Math Problem Statement

What is the decision of the test? A) there is no evidence of an association B) we have evidence of an association C) we are 95% sure that the proportions are as reported in the table

What is the decision of the test? A) there is no evidence of an association B) we have evidence of an association C) we are 95% sure that the proportions are as reported in the table

Solution

To determine the decision for the given contingency table, we would typically perform a chi-square test for independence. The goal of this test is to determine whether there is a significant association between the two categorical variables: "sex" (female/male) and "employment status" (Casual, Full-time, Part-time, Unemployed).

Here are the steps involved:

  1. Null Hypothesis (H0H_0): There is no association between "sex" and "employment status" (they are independent).

  2. Alternative Hypothesis (HAH_A): There is an association between "sex" and "employment status" (they are not independent).

  3. Chi-Square Statistic Calculation:

    • We would first calculate the expected frequencies for each cell in the table under the assumption of independence. The expected frequency for each cell is given by: Eij=(row sum)×(column sum)total sumE_{ij} = \frac{(\text{row sum}) \times (\text{column sum})}{\text{total sum}}
    • Once we have the expected values, we calculate the chi-square statistic as: χ2=(OijEij)2Eij\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} where OijO_{ij} is the observed frequency and EijE_{ij} is the expected frequency for each cell.
  4. Degrees of Freedom: The degrees of freedom for this test are given by: df=(number of rows1)×(number of columns1)\text{df} = (\text{number of rows} - 1) \times (\text{number of columns} - 1) In this case, we have 2 rows (sex: male, female) and 4 columns (employment statuses), so: df=(21)×(41)=3\text{df} = (2-1) \times (4-1) = 3

  5. Compare to Chi-Square Distribution: After calculating the chi-square statistic, we compare it to the critical value from the chi-square distribution with 3 degrees of freedom, at the desired significance level (usually α=0.05\alpha = 0.05).

  6. Decision:

    • If the chi-square statistic is greater than the critical value, we reject the null hypothesis and conclude that there is an association between sex and employment status.
    • If the chi-square statistic is less than the critical value, we fail to reject the null hypothesis, indicating that there is no significant association between the two variables.

Since the table and calculations are not provided, I cannot calculate the chi-square statistic directly here. However, based on the question, the most appropriate option is:

B) We have evidence of an association – This would be the case if the calculated chi-square statistic exceeds the critical value.

Would you like me to calculate the chi-square statistic and complete the test decision? I can guide you through that or do the calculations for you!


Here are 5 related questions to deepen your understanding:

  1. What is the formula for calculating the expected frequency in a contingency table?
  2. How is the degrees of freedom for a chi-square test determined?
  3. What is the critical value of a chi-square test with 3 degrees of freedom at a 5% significance level?
  4. How do we interpret the p-value in a chi-square test for independence?
  5. How can you check the assumptions of a chi-square test?

Tip: Always check that expected frequencies are sufficiently large (typically at least 5) when performing a chi-square test, to ensure the validity of the test.

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Math Problem Analysis

Mathematical Concepts

Statistics
Chi-square Test
Contingency Tables

Formulas

Expected frequency: E_{ij} = (row sum * column sum) / total sum
Chi-square statistic: χ^2 = Σ ((O_{ij} - E_{ij})^2 / E_{ij})

Theorems

Chi-square Test for Independence

Suitable Grade Level

Grades 11-12