Questions: The margin of error for a confidence interval in which the population standard deviation is known is: zc * (σ/√n) zc * (σ^2/n) zc * (σ^2/√n) zc * (σ/n)

Transcript text: The margin of error for a confidence interval in which the population standard deviation is known is: $z_{c} \cdot \frac{\sigma}{\sqrt{n}}$ $z_{c} \cdot \frac{\sigma^{2}}{n}$ $z_{c} \cdot \frac{\sigma^{2}}{\sqrt{n}}$ $z_{c} \cdot \frac{\sigma}{n}$

Solution

Solution Steps

Step 1: Understand the formula for the margin of error

The margin of error for a confidence interval when the population standard deviation ($\sigma$) is known is given by: \[ \text{Margin of Error} = z_{c} \cdot \frac{\sigma}{\sqrt{n}}, \] where:

$z_{c}$ is the critical value corresponding to the desired confidence level,
$\sigma$ is the population standard deviation,
$n$ is the sample size.

Step 2: Compare the given options

The correct formula for the margin of error is: \[ z_{c} \cdot \frac{\sigma}{\sqrt{n}}. \] This matches the first option provided in the question.

Step 3: Verify the incorrect options

The other options are incorrect because:

$z_{c} \cdot \frac{\sigma^{2}}{n}$ incorrectly squares $\sigma$ and divides by $n$ instead of $\sqrt{n}$.
$z_{c} \cdot \frac{\sigma^{2}}{\sqrt{n}}$ incorrectly squares $\sigma$.
$z_{c} \cdot \frac{\sigma}{n}$ incorrectly divides by $n$ instead of $\sqrt{n}$.