Question

Statistician: As a statistician, you wonder if there is an association between the subject and the likely hood of cheating on an exam (like the current one you are taking). You go out and conduct a random sample of 1,168 students from Sacramento City College who responded to this question, "How often do you use .com or another online site to cheat on exams? (never, sometimes, often, very often.) The results of the study are summarized in the table below. Task: 1. dentify the appropriate Hypothesis Test. Justify your choice. 2.State the Null and Alternative Hypotheses 3. Verify the conditions. 4. Compute the test statistic (round 4 decimal places- if necessary). 5. Demonstrates how to find the p-value. 6. Make a decision based on a significance level of 3%. 7. State and interpret the conclusion.

Answer 1

The appropriate hypothesis test for this study is a chi-square test of independence. The null hypothesis would be that there is no association between the subject and the likelihood of cheating on an exam. To verify the conditions, we need to ensure that the sample is random, the expected frequency in each cell is at least 5, and the sample size is large enough. The test statistic and p-value can be calculated using the chi-square formula, and a decision can be made based on the significance level.

Step-by-step explanation:

1. The appropriate hypothesis test for this study would be a chi-square test of independence. This test is used to determine if there is a relationship between two categorical variables.

2. The null hypothesis would be that there is no association between the subject and the likelihood of cheating on an exam. The alternative hypothesis would be that there is an association between the subject and the likelihood of cheating on an exam.

3. To verify the conditions, we need to ensure that the sample is random, the expected frequency in each cell is at least 5, and the sample size is large enough.

4. To compute the test statistic, we would use the formula chi-square = Σ (observed frequency - expected frequency)^2 / expected frequency. We would need to calculate the expected frequency for each cell in the table.

5. To find the p-value, we would compare the computed test statistic to the chi-square distribution with (number of rows - 1) * (number of columns - 1) degrees of freedom. The p-value represents the probability of observing a test statistic as extreme as the one calculated.

6. Based on a significance level of 3%, we would compare the p-value to 0.03. If the p-value is less than 0.03, we would reject the null hypothesis. If the p-value is greater than or equal to 0.03, we would fail to reject the null hypothesis.

7. The conclusion would be stated as follows: There is sufficient evidence to conclude that there is (or is not) an association between the subject and the likelihood of cheating on an exam at a significance level of 3%.