Questions: Is there an association between a new school policy and the position a person holds within the school? A single stratified random sample of students, teachers, and administrators (principals and team leaders) from the school population was The p-value for a chi-square test for independence on this association is = ?

Solution Steps

The given data can be represented in a contingency table:

| Position | Favor | Not Favor | Total | |---|---|---|---| | Student | 16 | 29 | 45 | | Teacher | 31 | 13 | 44 | | Administrator | 6 | 5 | 11 | | Total | 53 | 47 | 100 |

We calculate the expected values for each cell under the assumption of independence. The expected value for a cell is calculated as (row total * column total) / grand total.

| Position | Favor | Not Favor | Total | |---|---|---|---| | Student | (45_53)/100 = 23.85 | (45_47)/100 = 21.15 | 45 | | Teacher | (44_53)/100 = 23.32 | (44_47)/100 = 20.68 | 44 | | Administrator | (11_53)/100 = 5.83 | (11_47)/100 = 5.17 | 11 | | Total | 53 | 47 | 100 |

The chi-square statistic is calculated as the sum of (observed - expected)^2 / expected for each cell.

χ² = (16-23.85)²/23.85 + (29-21.15)²/21.15 + (31-23.32)²/23.32 + (13-20.68)²/20.68 + (6-5.83)²/5.83 + (5-5.17)²/5.17 χ² = 2.59 + 2.91 + 2.52 + 2.86 + 0.005 + 0.006 χ² ≈ 10.89

The degrees of freedom for a chi-square test of independence is calculated as (number of rows - 1) * (number of columns - 1). In this case, df = (3-1)*(2-1) = 2. Using a chi-square table or calculator with χ² = 10.89 and df = 2, we find that the p-value is between 0.001 and 0.01.

Specifically, using a chi-squared calculator with χ²=10.89 and df=2, you get p = 0.0042.

Final Answer