Question
The following table shows the results from a survey to compare the principal water source in three villages in North Africa. The numbers of households using a river, a pond, or a spring are given.
Village
Water Source Avillage Bevillage Cevillage
River 30 32 18
Pond 18 20 20
Spring 12 18 10
Test the hypothesis that there is no association between the villages and the number of households using different water sources.
Village
Water Source Avillage Bevillage Cevillage
River 30 32 18
Pond 18 20 20
Spring 12 18 10
Test the hypothesis that there is no association between the villages and the number of households using different water sources.
Answers
GPT-4o mini
To test the hypothesis that there is no association between the villages and the number of households using different water sources, you can perform a Chi-Square test of independence. Here’s how to do that step by step:
### Step 1: Set Up the Data
First, construct a contingency table based on the data provided.
| Water Source | A Village | B Village | C Village | Row Total |
|--------------|-----------|-----------|-----------|-----------|
| River | 30 | 32 | 18 | 80 |
| Pond | 18 | 20 | 20 | 58 |
| Spring | 12 | 18 | 10 | 40 |
| **Column Total** | **60** | **70** | **48** | **178** |
### Step 2: Define Hypotheses
- **Null Hypothesis (H0)**: There is no association between the village and the water source.
- **Alternative Hypothesis (H1)**: There is an association between the village and the water source.
### Step 3: Calculate Expected Frequencies
The expected frequency for each cell can be calculated using the formula:
\[
E = \frac{(Row Total) \times (Column Total)}{Grand Total}
\]
Here are the expected frequencies for each cell:
- For River in A Village:
\[
E = \frac{80 \times 60}{178} \approx 26.96
\]
- For River in B Village:
\[
E = \frac{80 \times 70}{178} \approx 39.32
\]
- For River in C Village:
\[
E = \frac{80 \times 48}{178} \approx 13.72
\]
- For Pond in A Village:
\[
E = \frac{58 \times 60}{178} \approx 19.53
\]
- For Pond in B Village:
\[
E = \frac{58 \times 70}{178} \approx 22.96
\]
- For Pond in C Village:
\[
E = \frac{58 \times 48}{178} \approx 15.51
\]
- For Spring in A Village:
\[
E = \frac{40 \times 60}{178} \approx 13.48
\]
- For Spring in B Village:
\[
E = \frac{40 \times 70}{178} \approx 15.73
\]
- For Spring in C Village:
\[
E = \frac{40 \times 48}{178} \approx 10.79
\]
### Step 4: Calculate Chi-Square Statistic
The Chi-Square statistic can be calculated using:
\[
\chi^2 = \sum \frac{(O - E)^2}{E}
\]
Where \( O \) is the observed frequency, and \( E \) is the expected frequency. Here’s the calculation for each cell:
- For River, A Village:
\[
\chi^2 = \frac{(30 - 26.96)^2}{26.96} \approx 0.0371
\]
- For River, B Village:
\[
\chi^2 = \frac{(32 - 39.32)^2}{39.32} \approx 1.1485
\]
- For River, C Village:
\[
\chi^2 = \frac{(18 - 13.72)^2}{13.72} \approx 0.8601
\]
- For Pond, A Village:
\[
\chi^2 = \frac{(18 - 19.53)^2}{19.53} \approx 0.1183
\]
- For Pond, B Village:
\[
\chi^2 = \frac{(20 - 22.96)^2}{22.96} \approx 0.3884
\]
- For Pond, C Village:
\[
\chi^2 = \frac{(20 - 15.51)^2}{15.51} \approx 0.8454
\]
- For Spring, A Village:
\[
\chi^2 = \frac{(12 - 13.48)^2}{13.48} \approx 0.1127
\]
- For Spring, B Village:
\[
\chi^2 = \frac{(18 - 15.73)^2}{15.73} \approx 0.4201
\]
- For Spring, C Village:
\[
\chi^2 = \frac{(10 - 10.79)^2}{10.79} \approx 0.0615
\]
Now, sum these contributions:
\[
\chi^2 \approx 0.0371 + 1.1485 + 0.8601 + 0.1183 + 0.3884 + 0.8454 + 0.1127 + 0.4201 + 0.0615 \approx 3.9371
\]
### Step 5: Determine Degrees of Freedom
The degrees of freedom for a Chi-Square test of independence is calculated as:
\[
df = (r - 1)(c - 1)
\]
Where \( r \) is the number of rows and \( c \) is the number of columns. In this case, \( r = 3 \) (water sources: river, pond, spring) and \( c = 3 \) (villages: A, B, C):
\[
df = (3 - 1)(3 - 1) = 2 \cdot 2 = 4
\]
### Step 6: Determine the Critical Value
Using a Chi-Square distribution table at a significance level (\( \alpha \)) of 0.05 with 4 degrees of freedom, the critical value is approximately \( 9.488 \).
