In a very large organisation, the director wanted to find out what proportions of the employees prefer to provide their own retirement benefits in lieu of a company – sponsored plan. A simple random sample of 75 employees was taken. It was found that 40%, that is, 0.4 of them are interested in providing their own retirement plans. The management requests that we use this sample to find an interval about which they can be 99 percent confident that it contains the true population proportion.

To find a 99% confidence interval for the true population proportion of employees who prefer to provide their own retirement benefits, we can use the formula for the confidence interval of a proportion:

\[
\text{Confidence Interval} = \hat{p} \pm Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}
\]

Where:
- \(\hat{p}\) is the sample proportion (0.4 in this case).
- \(Z_{\alpha/2}\) is the critical value for the desired confidence level (for 99% confidence, \(Z_{\alpha/2} = 2.576\)).
- \(n\) is the sample size (75 in this case).

### Step 1: Calculate the standard error (SE)
\[
SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.4(1-0.4)}{75}} = \sqrt{\frac{0.4 \times 0.6}{75}} = \sqrt{\frac{0.24}{75}} \approx \sqrt{0.0032} \approx 0.0566
\]

### Step 2: Calculate the margin of error (ME)
\[
ME = Z_{\alpha/2} \cdot SE = 2.576 \cdot 0.0566 \approx 0.1458
\]

### Step 3: Determine the confidence interval
\[
\text{Confidence Interval} = \hat{p} \pm ME = 0.4 \pm 0.1458
\]
\[
\text{Confidence Interval} = (0.4 - 0.1458, 0.4 + 0.1458) = (0.2542, 0.5458)
\]

### Conclusion
The 99% confidence interval for the true population proportion of employees who prefer to provide their own retirement benefits is approximately **(0.2542, 0.5458)**.

This means that management can be 99% confident that the true proportion of all employees in the organization who prefer to provide their own retirement benefits lies between 25.42% and 54.58%.

Would you like any more details or have any further questions?

Here are 5 related questions for further exploration:
1. How would the confidence interval change if we wanted a 95% confidence level instead of 99%?
2. What would happen to the confidence interval if the sample size increased to 150?
3. How does the sample proportion (\(\hat{p}\)) affect the width of the confidence interval?
4. Can you explain why a larger sample size results in a narrower confidence interval?
5. How would the confidence interval change if the sample proportion were 0.5 instead of 0.4?

**Tip:** Remember that the width of a confidence interval depends on the sample size, the level of confidence, and the sample proportion. Increasing the sample size decreases the margin of error, leading to a narrower confidence interval.

Learn how to calculate a 99% confidence interval for the true population proportion of employees preferring to provide their own retirement benefits using a simple random sample. Includes step-by-step formula application and interpretation of results.

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