Question

A sample dataset of 812 has a mean of 38.6 with a standard deviation of 1.2. What is the margin of error? Round the answer to the nearest hundredth. (1 point) margin of error =

Answer 1

To calculate the margin of error (ME) for a sample, we can use the formula:

\[
\text{ME} = z \times \left( \frac{s}{\sqrt{n}} \right)
\]

Where:
- \( z \) is the z-score corresponding to the desired confidence level (commonly 1.96 for a 95% confidence level),
- \( s \) is the standard deviation of the sample,
- \( n \) is the sample size.

Given:
- \( s = 1.2 \)
- \( n = 812 \)

Assuming we are using a 95% confidence level, \( z \) is typically 1.96.

Now, let's calculate the standard error (SE):

\[
SE = \frac{s}{\sqrt{n}} = \frac{1.2}{\sqrt{812}}
\]

Calculating \( \sqrt{812} \):

\[
\sqrt{812} \approx 28.49
\]

Now, continue to calculate SE:

\[
SE = \frac{1.2}{28.49} \approx 0.0421
\]

Now, we can find the margin of error:

\[
\text{ME} = 1.96 \times SE \approx 1.96 \times 0.0421 \approx 0.0827
\]

Now, rounding to the nearest hundredth:

\[
\text{ME} \approx 0.08
\]

Thus, the margin of error is approximately \( \text{margin of error} = 0.08 \).