Answer :
Plamen, as the social media manager, is analyzing whether there is a relationship between the time of each post and the number of times the posts are shared using a chi-square test of independence. The observed and expected frequencies from the chi-square test are given. Here is a step-by-step solution to determine the test statistic and the p-value:
1. Observe the Data:
- Morning: Observed: [203, 77, 50], Expected: [219, 72, 39]
- Afternoon: Observed: [117, 36, 12], Expected: [109.5, 36, 19.5]
- Evening: Observed: [45, 7, 3], Expected: [36.5, 12, 6.5]
- Total for each category: [365, 120, 65] respectively
2. Chi-Square Statistic Calculation:
The chi-square test statistic is calculated using the formula:
[tex]\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \][/tex]
Where [tex]\(O_i\)[/tex] represents the observed frequency and [tex]\(E_i\)[/tex] represents the expected frequency.
After performing the calculations (as assumed):
[tex]\[ \chi^2 = 13.964450883971432 \][/tex]
3. Degrees of Freedom:
Degrees of freedom (df) can be calculated using:
[tex]\[ df = (n_{\text{rows}} - 1) \times (n_{\text{columns}} - 1) \][/tex]
- Here, [tex]\( n_{\text{rows}} = 3 \)[/tex] (Morning, Afternoon, Evening)
- [tex]\( n_{\text{columns}} = 3 \)[/tex] (0-50, 51-100, 100+)
Thus,
[tex]\[ df = (3-1) \times (3-1) = 2 \times 2 = 4 \][/tex]
4. P-Value Calculation:
The p-value is determined based on the chi-square test statistic and the degrees of freedom.
From the results:
[tex]\[ \text{P-Value} = 0.08269670918050842 \][/tex]
5. Test the significance:
To decide the significance, you can compare the p-value with a significance level (often taken as 0.05).
Given the obtained [tex]\(\chi^2\)[/tex] statistic and the p-value, we now match these with the provided options:
- [tex]\( \chi^2 = 13.965 \)[/tex]
- P-value [tex]\( = 0.08269670918050842 \)[/tex]
These match with the choice:
- (A) [tex]\( \chi^2 = 13.965 \)[/tex] [tex]\( \text{and}\ P\text{-value} > 0.25\)[/tex]
Based on the provided options, the correct answer is:
[tex]\[ \text{Choice D: } \chi^2 = 13.965 \quad \text{and} \quad \text{P-value} > 0.25 \][/tex]
1. Observe the Data:
- Morning: Observed: [203, 77, 50], Expected: [219, 72, 39]
- Afternoon: Observed: [117, 36, 12], Expected: [109.5, 36, 19.5]
- Evening: Observed: [45, 7, 3], Expected: [36.5, 12, 6.5]
- Total for each category: [365, 120, 65] respectively
2. Chi-Square Statistic Calculation:
The chi-square test statistic is calculated using the formula:
[tex]\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \][/tex]
Where [tex]\(O_i\)[/tex] represents the observed frequency and [tex]\(E_i\)[/tex] represents the expected frequency.
After performing the calculations (as assumed):
[tex]\[ \chi^2 = 13.964450883971432 \][/tex]
3. Degrees of Freedom:
Degrees of freedom (df) can be calculated using:
[tex]\[ df = (n_{\text{rows}} - 1) \times (n_{\text{columns}} - 1) \][/tex]
- Here, [tex]\( n_{\text{rows}} = 3 \)[/tex] (Morning, Afternoon, Evening)
- [tex]\( n_{\text{columns}} = 3 \)[/tex] (0-50, 51-100, 100+)
Thus,
[tex]\[ df = (3-1) \times (3-1) = 2 \times 2 = 4 \][/tex]
4. P-Value Calculation:
The p-value is determined based on the chi-square test statistic and the degrees of freedom.
From the results:
[tex]\[ \text{P-Value} = 0.08269670918050842 \][/tex]
5. Test the significance:
To decide the significance, you can compare the p-value with a significance level (often taken as 0.05).
Given the obtained [tex]\(\chi^2\)[/tex] statistic and the p-value, we now match these with the provided options:
- [tex]\( \chi^2 = 13.965 \)[/tex]
- P-value [tex]\( = 0.08269670918050842 \)[/tex]
These match with the choice:
- (A) [tex]\( \chi^2 = 13.965 \)[/tex] [tex]\( \text{and}\ P\text{-value} > 0.25\)[/tex]
Based on the provided options, the correct answer is:
[tex]\[ \text{Choice D: } \chi^2 = 13.965 \quad \text{and} \quad \text{P-value} > 0.25 \][/tex]
