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

Triola - Elementary Statistics 14th Edition

Triola14th EditionElementary StatisticsISBN: 9780137366446당신이 사용하는 게 아니라요?교과서 변경

Triola 14th Edition

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

문제 11.RE.4

11장, 문제 11.RE.4

Does the Treatment Affect Success? The following table lists frequencies of successes and failures for different treatments used for a stress fracture in a foot bone (based on data from “Surgery Unfounded for Tarsal Navicular Stress Fracture,” by Bruce Jancin, Internal Medicine News, Vol. 42, No. 14). Use a 0.05 significance level to test the claim that success of the treatment is independent of the type of treatment. What does the result indicate about the increasing trend to use surgery?

검증된 단계별 안내

Step 1: Understand the problem. The goal is to test whether the success of the treatment is independent of the type of treatment. This involves performing a chi-square test for independence, which is used to determine if there is a significant association between two categorical variables.

Step 2: Organize the data into a contingency table. The table should display the frequencies of successes and failures for each type of treatment. Label the rows as 'Type of Treatment' and the columns as 'Success' and 'Failure'.

Step 3: State the null and alternative hypotheses. The null hypothesis (H₀) is that the success of the treatment is independent of the type of treatment. The alternative hypothesis (H₁) is that the success of the treatment is not independent of the type of treatment.

Step 4: Calculate the expected frequencies for each cell in the contingency table using the formula:

E = \frac{R \times C}{N}

, where E is the expected frequency, R is the row total, C is the column total, and N is the grand total.

Step 5: Compute the chi-square test statistic using the formula:

χ ² = \sum_{i}^{j} \frac{(O - E) ²}{E}

, where O is the observed frequency and E is the expected frequency. Compare the test statistic to the critical value from the chi-square distribution table at the 0.05 significance level to determine whether to reject the null hypothesis.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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 assumes no effect or relationship, and an alternative hypothesis (H1) that suggests a significant effect or relationship. The goal is to determine whether the observed data provides enough evidence to reject the null hypothesis at a specified significance level, such as 0.05.

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. In this context, it assesses whether the success of a treatment is independent of the type of treatment used. The test compares the observed frequencies of outcomes in a contingency table to the expected frequencies if the variables were independent, providing a p-value to evaluate significance.

Significance Level

The significance level, often denoted as alpha (α), is the threshold used to determine 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 (Type I error). If the p-value obtained from the test is less than or equal to the significance level, the null hypothesis is rejected, suggesting a statistically significant effect or relationship.

관련 실천

교과서 질문

Questions 6–10 refer to the sample data in the following table, which describes the fate of the passengers and crew aboard the Titanic when it sank on April 15, 1912. Assume that the data are a sample from a large population and we want to use a 0.05 significance level to test the claim that surviving is independent of whether the person is a man, woman, boy, or girl.

Given that the P-value for the hypothesis test is 0.000 when rounded to three decimal places, what do you conclude? What do the results indicate about the rule that women and children should be the first to be saved?

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Testing Goodness-of-Fit with a Normal Distribution Refer to Data Set 1 “Body Data” in Appendix B for the heights of females.

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a. Enter the observed frequencies in the table above.

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Cybersecurity The table below lists the frequency of leading digits of Internet traffic interarrival times for a computer, along with the percentages of each leading digit expected with Benford’s law.

a. Identify the general notation used for observed and expected values.

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교과서 질문

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.

Testing a Slot Machine The author purchased a slot machine (Bally Model 809) and tested it by playing it 1197 times. There are 10 different categories of outcomes, including no win, win jackpot, win with three bells, and so on. When testing the claim that the observed outcomes agree with the expected frequencies, the author obtained a test statistic of x2 = 8.815 Use a 0.05 significance level to test the claim that the actual outcomes agree with the expected frequencies. Does the slot machine appear to be functioning as expected?

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교과서 질문

Identifying Hypotheses Refer to the data given in Exercise 1 and assume that the requirements are all satisfied and we want to conduct a hypothesis test of independence using the methods of this section. Identify the null and alternative hypotheses.

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Weather-Related Deaths For the most recent year as of this writing, the numbers of weather-related U.S. deaths for each month were 61, 14, 22, 26, 29, 42, 93, 49, 47, 35, 96, 16, listed in order beginning with January (based on data from the National Weather Service). Use a 0.01 significance level to test the claim that weather-related deaths occur in the different months with the same frequency. Provide an explanation for the result.

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