In a random sample of 15 -year-olds, the mean height was 65 inches. The sampling distribution was normal with a standard error of 0.50 inches. A. [ 3 points] Construct a \( 90 \% \) confidence interval to estimate the population mean. Round your answer to 3 decimal places. Use the general form of a confidence interval: sample statistic \( \pm z^{*}(S E) \) Show your work.

To construct a \( 90\% \) confidence interval for the population mean, we start with the given values: - Sample mean (\( \bar{x} \)) = 65 inches - Standard error (SE) = 0.50 inches - Confidence level = \( 90\% \) First, we need to find the critical value (\( z^{*} \)) for a \( 90\% \) confidence level. This value can be found using a standard normal distribution table or calculator. The \( z^{*} \) value for a \( 90\% \) confidence level (which corresponds to \( 5\% \) in each tail) is approximately \( 1.645 \). Now, we can construct the confidence interval using the formula: \[ CI = \bar{x} \pm z^{*} \times SE \] Substituting in the known values: \[ CI = 65 \pm 1.645 \times 0.50 \] Calculating the margin of error: \[ Margin \ of \ Error = 1.645 \times 0.50 = 0.8225 \] Now we can calculate the upper and lower bounds of the confidence interval: Lower bound: \[ 65 - 0.8225 = 64.1775 \quad (rounded \ to \ 64.178) \] Upper bound: \[ 65 + 0.8225 = 65.8225 \quad (rounded \ to \ 65.823) \] Thus, the \( 90\% \) confidence interval to estimate the population mean is: \[ (64.178, 65.823) \] So, we can summarize: \[ (64.178, 65.823) \]