**Research question 2:** we are interested in studying the association between the age of students and the way they watched the lecture (livestream or in class) in the population under consideration.
QUESTIONS:

Report the number of the degrees of freedom [integer]:

Answer 5 Question 1

Report the value of the test statistic [use at least 2dp]:

Answer 6 Question 1

Report the 𝑝𝑝-value [use at least 3dp]:

Answer 7 Question 1

What is the decision of the test?

To address this question, it sounds like you're analyzing a categorical association between student age groups and lecture-watching mode (livestream vs. in-person). This setup suggests using a **chi-square test of independence**, which evaluates if there's a significant association between two categorical variables. Here's a general breakdown for each part of the question:

### Step-by-Step Explanation:

1. **Degrees of Freedom**: For a chi-square test of independence, the degrees of freedom (df) are calculated using:
   \[
   \text{df} = (r - 1)(c - 1)
   \]
   where \( r \) is the number of age groups (rows) and \( c \) is the number of lecture-watching methods (columns). If we assume two age groups and two modes (livestream and in-person), this would yield:
   \[
   \text{df} = (2 - 1)(2 - 1) = 1
   \]
   However, substitute in the actual number of groups to confirm. If it’s 2x2, the answer will be **df = 1**.

2. **Test Statistic**: The chi-square test statistic is calculated based on the observed and expected counts. The formula is:
   \[
   \chi^2 = \sum \frac{(O - E)^2}{E}
   \]
   where \( O \) is the observed frequency, and \( E \) is the expected frequency. Using the observed counts and calculating the expected counts based on marginal totals can yield the chi-square statistic value to two decimal places.

3. **p-value**: The p-value is obtained from the chi-square distribution with the calculated degrees of freedom. After finding the chi-square test statistic, consult a chi-square table or calculator to obtain the p-value to three decimal places.

4. **Decision of the Test**: If the p-value is less than the significance level (commonly 0.05), reject the null hypothesis, concluding a statistically significant association between age and lecture mode. Otherwise, retain the null hypothesis if \( p > 0.05 \).

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Please share any specific values for observed frequencies or further information if available, so I can help calculate exact answers for the test statistic, p-value, and decision. 

Would you like me to go through this with specific values, or do you need further clarification on any steps?

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### Relative Questions
1. What are the assumptions required for a chi-square test of independence?
2. How would results differ if you used an ANOVA or t-test instead of a chi-square test for this setup?
3. What is the impact of sample size on the reliability of the chi-square test?
4. Can the chi-square test determine causation between age and lecture-watching mode?
5. How could a post-hoc test be used if multiple age groups were compared?

### Tip
When conducting a chi-square test, ensure all expected counts are at least 5 to satisfy test validity requirements.

We are interested in studying the association between the age of students and the way they watched the lecture (livestream or in class) in the population under consideration. Report the number of the degrees of freedom, the test statistic value, the p-value, and the decision of the test.

Learn how to analyze the association between student age and lecture mode (livestream or in-class) using a Chi-Square test of independence. This includes calculating degrees of freedom, test statistics, and p-values for a statistical decision. Suitable for students in Grades 10-12.

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