Question 1200633: Gaynor Advertising conducted a survey to determine individual’s main source of news as well as their educational level. 2500 individuals were surveyed and the results are given below:
Source of News Primary Secondary Graduate Post Graduate Total
Newspapers 95 200 121 400 816
Television 136 127 135 211 609
Social Media 155 99 170 146 570
Grapevine 250 155 95 25 505
Total 636 561 521 782 2500
a. State both the null and alternative hypotheses. [2 marks]
b. Provide the decision rule for making this decision. Use an alpha level of 5%. [2 marks]
c. Show all of the work necessary to calculate the appropriate statistic. [5 marks]
d. What conclusion are you allowed to draw? [2 marks]
e. Would your conclusion change at the 10% level of significance? [2 marks]
f. Confirm test results in part (c) using JASP. Note: All JASP input files and output tables should be provided [3 marks]
Answer by GingerAle(43)   (Show Source):
You can put this solution on YOUR website! **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!
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