We want to conduct the hypothesis test Ho: Attachment pattern and child care time are independent, versus H: Attachment pattern and child care time are dependent. Recall that the test statistic X2 for the chi-square test of independence is given by the following. _Σ(0,-Ê,² ij' The observed cell counts, O., are given in row i and column j of the given contingency table. The estimated expected cell counts, Ê, can be calculated based on the data assuming that the rows and columns are independent, which is claimed by the null hypothesis. If Ho is assumed to be true, then the estimate of the expected cell count in row i and column j is given by the following. Secure X² Anxious Total 65 The total number of observations is n, r, is the total for row i, and c, is the total for column j. First determine the row and column totals, r, and c,, for the given contingency table, as well as the total of all observations, n. Low (0-3 hours) Moderate (4-19 hours) High (20-54 hours) = rcj n 23 9 C₁ = 23 +9 C₂ = 34 8 4 6 Total ₁23 +34 + 4 = 2 = n = ₁ + ₂ = C₁ + ₂ + C3
We want to conduct the hypothesis test Ho: Attachment pattern and child care time are independent, versus H: Attachment pattern and child care time are dependent. Recall that the test statistic X2 for the chi-square test of independence is given by the following. _Σ(0,-Ê,² ij' The observed cell counts, O., are given in row i and column j of the given contingency table. The estimated expected cell counts, Ê, can be calculated based on the data assuming that the rows and columns are independent, which is claimed by the null hypothesis. If Ho is assumed to be true, then the estimate of the expected cell count in row i and column j is given by the following. Secure X² Anxious Total 65 The total number of observations is n, r, is the total for row i, and c, is the total for column j. First determine the row and column totals, r, and c,, for the given contingency table, as well as the total of all observations, n. Low (0-3 hours) Moderate (4-19 hours) High (20-54 hours) = rcj n 23 9 C₁ = 23 +9 C₂ = 34 8 4 6 Total ₁23 +34 + 4 = 2 = n = ₁ + ₂ = C₁ + ₂ + C3
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Author:Amos Gilat
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![To conduct a hypothesis test where \( H_0 \) suggests the attachment pattern and childcare time are independent (versus \( H_a \), where they are dependent), we use the chi-square test of independence.
### Chi-Square Test Statistic Formula
The test statistic \( \chi^2 \) is calculated as:
\[
\chi^2 = \sum \frac{(O_{ij} - \hat{E}_{ij})^2}{\hat{E}_{ij}}
\]
### Explanation of Terms
- \( O_{ij} \): Observed cell count in row \( i \) and column \( j \).
- \( \hat{E}_{ij} \): Estimated expected cell count, calculated as:
\[
\hat{E}_{ij} = \frac{r_i c_j}{n}
\]
Where:
- \( n \) is the total number of observations.
- \( r_i \) is the total for row \( i \).
- \( c_j \) is the total for column \( j \).
### Contingency Table
The table displays the observed frequencies:
| | Low (0-3 hours) | Moderate (4-19 hours) | High (20-54 hours) | Total |
|----------------------|-----------------|-----------------------|--------------------|---------|
| **Secure** | 23 | 34 | 4 | \( r_1 = 23 + 34 + 4 = \) |
| **Anxious** | 9 | 8 | 6 | \( r_2 = \) |
| **Total** | \( c_1 = 23 + 9 \) | \( c_2 = \) | \( c_3 = \) | \( n = r_1 + r_2 = \) |
- **Secure**: The counts for low, moderate, and high childcare hours are 23, 34, and 4, respectively.
- **Anxious**: The counts for low, moderate, and high childcare hours are 9, 8, and 6, respectively.
To find the totals:
- Calculate the row totals (\( r_1, r_2 \)).
- Calculate the column totals (\( c_1, c_2, c_3 \)).
- Calculate the total number of observations (\( n \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93e02f2d-b0b4-40f2-b59c-ee2919cc5490%2Fc3432c89-83a9-48ea-a43b-a6d6bab8cd87%2Fl9hdxc6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:To conduct a hypothesis test where \( H_0 \) suggests the attachment pattern and childcare time are independent (versus \( H_a \), where they are dependent), we use the chi-square test of independence.
### Chi-Square Test Statistic Formula
The test statistic \( \chi^2 \) is calculated as:
\[
\chi^2 = \sum \frac{(O_{ij} - \hat{E}_{ij})^2}{\hat{E}_{ij}}
\]
### Explanation of Terms
- \( O_{ij} \): Observed cell count in row \( i \) and column \( j \).
- \( \hat{E}_{ij} \): Estimated expected cell count, calculated as:
\[
\hat{E}_{ij} = \frac{r_i c_j}{n}
\]
Where:
- \( n \) is the total number of observations.
- \( r_i \) is the total for row \( i \).
- \( c_j \) is the total for column \( j \).
### Contingency Table
The table displays the observed frequencies:
| | Low (0-3 hours) | Moderate (4-19 hours) | High (20-54 hours) | Total |
|----------------------|-----------------|-----------------------|--------------------|---------|
| **Secure** | 23 | 34 | 4 | \( r_1 = 23 + 34 + 4 = \) |
| **Anxious** | 9 | 8 | 6 | \( r_2 = \) |
| **Total** | \( c_1 = 23 + 9 \) | \( c_2 = \) | \( c_3 = \) | \( n = r_1 + r_2 = \) |
- **Secure**: The counts for low, moderate, and high childcare hours are 23, 34, and 4, respectively.
- **Anxious**: The counts for low, moderate, and high childcare hours are 9, 8, and 6, respectively.
To find the totals:
- Calculate the row totals (\( r_1, r_2 \)).
- Calculate the column totals (\( c_1, c_2, c_3 \)).
- Calculate the total number of observations (\( n \
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