Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

Plamen is a social media manager for a large company. He takes a random sample of their posts to see if there is a relationship between the time of each post and the number of times the post gets shared. Below are the outcomes and partial results of a chi-square test (expected counts appear below observed counts):

Chi-square test: Time of post vs. number of shares
[tex]\[
\begin{array}{lrrr}
& 0-50 & 51-100 & 100+ & \text{Total} \\
\text{Morning} & 203 & 77 & 50 & 330 \\
& (219) & (72) & (39) & \\
\text{Afternoon} & 117 & 36 & 12 & 165 \\
& (109.5) & (36) & (19.5) & \\
\text{Evening} & 45 & 7 & 3 & 55 \\
& (36.5) & (12) & (6.5) & \\
\text{Total} & 365 & 120 & 65 & 550 \\
\end{array}
\][/tex]

They want to use these results to carry out a chi-square test of independence. Assume that all conditions for inference were met.

What are the values of the test statistic and P-value for their test?

Choose one answer:
A. [tex]\( \chi^2 = 13.965 \)[/tex], [tex]\(0.005 \ \textless \ \text{P-value} \ \textless \ 0.01\)[/tex]
B. [tex]\( \chi^2 = 18.435 \)[/tex], [tex]\(0.0005 \ \textless \ \text{P-value} \ \textless \ 0.001\)[/tex]
C. [tex]\( \chi^2 = 13.965 \)[/tex], [tex]\( \text{P-value} \ \textgreater \ 0.25 \)[/tex]
D. [tex]\( \chi^2 = 18.435 \)[/tex], [tex]\(0.005 \ \textless \ \text{P-value} \ \textless \ 0.01\)[/tex]


Sagot :

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]