In a survey, 22 parents were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of \( \$ 46 \) and standard deviation of \( \$ 2 \). 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 use the **t-distribution** because the sample size is small (n = 22) and the population standard deviation is unknown. ### Given: - **Sample Mean (\(\bar{x}\))**: \$46 - **Sample Standard Deviation (s)**: \$2 - **Sample Size (n)**: 22 - **Confidence Level**: 98% ### Steps to Calculate the Confidence Interval: 1. **Determine the Degrees of Freedom (df)**: \[ df = n - 1 = 22 - 1 = 21 \] 2. **Find the Critical t-Value (\(t^*\))**: For a 98% confidence level and 21 degrees of freedom, the critical t-value (\(t_{0.99}\)) is approximately **2.518**. (This can be found using a t-table or statistical software.) 3. **Calculate the Standard Error (SE)**: \[ SE = \frac{s}{\sqrt{n}} = \frac{2}{\sqrt{22}} \approx 0.426 \] 4. **Determine the Margin of Error (ME)**: \[ ME = t^* \times SE \approx 2.518 \times 0.426 \approx 1.075 \] Rounded to one decimal place, **ME ≈ 1.1** 5. **Construct the Confidence Interval**: \[ \text{Lower Limit} = \bar{x} - ME = 46 - 1.1 = 44.9 \] \[ \text{Upper Limit} = \bar{x} + ME = 46 + 1.1 = 47.1 \] ### **Final 98% Confidence Interval**: \[ \$44.9 \leq \mu \leq \$47.1 \] **Interpretation**: We are 98% confident that the true average amount parents spent on their child's birthday gift lies between \$44.9 and \$47.1. **Answer:** A 98 % confidence interval for the mean is ($44.9, $47.1).