In a survey, 16 parents were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of \( \$ 35 \) and standard deviation of \( \$ 16 \). Construct a confidence interval at a \( 98 \% \) confidence level to estimate the average amount that parents spent on their child's birthday gift. Give your answers to one decimal place.

To construct a 98% confidence interval for the average amount parents spent on their child's birthday gift, we'll follow these steps: 1. **Identify the Given Information:** - Sample size (\( n \)) = 16 - Sample mean (\( \bar{x} \)) = \$35 - Sample standard deviation (\( s \)) = \$16 - Confidence level = 98% 2. **Determine the Appropriate Distribution:** - Since the sample size is small (\( n < 30 \)) and the population is roughly bell-shaped, we use the **t-distribution**. - Degrees of freedom (\( df \)) = \( n - 1 = 15 \) 3. **Find the Critical t-Value:** - For a 98% confidence level and \( df = 15 \), the critical t-value (\( t_{0.99,15} \)) is approximately **2.602**. 4. **Calculate the Margin of Error (ME):** \[ ME = t \times \left( \frac{s}{\sqrt{n}} \right) = 2.602 \times \left( \frac{16}{\sqrt{16}} \right) = 2.602 \times 4 = 10.408 \] 5. **Construct the Confidence Interval:** \[ \text{Lower bound} = \bar{x} - ME = 35 - 10.4 = 24.6 \] \[ \text{Upper bound} = \bar{x} + ME = 35 + 10.4 = 45.4 \] 6. **Final Confidence Interval:** \[ \$24.6 \text{ to } \$45.4 \] **Therefore, the 98% confidence interval for the average amount spent is \$24.6 to \$45.4.** **Answer:** At 98 % confidence, the average spent is between \$ 24.6 and \$ 45.4