Question

1. a sample of 50 employees shows an average monthly commute - time of 30 minutes with a standard deviation of 5 minutes. calculate the margin of error (me) for a 99% confidence level (z = 2.58).

a. 1.93 minutes
b. 1.50 minutes
c. 1.82 minutes
d. 2.00 minutes

Explanation:

Step1: Recall margin - of - error formula

The formula for the margin of error (ME) when the population standard deviation $\sigma$ is known is $ME = z\times\frac{\sigma}{\sqrt{n}}$, where $z$ is the z - score, $\sigma$ is the population standard deviation, and $n$ is the sample size.

Step2: Identify given values

We are given that $z = 2.58$ (for a 99% confidence level), $\sigma=5$ minutes, and $n = 50$.

Step3: Substitute values into formula

$ME=2.58\times\frac{5}{\sqrt{50}}$. First, calculate $\sqrt{50}\approx7.071$. Then $\frac{5}{\sqrt{50}}=\frac{5}{7.071}\approx0.707$. Finally, $ME = 2.58\times0.707\approx1.82$ minutes.

Answer:

C. 1.82 minutes