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Part 1
A large company employing several thousand workers in a large city in the southeastern United States would like to estimate the proportion of its employees who prefer to work from home rather than commute to its campus. The company hopes to use the information to develop a proposal to determine the best way to retain current and future employees. A pilot study of 5050 randomly sampled employees found that 1919 would prefer to work from home. Complete parts a and b.
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Part 1
a. How many more employees must the company randomly sample to be able to estimate the true population of employees who prefer to work from home with a margin of error of 0.030.03 and a level of confidence of 9999​%?
The company must sample enter your response here more employees.

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.

A large company employing several thousand workers in a large city in the southeastern United States would like to estimate the proportion of its employees who prefer to work from home rather than commute to its campus. The company hopes to use the information to develop a proposal to determine the best way to retain current and future employees. A pilot study of 5050 randomly sampled employees found that 1919 would prefer to work from home. Complete parts a and b. a. How many more employees must the company randomly sample to be able to estimate the true population of employees who prefer to work from home with a margin of error of 0.03 and a level of confidence of 99%?

Calculate the required sample size for estimating the proportion of employees preferring to work from home with a 99% confidence level and a 0.03 margin of error. The problem applies concepts in statistics such as confidence intervals and uses the sample proportion from a pilot study.

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