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 = [tex]\sigma = 1.4[/tex]
z = 1.645
Formula of 90% confidence level margin of error for this sample mean=[tex]Z \times \frac{\sigma}{\sqrt{n}}[/tex]
So, margin of error for this sample mean=[tex]1.645 \times \frac{1.4}{\sqrt{30}}[/tex]
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
