To calculate the 90% confidence interval for the proportion of satisfied workers, we will follow these steps:
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Identify the sample proportion (\(\hat{p}\)): \[ \hat{p} = \frac{x}{n} \] where \(x\) is the number of satisfied workers and \(n\) is the total number of workers in the sample. Here, \(x = 100\) and \(n = 150\): \[ \hat{p} = \frac{100}{150} = 0.6667 \]
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Calculate the standard error (SE) of the sample proportion: \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.6667 \times (1 - 0.6667)}{150}} \] \[ SE = \sqrt{\frac{0.6667 \times 0.3333}{150}} = \sqrt{\frac{0.2222}{150}} \approx \sqrt{0.001481} \approx 0.0385 \]
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Determine the z-score for a 90% confidence level. The z-score for a 90% confidence level is approximately 1.645.
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Calculate the margin of error (ME): \[ ME = z \times SE = 1.645 \times 0.0385 \approx 0.0632 \]
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Construct the confidence interval: \[ \text{Lower limit} = \hat{p} - ME = 0.6667 - 0.0632 \approx 0.6035 \] \[ \text{Upper limit} = \hat{p} + ME = 0.6667 + 0.0632 \approx 0.7299 \]
Thus, the 90% confidence interval for the proportion of satisfied workers is approximately (0.6035, 0.7299).
In summary, we can conclude that we are 90% confident that the true proportion of satisfied workers lies between 60.35% and 72.99%.