic for the difference between two proportions are related by the formula z² = χ². This means that the square of the z-test statistic is equal to the chi-square test statistic. We will verify this relationship step by step.

Step 1: Write the formula for the chi-square test statistic for a 2x2 table. The formula is:

χ^{2} = Σ ((O - E)

² / E), where O represents the observed frequencies, E represents the expected frequencies, and the summation is over all cells in the table.

Step 2: Write the formula for the z-test statistic for the difference between two proportions. The formula is:

z = (p₁ - p₂) / sqrt(p(1-p)(1/n₁ + 1/n₂)),

where p₁ and p₂ are the sample proportions, p is the pooled proportion, and n₁ and n₂ are the sample sizes.

Step 3: Show the equivalence: Substitute the expressions for the observed and expected frequencies in the chi-square formula and simplify. You will find that the chi-square statistic can be expressed in terms of the z-test statistic as

z^{2} = χ^{2} .

Step 4: Verify the critical values: The critical value for the chi-square test with 1 degree of freedom is the square of the critical value for the z-test. For example, if the critical value for the z-test is ±1.96 (for a 5% significance level, two-tailed), the critical value for the chi-square test is (1.96)² = 3.841. This confirms that the critical values also follow the relationship z² = χ².

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Key Concepts

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

Chi-Squared Test

The Chi-Squared (x^2) test is a statistical method used to determine if there is a significant association between categorical variables. In a 2 x 2 table, it compares the observed frequencies of occurrences in each category to the expected frequencies under the null hypothesis. A higher x^2 value indicates a greater deviation from the expected distribution, suggesting a potential relationship between the variables.

Z-Test for Proportions

The Z-test for proportions is used to compare the proportions of two groups to determine if they are significantly different from each other. It calculates a z statistic based on the difference between the sample proportions, adjusted for the standard error. This test is particularly useful when dealing with large sample sizes, as it relies on the normal approximation of the binomial distribution.

Relationship Between Chi-Squared and Z-Test

The relationship between the Chi-Squared test and the Z-test for proportions is established through the formula z^2 = x^2. This means that the square of the z statistic from the Z-test is equivalent to the Chi-Squared statistic from the Chi-Squared test. This equivalence allows researchers to use either test depending on the context, while still drawing the same conclusions about the significance of the observed differences.

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Step 2: Calculate Test Statistic

Related Practice

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.

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Textbook Question

Dogs Detecting Malaria The following table lists results from an experiment designed to test the ability of dogs to use their extraordinary sense of smell to detect malaria in samples of children’s socks (based on data presented at an annual meeting of the American Society of Tropical Medicine, by principal investigator Steve Lindsay). Assuming that the dog being correct is independent of whether malaria is present, find the expected value for the observed frequency of 123.

Textbook Question

Benford’s Law

According to Benford’s law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. In Exercises 21–24, test for goodness-of-fit with the distribution described by Benford’s law.

Detecting Fraud When working for the Brooklyn district attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 15, 0, 76, 479, 183, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford’s law, the check amounts appear to result from fraud. Use a 0.01 significance level to test for goodness-of-fit with Benford’s law. Does it appear that the checks are the result of fraud?

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

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.