A chi-square test of independence was performed to determine whether the pass/fail rates are dependent on school. The p-value was 0.9373, so we cannot conclude that the pass/fail rates are dependent on school correct option is c.
The chi-square statistic is calculated as follows:
χ² = Σ (O - E)² / E
where O is the observed number of students in each category, and E is the expected number of students in each category. The expected number of students in each category is calculated under the assumption of independence.
The degrees of freedom for the chi-square test of independence are (r - 1)(c - 1), where r is the number of rows in the contingency table and c is the number of columns. In this case, r = 2 and c = 3, so the degrees of freedom are (2 - 1)(3 - 1) = 2.
The p-value for the chi-square test is the probability of obtaining a chi-square statistic as large as or larger than the observed chi-square statistic, assuming that the null hypothesis is true. The p-value can be calculated using a chi-square distribution table or software.
The following table shows the expected number of students in each category, under the assumption of independence:
| School | Pass | Fail |
| A | 41.42 | 57.58 |
| B | 32.63 | 45.37 |
| C | 48.95 | 68.05 |
The chi-square statistic is calculated as follows:
χ² = (40 - 41.42)² / 41.42 + (33 - 32.63)² / 32.63 + (50 - 48.95)² / 48.95 + (59 - 57.58)² / 57.58 + (45 - 45.37)² / 45.37 + (67 - 68.05)² / 68.05 = 0.1294
The p-value for the chi-square test is 0.9373.
Since the p-value is greater than 0.10, we do not reject the null hypothesis. Therefore, we cannot conclude that the pass/fail rates are dependent on school.
The correct answer is C. No, it cannot be concluded that pass/fail rates are dependent on school because the p-value = 0.9373.