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.
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To construct a confidence interval for the average amount spent on a child's birthday gift, you can use the formula for the confidence interval when the population standard deviation is known: \[ \text{Confidence Interval} = \text{mean} \pm z \times \left(\frac{\sigma}{\sqrt{n}}\right) \] Where: - Mean (\( \mu \)) = \$46 - Standard deviation (\( \sigma \)) = \$2 - Sample size (\( n \)) = 22 - \( z \) value for a 98% confidence level ≈ 2.33 (You can find this value from z-tables or calculators) Now, let's calculate the margin of error: 1. Calculate the standard error (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{22}} \approx \frac{2}{4.690} \approx 0.426 \] 2. Calculate the margin of error: \[ \text{Margin of Error} = z \times SE \approx 2.33 \times 0.426 \approx 0.994 \] 3. Determine the confidence interval: \[ \text{Lower Limit} = 46 - 0.994 \approx 45.0 \] \[ \text{Upper Limit} = 46 + 0.994 \approx 47.0 \] Thus, the 98% confidence interval to estimate the average amount that parents spent on their child's birthday gift is approximately \( \text{(45.0, 47.0)} \).
