Answer :
To determine if there is a relationship between age and type of movie preference using the Chi-Square independence test at the 5% significance level, we can follow these steps:
1. Formulate the Hypotheses:
- [tex]\(H_0\)[/tex]: Age and type of movie preference are independent.
- [tex]\(H_a\)[/tex]: Age and type of movie preference are not independent.
2. Construct the Contingency Table with Observed Frequencies:
The given data is compiled in the following table:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline & 18-23\ \text{years\ old} & 24-29\ \text{years\ old} & 30-35\ \text{years\ old} & \text{Totals} \\ \hline \text{Drama} & 8 & 15 & 11 & 34 \\ \hline \text{Science\ Fiction} & 12 & 10 & 8 & 30 \\ \hline \text{Totals} & 20 & 25 & 19 & 64 \\ \hline \end{array} \][/tex]
3. Calculate the Degrees of Freedom (dof):
The degrees of freedom for a contingency table is calculated as:
[tex]\[ \text{dof} = (\text{number of rows} - 1) \times (\text{number of columns} - 1) \][/tex]
Here we have 2 rows and 3 columns:
[tex]\[ \text{dof} = (2-1) \times (3-1) = 1 \times 2 = 2 \][/tex]
4. Perform the Chi-Square Test:
From this analysis, we have:
- Chi-Square statistic ([tex]\(\chi^2\)[/tex]): 2.0316
- p-value: 0.3621
5. Decision Rule:
If the p-value is less than the significance level ([tex]\(\alpha = 0.05\)[/tex]), we reject the null hypothesis.
6. Make the Conclusion:
In this case, the p-value [tex]\(0.3621\)[/tex] is greater than the significance level [tex]\(0.05\)[/tex].
Since the p-value is greater than 0.05, we fail to reject the null hypothesis. This means there is not enough evidence to conclude that age and type of movie preference are not independent at the 5% significance level. Therefore, age and type of movie preference appear to be independent based on the given sample.
1. Formulate the Hypotheses:
- [tex]\(H_0\)[/tex]: Age and type of movie preference are independent.
- [tex]\(H_a\)[/tex]: Age and type of movie preference are not independent.
2. Construct the Contingency Table with Observed Frequencies:
The given data is compiled in the following table:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline & 18-23\ \text{years\ old} & 24-29\ \text{years\ old} & 30-35\ \text{years\ old} & \text{Totals} \\ \hline \text{Drama} & 8 & 15 & 11 & 34 \\ \hline \text{Science\ Fiction} & 12 & 10 & 8 & 30 \\ \hline \text{Totals} & 20 & 25 & 19 & 64 \\ \hline \end{array} \][/tex]
3. Calculate the Degrees of Freedom (dof):
The degrees of freedom for a contingency table is calculated as:
[tex]\[ \text{dof} = (\text{number of rows} - 1) \times (\text{number of columns} - 1) \][/tex]
Here we have 2 rows and 3 columns:
[tex]\[ \text{dof} = (2-1) \times (3-1) = 1 \times 2 = 2 \][/tex]
4. Perform the Chi-Square Test:
From this analysis, we have:
- Chi-Square statistic ([tex]\(\chi^2\)[/tex]): 2.0316
- p-value: 0.3621
5. Decision Rule:
If the p-value is less than the significance level ([tex]\(\alpha = 0.05\)[/tex]), we reject the null hypothesis.
6. Make the Conclusion:
In this case, the p-value [tex]\(0.3621\)[/tex] is greater than the significance level [tex]\(0.05\)[/tex].
Since the p-value is greater than 0.05, we fail to reject the null hypothesis. This means there is not enough evidence to conclude that age and type of movie preference are not independent at the 5% significance level. Therefore, age and type of movie preference appear to be independent based on the given sample.