Questions: If n=32, x̅=42, and s=5, find the margin of error E at a 99% confidence level. Give your answer to two decimal places.

Solution Steps

We are given the following values:

• Sample size \( n = 32 \)
• Sample mean \( \bar{x} = 42 \)
• Sample standard deviation \( s = 5 \)
• Confidence level \( 99\% \)

For a \( 99\% \) confidence level, the Z-score \( Z \) is approximately \( 2.58 \).

The formula for the margin of error \( E \) is given by:

\[ E = \frac{Z \times s}{\sqrt{n}} \]

Substituting the known values:

\[ E = \frac{2.58 \times 5}{\sqrt{32}} \]

Calculating the margin of error:

\[ E = \frac{2.58 \times 5}{\sqrt{32}} \approx 2.28 \]

Final Answer