A random sample of 200 individuals working in a large city indicated that 40 are dissatisfied with their working conditions. Based upon this, compute a 99% confidence interval for the proportion of all individuals in this city who are dissatisfied with their working conditions. Then find the lower limit and upper limit of

the 99% confidence interval.

Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.)

Lower limit:

Upper limit:

To calculate the 99% confidence interval for the proportion of individuals dissatisfied with their working conditions, follow these steps:

### Step 1: Identify the Given Information
- Sample size (\(n\)) = 200
- Number of dissatisfied individuals (\(x\)) = 40
- Confidence level = 99%

### Step 2: Calculate the Sample Proportion (\(\hat{p}\))
The sample proportion \(\hat{p}\) is calculated as:

\[
\hat{p} = \frac{x}{n} = \frac{40}{200} = 0.2
\]

### Step 3: Find the Critical Value (Z)
For a 99% confidence level, the critical value \(Z_{\alpha/2}\) is approximately 2.576 (from the Z-table).

### Step 4: Calculate the Standard Error (SE)
The standard error of the sample proportion is calculated using the formula:

\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]

Substitute the values:

\[
SE = \sqrt{\frac{0.2(1 - 0.2)}{200}} = \sqrt{\frac{0.2 \times 0.8}{200}} = \sqrt{\frac{0.16}{200}} = \sqrt{0.0008} \approx 0.0283
\]

### Step 5: Calculate the Margin of Error (ME)
The margin of error is calculated as:

\[
ME = Z_{\alpha/2} \times SE = 2.576 \times 0.0283 \approx 0.0729
\]

### Step 6: Calculate the Confidence Interval
Finally, the confidence interval for the population proportion is given by:

\[
\text{Lower limit} = \hat{p} - ME = 0.2 - 0.0729 \approx 0.1271
\]

\[
\text{Upper limit} = \hat{p} + ME = 0.2 + 0.0729 \approx 0.2729
\]

### Final Answer:
- **Lower limit:** \(0.13\)
- **Upper limit:** \(0.27\)

This means we are 99% confident that the true proportion of individuals in this city who are dissatisfied with their working conditions lies between 13% and 27%.

Would you like more details or have any other questions? Here are some related questions you might find useful:

1. How does the sample size affect the confidence interval for proportions?
2. What are the assumptions for using the Z-distribution in proportion confidence intervals?
3. How would a different confidence level affect the width of the confidence interval?
4. How does the observed proportion influence the margin of error?
5. What are the implications if the confidence interval for the proportion is very wide?

**Tip:** When calculating confidence intervals for proportions, ensure that both \(n\hat{p}\) and \(n(1-\hat{p})\) are at least 5 to satisfy the normal approximation conditions.

Learn how to calculate a 99% confidence interval for the proportion of dissatisfied workers based on a sample. Understand the formulas for sample proportion, standard error, margin of error, and confidence interval. Suitable for advanced high school students or college-level learners.

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