Question
A ramdom sample of 150 workers showed that 90 workers are satisfied with the new medical scheme. Find a 90% confidence interval for all satisfied workers.
Answers
GPT-4o mini
To find a 90% confidence interval for the proportion of satisfied workers based on the sample, we can use the formula for a confidence interval for a population proportion.
Here are the steps to calculate the confidence interval:
1. **Calculate the sample proportion (\( \hat{p} \))**:
\[
\hat{p} = \frac{x}{n} = \frac{90}{150} = 0.6
\]
where \( x \) is the number of satisfied workers (90) and \( n \) is the total number of workers (150).
2. **Find the Z-score for the desired confidence level**:
For a 90% confidence interval, the Z-score (from the Z-table) is approximately 1.645.
3. **Calculate the standard error (SE)**:
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.6(1 - 0.6)}{150}} = \sqrt{\frac{0.6 \times 0.4}{150}} = \sqrt{\frac{0.24}{150}} \approx \sqrt{0.0016} \approx 0.04
\]
4. **Calculate the margins of error (ME)**:
\[
ME = Z \cdot SE = 1.645 \cdot 0.04 \approx 0.0658
\]
5. **Calculate the confidence interval**:
\[
CI = \hat{p} \pm ME = 0.6 \pm 0.0658
\]
This gives us:
\[
CI = (0.6 - 0.0658, 0.6 + 0.0658) \approx (0.5342, 0.6658)
\]
Thus, the 90% confidence interval for the proportion of all satisfied workers is approximately **(0.5342, 0.6658)**. This means we are 90% confident that the true proportion of satisfied workers in the entire population lies within this interval.
Here are the steps to calculate the confidence interval:
1. **Calculate the sample proportion (\( \hat{p} \))**:
\[
\hat{p} = \frac{x}{n} = \frac{90}{150} = 0.6
\]
where \( x \) is the number of satisfied workers (90) and \( n \) is the total number of workers (150).
2. **Find the Z-score for the desired confidence level**:
For a 90% confidence interval, the Z-score (from the Z-table) is approximately 1.645.
3. **Calculate the standard error (SE)**:
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.6(1 - 0.6)}{150}} = \sqrt{\frac{0.6 \times 0.4}{150}} = \sqrt{\frac{0.24}{150}} \approx \sqrt{0.0016} \approx 0.04
\]
4. **Calculate the margins of error (ME)**:
\[
ME = Z \cdot SE = 1.645 \cdot 0.04 \approx 0.0658
\]
5. **Calculate the confidence interval**:
\[
CI = \hat{p} \pm ME = 0.6 \pm 0.0658
\]
This gives us:
\[
CI = (0.6 - 0.0658, 0.6 + 0.0658) \approx (0.5342, 0.6658)
\]
Thus, the 90% confidence interval for the proportion of all satisfied workers is approximately **(0.5342, 0.6658)**. This means we are 90% confident that the true proportion of satisfied workers in the entire population lies within this interval.