### Hypothesis Testing: Chi-Square Test of Independence**Problem Statement**A Superintendent of Education seeks to determine if test scores are independent of school location. The table below presents the number of students achieving a basic level in three subjects (Reading, Math, and Science). The claim is tested at a significance level of α = 0.01 to evaluate if location of school and academic achievement are independent.**Data Table**| Location of School | Subject | ||--------------------|----------|------------------------|| | Reading | Math | Science || Urban | 43 | 42 | 38 || Suburban | 63 | 66 | 65 |**Hypothesis Testing Steps****A. State Both Hypotheses**- **Null Hypothesis ($H_0$)**: The location of the school (Urban or Suburban) and student performance in subjects (Reading, Math, Science) are independent.- **Alternative Hypothesis ($H_1$)**: The location of the school (Urban or Suburban) and student performance in subjects (Reading, Math, Science) are not independent.**B. Test Statistic (TV) and Critical Value (CV) or P-Value (PV) and Alpha ($\alpha$)**1. **Calculating the Expected Frequencies:** Expected Frequency = $\frac{\text{(Row Total) \(\times$ (Column Total)}}{\text{Grand Total}}\)2. **Calculate the Chi-Square Test Statistic ($\chi^2$):** $\chi^2 = \sum \frac{(O - E)^2}{E}$ where $O$ is the observed frequency, and $E$ is the expected frequency.3. **Degrees of Freedom (df):** $df = (r - 1) \times (c - 1)$ where $r$ is the number of rows and $c$ is the number of columns.4. **Find the Critical Value ($CV$) for $\alpha = 0.01$ from the Chi-Square distribution table.5. **Compare the Test Statistic ($\chi^2$) with the Critical Value ($CV$), or use the P-Value approach.**C. Decision, Summary, and Other Requests

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A Superintendent of Education wants to determine if test scores are independent of school location. The following table shows the number of students achieving a basic level in the following subjects. Test the claim at a = 0.01 that location of school and academic achievement are independent. Location of School Subject Reading Math Science Urban 43 42 38 Suburban 63 66 65

A Superintendent of Education wants to determine if test scores are independent of school location. The following table shows the number of students achieving a basic level in the following subjects. Test the claim at a = 0.01 that location of school and academic achievement are independent. Location of School Subject Reading Math Science Urban 43 42 38 Suburban 63 66 65

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### Hypothesis Testing: Chi-Square Test of Independence

**Problem Statement**
A Superintendent of Education seeks to determine if test scores are independent of school location. The table below presents the number of students achieving a basic level in three subjects (Reading, Math, and Science). The claim is tested at a significance level of α = 0.01 to evaluate if location of school and academic achievement are independent.

**Data Table**

| Location of School | Subject | |
|--------------------|----------|------------------------|
| | Reading | Math | Science |
| Urban | 43 | 42 | 38 |
| Suburban | 63 | 66 | 65 |

**Hypothesis Testing Steps**

**A. State Both Hypotheses**

- **Null Hypothesis ($H_0$)**: The location of the school (Urban or Suburban) and student performance in subjects (Reading, Math, Science) are independent.
- **Alternative Hypothesis ($H_1$)**: The location of the school (Urban or Suburban) and student performance in subjects (Reading, Math, Science) are not independent.

**B. Test Statistic (TV) and Critical Value (CV) or P-Value (PV) and Alpha ($\alpha$)**

1. **Calculating the Expected Frequencies:**
Expected Frequency = $\frac{\text{(Row Total) \(\times$ (Column Total)}}{\text{Grand Total}}\)

2. **Calculate the Chi-Square Test Statistic ($\chi^2$):**
$\chi^2 = \sum \frac{(O - E)^2}{E}$
where $O$ is the observed frequency, and $E$ is the expected frequency.

3. **Degrees of Freedom (df):**
$df = (r - 1) \times (c - 1)$
where $r$ is the number of rows and $c$ is the number of columns.

4. **Find the Critical Value ($CV$) for $\alpha = 0.01$ from the Chi-Square distribution table.

5. **Compare the Test Statistic ($\chi^2$) with the Critical Value ($CV$), or use the P-Value approach.

**C. Decision, Summary, and Other Requests

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