Question 1193979
**1. Set up Hypotheses**

* **Null Hypothesis (H0):** There is no association between academic performance and self-esteem in the population of ten-year-old children. 
* **Alternative Hypothesis (H1):** There is an association between academic performance and self-esteem in the population of ten-year-old children.

**2. Calculate Expected Frequencies**

* **Create a table of expected frequencies:**
    | Level of Academic Performance | High Self-Esteem | Medium Self-Esteem | Low Self-Esteem | Row Total |
    |---|---|---|---|---|
    | High | (60*60)/150 = 24 | (60*75)/150 = 30 | (60*15)/150 = 18 | 60 |
    | Low | (90*60)/150 = 36 | (90*75)/150 = 45 | (90*15)/150 = 27 | 90 |
    | Column Total | 60 | 75 | 15 | 150 |

* **Where:**
    * Row Total: Total number of children in each academic performance level.
    * Column Total: Total number of children in each self-esteem level.
    * Grand Total: Total number of children in the sample (150).

**3. Calculate the Chi-Square Test Statistic**

* **Formula:** χ² = Σ [(O - E)² / E] 
    * Where:
        * O = Observed frequency in each cell
        * E = Expected frequency in each cell

* **Calculate for each cell:**
    * (17-24)²/24 = 2.04
    * (32-30)²/30 = 0.13
    * (11-18)²/18 = 2.72
    * (13-36)²/36 = 12.5
    * (43-45)²/45 = 0.09
    * (34-27)²/27 = 1.70

* **Sum the values:** χ² = 2.04 + 0.13 + 2.72 + 12.5 + 0.09 + 1.70 = 19.18

**4. Determine the Degrees of Freedom**

* Degrees of Freedom (df) = (Number of rows - 1) * (Number of columns - 1) = (2 - 1) * (3 - 1) = 2

**5. Find the Critical Value**

* Using a chi-square distribution table, find the critical value for α = 0.05 and df = 2. 
* The critical value is approximately 5.991.

**6. Make a Decision**

* **Compare the calculated chi-square statistic to the critical value:**
    * 19.18 > 5.991

* **Decision:** Since the calculated chi-square statistic (19.18) is greater than the critical value (5.991), we reject the null hypothesis.

**7. Conclusion**

* There is sufficient evidence at the 0.05 level of significance to conclude that there is an association between academic performance and self-esteem in the population of ten-year-old children.

**Note:**

* This analysis assumes that the expected frequencies in each cell are greater than 5. 
* You can use statistical software (like R or SPSS) to perform the chi-square test for independence more easily and accurately.

This analysis provides a basic framework for conducting a chi-square test for independence.