In a survey, 17 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 \( \$ 5 \). Construct a confidence interval at a \( 80 \% \) 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 an \( 80\% \) confidence interval for the mean, we will use the formula: \[ \text{Confidence Interval} = \bar{X} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \] Where: - \(\bar{X} = 46\) (mean) - \(z\) is the z-value corresponding to \(80\%\) confidence level (which is approximately \(1.28\)) - \(\sigma = 5\) (standard deviation) - \(n = 17\) (sample size) Now, calculate the standard error (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{17}} \approx 1.21 \] Next, calculate the margin of error (ME): \[ ME = z \times SE = 1.28 \times 1.21 \approx 1.55 \] Now, we can find the confidence interval: \[ \text{Lower Limit} = \bar{X} - ME = 46 - 1.55 \approx 44.5 \] \[ \text{Upper Limit} = \bar{X} + ME = 46 + 1.55 \approx 47.6 \] Thus, the \( 80\% \) confidence interval for the average amount that parents spent on their child's birthday gift is approximately \( (44.5, 47.6) \).