To calculate the 90% confidence interval for the true proportion of club members who use compost, we can use the formula for the confidence interval for a population proportion.
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Determine the sample proportion (\( \hat{p} \)): \[ \hat{p} = \frac{\text{number of favorable responses}}{\text{total responses}} = \frac{0.45 \times 200}{200} = 0.45 \]
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Determine the sample size (\( n \)): \[ n = 200 \]
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Calculate the standard error (SE) of the proportion: \[ SE = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}} = \sqrt{\frac{0.45 \times (1 - 0.45)}{200}} = \sqrt{\frac{0.45 \times 0.55}{200}} \approx \sqrt{\frac{0.2475}{200}} \approx \sqrt{0.0012375} \approx 0.0352 \]
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Find the z-score for a 90% confidence level: The critical z-value for a 90% confidence interval is approximately 1.645 (you can find this from a z-table or standard normal distribution).
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Calculate the margin of error (ME): \[ ME = z \cdot SE = 1.645 \cdot 0.0352 \approx 0.0579 \]
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Construct the confidence interval: \[ \text{Confidence Interval} = \hat{p} \pm ME = 0.45 \pm 0.0579 \] This gives us: \[ \text{Lower limit} = 0.45 - 0.0579 \approx 0.3921 \] \[ \text{Upper limit} = 0.45 + 0.0579 \approx 0.5079 \]
Thus, the 90% confidence interval for the true proportion of club members who use compost is approximately \((0.3921, 0.5079)\).
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