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.
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To construct a \( 90\% \) confidence interval for the population mean height of 15-year-olds, we start with the sample statistics. The sample mean (\( \bar{x} \)) is \( 65 \) inches and the standard error (SE) is \( 0.50 \) inches. First, we need to find the critical value (\( z^{*} \)) for a \( 90\% \) confidence level. For a \( 90\% \) confidence interval, the area in each tail of the standard normal distribution is \( 0.05 \). Using the standard normal (Z) table or calculator, we find that \( z^{*} \approx 1.645 \). Now we can calculate the confidence interval using the formula: \[ \text{Confidence Interval} = \bar{x} \pm z^{*} \times SE \] Substituting in the values: \[ \text{Confidence Interval} = 65 \pm 1.645 \times 0.50 \] Calculating the margin of error: \[ 1.645 \times 0.50 = 0.8225 \] Now we can construct the interval: \[ \text{Lower Limit} = 65 - 0.8225 = 64.1775 \] \[ \text{Upper Limit} = 65 + 0.8225 = 65.8225 \] Finally, rounding to three decimal places gives us: \[ \text{Confidence Interval} = (64.178, 65.823) \] Thus, the \( 90\% \) confidence interval for the population mean height of 15-year-olds is \( (64.178, 65.823) \).
