To determine whether the proportion of households that regularly recycle waste in Jozi has changed after the campaign, we can perform a hypothesis test for a proportion.
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Define the Hypotheses:
- Null Hypothesis (\(H_0\)): The proportion of households recycling has not changed, \( p = 0.60 \).
- Alternative Hypothesis (\(H_a\)): The proportion of households recycling has changed, \( p \neq 0.60 \).
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Sample Proportion:
- Number of households surveyed = 650
- Number of households recycling = 450
- Sample proportion (\( \hat{p} \)) = \( \frac{450}{650} = \frac{45}{65} \approx 0.6923 \)
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Calculate the Test Statistic: We will use the formula for the z-test for proportions:
\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]
Where:
- \( \hat{p} \) = sample proportion = 0.6923
- \( p_0 \) = hypothesized population proportion = 0.60
- \( n \) = sample size = 650
Calculate the standard error (SE):
\[ SE = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.60 \times 0.40}{650}} = \sqrt{\frac{0.24}{650}} \approx \sqrt{0.0003692307} \approx 0.0192 \]
Now substitute into the z formula:
\[ z = \frac{0.6923 - 0.60}{0.0192} \approx \frac{0.0923}{0.0192} \approx 4.81 \]
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Determine the Level of Significance: Since we are testing at the 90% level of significance for a two-tailed test, the critical z-values would be approximately -1.645 and +1.645.
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Decision:
- If the computed z-value is beyond the critical values (either less than -1.645 or greater than +1.645), we reject the null hypothesis.
- Here, our computed z-value is approximately 4.81, which exceeds 1.645.
Given the value of the sample test statistic \( z \approx 4.81 \), we reject the null hypothesis. There is sufficient evidence at the 90% level of significance to conclude that the proportion of households regularly recycling waste in Jozi has changed after the campaign.
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