Question 12
In a survey, 13 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $\$ 47$ and standard deviation of $\$ 16$. Construct a confidence interval at a $99 \%$ confidence level.
Give your answers to one decimal place.
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Step 1 ：Given that the mean is \$47, the standard deviation is \$16, the sample size is 13, and the Z-score for a 99% confidence level is approximately 2.576.

Step 2 ：We use the formula for the confidence interval: \(\text{Confidence Interval} = \text{mean} \pm Z \times \frac{\text{standard deviation}}{\sqrt{\text{sample size}}}\)

Step 3 ：Substitute the given values into the formula: \(\text{Confidence Interval} = 47 \pm 2.576 \times \frac{16}{\sqrt{13}}\)

Step 4 ：Calculate the margin of error: \(\text{margin of error} = 2.576 \times \frac{16}{\sqrt{13}} = 11.4\)

Step 5 ：Subtract the margin of error from the mean to get the lower bound of the confidence interval: \(\text{lower bound} = 47 - 11.4 = 35.6\)

Step 6 ：Add the margin of error to the mean to get the upper bound of the confidence interval: \(\text{upper bound} = 47 + 11.4 = 58.4\)

Step 7 ：Final Answer: The 99% confidence interval for the amount spent on their child's last birthday gift is \(\boxed{(\$35.6, \$58.4)}\)