SOLUTION: A researcher is investigating the relationship between academic performance and self-esteem. A sample of 150 ten-year old children is obtained, and each child is classified by lev

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Question 1193979: A researcher is investigating the relationship between academic performance and self-esteem. A sample of 150 ten-year old
children is obtained, and each child is classified by level of academic performance and level of self-esteem. Test at 0.05 level of
significance.
Level of acad
performance/ level
of self esteem. High. Medium. Low
High 17. 32. 11
Low. 13. 43. 34
Answer by parmen(42) (Show Source):
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**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.