Questions: In a survey, 21 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of 41 and standard deviation of 9. Find the margin of error at a 99% confidence level. Give your answer to two decimal places.

Solution Steps

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

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

\[ \text{Margin of Error} = \frac{Z \times \sigma}{\sqrt{n}} \]

Where:

• \( Z = 2.58 \)
• \( \sigma = 9 \) (standard deviation)
• \( n = 21 \) (sample size)

Substituting the values into the formula:

\[ \text{Margin of Error} = \frac{2.58 \times 9}{\sqrt{21}} \]

Calculating the denominator:

\[ \sqrt{21} \approx 4.5826 \]

Now substituting back into the margin of error calculation:

\[ \text{Margin of Error} = \frac{2.58 \times 9}{4.5826} \approx \frac{23.22}{4.5826} \approx 5.06 \]

Final Answer