### Step 7: Compare and Make a Decision
- Calculate \( \chi^2 \) value obtained: \( \chi^2 = 3.9371 \).
- Critical value at \( df = 4 \) and \( \alpha = 0.05 \): \( 9.488 \).
Since \( 3.9371 < 9.488 \), we fail to reject the null hypothesis.
### Conclusion
There is not enough evidence to conclude that there is an association between the villages and the number of households using different water sources.
### Step 1: Set Up the Data
First, construct a contingency table based on the data provided.
| Water Source | A Village | B Village | C Village | Row Total |
|--------------|-----------|-----------|-----------|-----------|
| River | 30 | 32 | 18 | 80 |
| Pond | 18 | 20 | 20 | 58 |
| Spring | 12 | 18 | 10 | 40 |
| **Column Total** | **60** | **70** | **48** | **178** |
### Step 2: Define Hypotheses
- **Null Hypothesis (H0)**: There is no association between the village and the water source.
- **Alternative Hypothesis (H1)**: There is an association between the village and the water source.
### Step 3: Calculate Expected Frequencies
The expected frequency for each cell can be calculated using the formula:
\[
E = \frac{(Row Total) \times (Column Total)}{Grand Total}
\]
Here are the expected frequencies for each cell:
- For River in A Village:
\[
E = \frac{80 \times 60}{178} \approx 26.96
\]
- For River in B Village:
\[
E = \frac{80 \times 70}{178} \approx 39.32
\]
- For River in C Village:
\[
E = \frac{80 \times 48}{178} \approx 13.72
\]
- For Pond in A Village:
\[
E = \frac{58 \times 60}{178} \approx 19.53
\]
- For Pond in B Village:
\[
E = \frac{58 \times 70}{178} \approx 22.96
\]
- For Pond in C Village:
\[
E = \frac{58 \times 48}{178} \approx 15.51
\]
- For Spring in A Village:
\[
E = \frac{40 \times 60}{178} \approx 13.48
\]
- For Spring in B Village:
\[
E = \frac{40 \times 70}{178} \approx 15.73
\]
- For Spring in C Village:
\[
E = \frac{40 \times 48}{178} \approx 10.79
\]
### Step 4: Calculate Chi-Square Statistic
The Chi-Square statistic can be calculated using:
\[
\chi^2 = \sum \frac{(O - E)^2}{E}
\]
Where \( O \) is the observed frequency, and \( E \) is the expected frequency. Here’s the calculation for each cell:
- For River, A Village:
\[
\chi^2 = \frac{(30 - 26.96)^2}{26.96} \approx 0.0371
\]
- For River, B Village:
\[
\chi^2 = \frac{(32 - 39.32)^2}{39.32} \approx 1.1485
\]
- For River, C Village:
\[
\chi^2 = \frac{(18 - 13.72)^2}{13.72} \approx 0.8601
\]
- For Pond, A Village:
\[
\chi^2 = \frac{(18 - 19.53)^2}{19.53} \approx 0.1183
\]
- For Pond, B Village:
\[
\chi^2 = \frac{(20 - 22.96)^2}{22.96} \approx 0.3884
\]
- For Pond, C Village:
\[
\chi^2 = \frac{(20 - 15.51)^2}{15.51} \approx 0.8454
\]
- For Spring, A Village:
\[
\chi^2 = \frac{(12 - 13.48)^2}{13.48} \approx 0.1127
\]
- For Spring, B Village:
\[
\chi^2 = \frac{(18 - 15.73)^2}{15.73} \approx 0.4201
\]
- For Spring, C Village:
\[
\chi^2 = \frac{(10 - 10.79)^2}{10.79} \approx 0.0615
\]
Now, sum these contributions:
\[
\chi^2 \approx 0.0371 + 1.1485 + 0.8601 + 0.1183 + 0.3884 + 0.8454 + 0.1127 + 0.4201 + 0.0615 \approx 3.9371
\]
### Step 5: Determine Degrees of Freedom
The degrees of freedom for a Chi-Square test of independence is calculated as:
\[
df = (r - 1)(c - 1)
\]
Where \( r \) is the number of rows and \( c \) is the number of columns. In this case, \( r = 3 \) (water sources: river, pond, spring) and \( c = 3 \) (villages: A, B, C):
\[
df = (3 - 1)(3 - 1) = 2 \cdot 2 = 4
\]
### Step 6: Determine the Critical Value
Using a Chi-Square distribution table at a significance level (\( \alpha \)) of 0.05 with 4 degrees of freedom, the critical value is approximately \( 9.488 \).
### Step 7: Compare and Make a Decision
- Calculate \( \chi^2 \) value obtained: \( \chi^2 = 3.9371 \).
- Critical value at \( df = 4 \) and \( \alpha = 0.05 \): \( 9.488 \).
Since \( 3.9371 < 9.488 \), we fail to reject the null hypothesis.
### Conclusion
There is not enough evidence to conclude that there is an association between the villages and the number of households using different water sources.