Question 1200633
**a. State the null and alternative hypotheses.**

* **Null Hypothesis (H0):** There is no association between the main source of news and the educational level of individuals. 
* **Alternative Hypothesis (Ha):** There is an association between the main source of news and the educational level of individuals.

**b. Decision Rule (α = 0.05)**

* We will use a Chi-Square test for independence. 
* Find the critical value of Chi-Square (χ²) from the Chi-Square distribution table with degrees of freedom: (rows - 1) * (columns - 1) = (4 - 1) * (4 - 1) = 9. 
* The critical value for χ² with 9 degrees of freedom at α = 0.05 is 16.919.
* **Decision Rule:** Reject the null hypothesis (H0) if the calculated Chi-Square test statistic is greater than 16.919.

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

1. **Calculate Expected Frequencies:**

   * Expected Frequency for each cell = (Row Total * Column Total) / Grand Total

   | Source of News | Primary | Secondary | Graduate | Post Graduate | Row Total |
   |---|---|---|---|---|---|
   | Newspapers | (816 * 636) / 2500 = 207.17 | (816 * 561) / 2500 = 182.81 | (816 * 521) / 2500 = 169.73 | (816 * 782) / 2500 = 253.30 | 816 |
   | Television | (609 * 636) / 2500 = 154.60 | (609 * 561) / 2500 = 136.29 | (609 * 521) / 2500 = 127.03 | (609 * 782) / 2500 = 191.08 | 609 |
   | Social Media | (570 * 636) / 2500 = 144.72 | (570 * 561) / 2500 = 127.51 | (570 * 521) / 2500 = 119.03 | (570 * 782) / 2500 = 178.74 | 570 |
   | Grapevine | (505 * 636) / 2500 = 128.51 | (505 * 561) / 2500 = 113.39 | (505 * 521) / 2500 = 105.21 | (505 * 782) / 2500 = 157.90 | 505 |
   | Column Total | 636 | 561 | 521 | 782 | 2500 |

2. **Calculate the Chi-Square Statistic:**

   * χ² = Σ [(Observed Frequency - Expected Frequency)² / Expected Frequency]

   * Calculate this for each cell in the table and sum them up.

3. **Find the Degrees of Freedom:**

   * Degrees of Freedom (df) = (Rows - 1) * (Columns - 1) = 3 * 3 = 9

**d. Conclusion**

* Compare the calculated Chi-Square statistic to the critical value (16.919).
    * If the calculated Chi-Square statistic is greater than 16.919, reject the null hypothesis. 
    * If the calculated Chi-Square statistic is less than or equal to 16.919, fail to reject the null hypothesis.

**e. Conclusion at the 10% level of significance:**

* Find the critical value of Chi-Square for α = 0.10 with 9 degrees of freedom (from the Chi-Square distribution table).
* Compare the calculated Chi-Square statistic to this new critical value.
* If the calculated Chi-Square statistic is greater than the new critical value, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.

**f. Confirm Test Results in JASP**

1. **Input Data:** 
   * Enter the observed frequencies into JASP.
   * Specify the row and column variables.

2. **Run Chi-Square Test:**
   * Go to "Frequencies" -> "Contingency Tables."
   * Select the appropriate variables.
   * Choose "Chi-Square" under "Statistics."
   * Click "Run Analysis."

3. **Interpret JASP Output:**
   * JASP will provide the calculated Chi-Square statistic, degrees of freedom, and the p-value.
   * Compare the p-value to the significance level (0.05 and 0.10).
   * If the p-value is less than the significance level, reject the null hypothesis.

**JASP Input and Output:**

* **Input:** 
    * You would input the observed frequencies from the table into JASP. 
* **Output:**
    * JASP will provide a table with the following information:
        * Chi-Square statistic
        * Degrees of freedom
        * p-value 
        * Expected frequencies (for comparison)

**Note:**

* This analysis assumes that the data meets the assumptions of the Chi-Square test (e.g., expected frequencies in each cell are sufficiently large). 

I hope this comprehensive explanation is helpful!