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To determine if the distribution of household sizes among welfare recipients is the same as the distribution provided by the 2000 census data, we can use a chi-square test for goodness of fit.
1. State the hypotheses:
- Null Hypothesis (H₀): The distribution of household sizes among welfare recipients is the same as that provided by the census data.
- Alternative Hypothesis (Hᴀ): The distribution of household sizes among welfare recipients is different from that provided by the census data.
2. Observed frequencies from the study sample:
- 1-person households: 513
- 2-person households: 550
- 3-person households: 301
- 4-person households: 277
- 5-or-more-person households: 203
3. Expected frequencies based on census data (percentages):
- Total number of observations: 1844
- Expected percentages:
- 1-person households: [tex]\( 25.81\% \)[/tex]
- 2-person households: [tex]\( 32.63\% \)[/tex]
- 3-person households: [tex]\( 16.53\% \)[/tex]
- 4-person households: [tex]\( 14.20\% \)[/tex]
- 5-or-more-person households: [tex]\( 10.83\% \)[/tex]
4. Calculate expected frequencies:
- 1-person households: [tex]\( 1844 \times \frac{25.81}{100} = 475.61 \)[/tex]
- 2-person households: [tex]\( 1844 \times \frac{32.63}{100} = 601.57 \)[/tex]
- 3-person households: [tex]\( 1844 \times \frac{16.53}{100} = 304.77 \)[/tex]
- 4-person households: [tex]\( 1844 \times \frac{14.20}{100} = 261.85 \)[/tex]
- 5-or-more-person households: [tex]\( 1844 \times \frac{10.83}{100} = 199.20 \)[/tex]
5. Perform the chi-square test for goodness of fit:
- The chi-square statistic (χ²) is calculated as follows:
[tex]\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \][/tex]
where [tex]\( O_i \)[/tex] is the observed frequency and [tex]\( E_i \)[/tex] is the expected frequency for each category.
6. Calculate the chi-square statistic and p-value:
- From the given solution:
[tex]\[ \chi^2 = 8.31 \quad (\text{rounded to two decimal places}) \][/tex]
[tex]\[ p = 0.081 \quad (\text{rounded to three decimal places}) \][/tex]
7. Decision:
- Typically, we use a significance level of 0.05. If the p-value is less than the significance level, we reject the null hypothesis.
- In this case, [tex]\( p = 0.081 \)[/tex], which is greater than 0.05.
8. Conclusion:
- Since the p-value is greater than the significance level, we do not reject the null hypothesis. There is not enough evidence to conclude that the distribution of household sizes among welfare recipients is different from the distribution provided by the census data.
Therefore, the most appropriate null hypothesis is:
- "The distribution of household sizes among welfare recipients is the same as that provided by the census data."
1. State the hypotheses:
- Null Hypothesis (H₀): The distribution of household sizes among welfare recipients is the same as that provided by the census data.
- Alternative Hypothesis (Hᴀ): The distribution of household sizes among welfare recipients is different from that provided by the census data.
2. Observed frequencies from the study sample:
- 1-person households: 513
- 2-person households: 550
- 3-person households: 301
- 4-person households: 277
- 5-or-more-person households: 203
3. Expected frequencies based on census data (percentages):
- Total number of observations: 1844
- Expected percentages:
- 1-person households: [tex]\( 25.81\% \)[/tex]
- 2-person households: [tex]\( 32.63\% \)[/tex]
- 3-person households: [tex]\( 16.53\% \)[/tex]
- 4-person households: [tex]\( 14.20\% \)[/tex]
- 5-or-more-person households: [tex]\( 10.83\% \)[/tex]
4. Calculate expected frequencies:
- 1-person households: [tex]\( 1844 \times \frac{25.81}{100} = 475.61 \)[/tex]
- 2-person households: [tex]\( 1844 \times \frac{32.63}{100} = 601.57 \)[/tex]
- 3-person households: [tex]\( 1844 \times \frac{16.53}{100} = 304.77 \)[/tex]
- 4-person households: [tex]\( 1844 \times \frac{14.20}{100} = 261.85 \)[/tex]
- 5-or-more-person households: [tex]\( 1844 \times \frac{10.83}{100} = 199.20 \)[/tex]
5. Perform the chi-square test for goodness of fit:
- The chi-square statistic (χ²) is calculated as follows:
[tex]\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \][/tex]
where [tex]\( O_i \)[/tex] is the observed frequency and [tex]\( E_i \)[/tex] is the expected frequency for each category.
6. Calculate the chi-square statistic and p-value:
- From the given solution:
[tex]\[ \chi^2 = 8.31 \quad (\text{rounded to two decimal places}) \][/tex]
[tex]\[ p = 0.081 \quad (\text{rounded to three decimal places}) \][/tex]
7. Decision:
- Typically, we use a significance level of 0.05. If the p-value is less than the significance level, we reject the null hypothesis.
- In this case, [tex]\( p = 0.081 \)[/tex], which is greater than 0.05.
8. Conclusion:
- Since the p-value is greater than the significance level, we do not reject the null hypothesis. There is not enough evidence to conclude that the distribution of household sizes among welfare recipients is different from the distribution provided by the census data.
Therefore, the most appropriate null hypothesis is:
- "The distribution of household sizes among welfare recipients is the same as that provided by the census data."
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