Question

. A random sample of 150 workers showed that 100 are satisfied with their new medical scheme. Construct a 90% confidence interval for the proportion of satisfied workers.

Answer 1

To calculate the 90% confidence interval for the proportion of satisfied workers, we will follow these steps:

1. **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 \(n = 150\):
\[
\hat{p} = \frac{100}{150} = 0.6667
\]

2. **Calculate the standard error (SE)** of the sample proportion:
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.6667 \times (1 - 0.6667)}{150}}
\]
\[
SE = \sqrt{\frac{0.6667 \times 0.3333}{150}} = \sqrt{\frac{0.2222}{150}} \approx \sqrt{0.001481} \approx 0.0385
\]

3. **Determine the z-score** for a 90% confidence level. The z-score for a 90% confidence level is approximately 1.645.

4. **Calculate the margin of error (ME)**:
\[
ME = z \times SE = 1.645 \times 0.0385 \approx 0.0632
\]

5. **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%.