To compute a 99% confidence interval for the proportion of individuals dissatisfied with their working conditions, we use the formula for a confidence interval for a proportion:

$\hat{p} \pm z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Steps:

1. Determine the sample proportion ($\hat{p}$): $\hat{p} = \frac{x}{n} = \frac{85}{275}$

2. Find the critical value $z$ for a 99% confidence level. For 99%, $z = 2.576$ (from the standard normal table).

3. Compute the standard error (SE): $SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

4. Construct the interval: $\text{Lower limit} = \hat{p} - z \cdot SE$ $\text{Upper limit} = \hat{p} + z \cdot SE$

Let me compute this step-by-step. Given values x = 85 # number of dissatisfied individuals n = 275 # sample size z = 2.576 # z-score for 99% confidence

Step 1: Calculate sample proportion (p-hat)

p_hat = x / n

Step 2: Compute standard error (SE)

SE = (p_hat * (1 - p_hat) / n) ** 0.5

Step 3: Calculate confidence interval

lower_limit = p_hat - z * SE upper_limit = p_hat + z * SE

Round results to two decimal places

p_hat, SE, lower_limit, upper_limit = round(p_hat, 3), round(SE, 3), round(lower_limit, 3), round(upper_limit, 3) p_hat, SE, lower_limit, upper_limit### Calculations:

1. Sample proportion ($\hat{p}$):
   $\hat{p} = 0.309$

2. Standard error (SE):
   $SE = 0.028$

3. 99% Confidence Interval:

   • Lower limit: $0.237$
   • Upper limit: $0.381$

Final Answer:

• Lower limit: 0.24
• Upper limit: 0.38

Would you like further explanation or help with other statistical concepts?

Related Questions:

1. How does the width of a confidence interval change with the sample size?
2. Why is a 99% confidence interval wider than a 95% confidence interval?
3. What assumptions are required to calculate this confidence interval for a proportion?
4. How does the choice of $z$-score affect the confidence interval?
5. Can we use this method if the sample size or proportion is very small?

Tip:

When calculating confidence intervals, always ensure the sample size is large enough for the approximation to be valid ($n \cdot \hat{p} \geq 5$ and $n \cdot (1 - \hat{p}) \geq 5$).