Question

3. The Human Resources department of a large corporation wants to estimate the mean number

Answer 1

Answer:

The 90% confidence level margin of error for this sample mean is 0.42

Step-by-step explanation:

The Human Resources department conduct a random sample of 30  employees

n = 30

Standard deviation = $\sigma = 1.4$

z = 1.645

Formula of 90% confidence level margin of error for this sample mean=$Z \times \frac{\sigma}{\sqrt{n}}$

So, margin of error for this sample mean=$1.645 \times \frac{1.4}{\sqrt{30}}$

So, margin of error for this sample mean=0.42

Hence the 90% confidence level margin of error for this sample mean is 0.42