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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 11.2.15

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?

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Step 1: Define the null and alternative hypotheses. The null hypothesis (H₀) states that getting a cold is independent of the treatment group, while the alternative hypothesis (H₁) states that getting a cold is dependent on the treatment group.
Step 2: Organize the data into a contingency table. The table provided already shows the observed frequencies for each treatment group and outcome (Got a Cold vs. Did Not Get a Cold).
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 'Placebo and Got a Cold' as: E = ((88 + 15) × (88 + 48 + 42)) / grand total.
Step 4: Compute the chi-square test statistic using the formula: χ² = Σ((O - E)² / E), where O represents the observed frequency and E represents the expected frequency for each cell. Perform this calculation for all cells in the table.
Step 5: Compare the calculated chi-square test statistic to the critical value from the chi-square distribution table at a significance level of 0.05 with the appropriate degrees of freedom (df = (number of rows - 1) × (number of columns - 1)). If the test statistic exceeds the critical value, reject the null hypothesis; otherwise, fail to reject it.

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Key Concepts

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

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions about a population based on sample data. It involves formulating a null hypothesis (H0) that represents no effect or no difference, and an alternative hypothesis (H1) that indicates the presence of an effect. In this case, the null hypothesis would state that getting a cold is independent of the treatment group, while the alternative would suggest a dependence.
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Step 1: Write Hypotheses

Chi-Square Test of Independence

The Chi-Square Test of Independence is a statistical test used to determine if there is a significant association between two categorical variables. It compares the observed frequencies in each category of a contingency table to the frequencies expected under the null hypothesis. In this scenario, it will help assess whether the incidence of colds is related to the type of treatment received (placebo vs. echinacea).
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Probability of Multiple Independent Events

Significance Level

The significance level, often denoted as alpha (α), is the threshold for determining whether to reject the null hypothesis in hypothesis testing. A common significance level is 0.05, which indicates a 5% risk of concluding that a difference exists when there is none. In this clinical trial, using a 0.05 significance level means that if the p-value from the Chi-Square test is less than 0.05, the results would suggest that echinacea treatment affects the likelihood of getting a cold.
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Step 4: State Conclusion Example 4
Related Practice
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

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

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.


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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