To determine how many more employees the company must sample to estimate the true population proportion with a specified margin of error and confidence level, we use the formula for the sample size needed for estimating a population proportion:

$n = \left( \frac{Z^2 \cdot \hat{p} \cdot (1 - \hat{p})}{E^2} \right)$

Where:

• $Z$ is the z-value corresponding to the desired confidence level,
• $\hat{p}$ is the sample proportion of employees who prefer to work from home,
• $E$ is the desired margin of error.

Step 1: Identify the parameters

• Confidence level: 99% (so the z-value is $Z = 2.576$),
• Margin of error: $E = 0.03$,
• Sample proportion: $\hat{p} = \frac{1919}{5050} = 0.38$.

Step 2: Calculate the sample size

Now, let's compute the sample size $n$.

$n = \left( \frac{(2.576)^2 \cdot (0.38) \cdot (1 - 0.38)}{(0.03)^2} \right)$

First, calculate the numerator: $2.576^2 = 6.635$ $\hat{p}(1 - \hat{p}) = 0.38 \cdot 0.62 = 0.2356$ Now multiply: $6.635 \cdot 0.2356 = 1.5623$

Now divide by the square of the margin of error: $n = \frac{1.5623}{0.03^2} = \frac{1.5623}{0.0009} \approx 1735.89$

Step 3: Adjust for the already sampled employees

Since the company already sampled 5050 employees, the additional number of employees required is: $n_{\text{additional}} = 1736 - 5050$

As the sample size already exceeds the required value, the company does not need to sample more employees.

Would you like more details or have questions about this? Here are 5 follow-up questions:

1. What does the margin of error represent in a survey?
2. Why is the z-value for 99% confidence higher than for 95%?
3. How does increasing the sample size affect the margin of error?
4. What is the significance of the sample proportion $\hat{p}$ in this context?
5. What would happen if the company wanted a smaller margin of error?

Tip: The margin of error decreases as the sample size increases, but the reduction effect diminishes with large sample sizes.