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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. B. [2 points] Interpret your confidence interval (i.e., "I am 90\% confident that..." ) Remember to use the word "population" in your interpretation.

Ask by Daniel Logan. in the United States
Mar 11,2025

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**A. Confidence Interval** - **Sample Mean (\( \overline{x} \))**: 65 inches - **Standard Error (SE)**: 0.50 inches - **Critical Value (\( z^* \))**: 1.645 (for 90% confidence) - **Margin of Error**: \( 1.645 \times 0.50 = 0.823 \) inches - **Confidence Interval**: \( 65 \pm 0.823 \) inches \[ (64.177, 65.823) \text{ inches} \] **B. Interpretation** I am 90% confident that the population mean height of 15-year-olds is between 64.177 inches and 65.823 inches.

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To construct the 90% confidence interval for the population mean, we will use the formula: **Confidence Interval = sample statistic ± z*(SE)** **A. Calculation:** 1. **Sample Statistic (mean height)**: 65 inches 2. **Standard Error (SE)**: 0.50 inches 3. **Z*-value for 90% confidence**: The z-value for a 90% confidence level is approximately 1.645 (you can find this on a Z-table). Now we can plug in these numbers into the formula: **Confidence Interval = 65 ± (1.645 * 0.50)** **Confidence Interval = 65 ± 0.8225** This results in: **Lower Limit**: 65 - 0.8225 = 64.1775 (rounded to 64.178) **Upper Limit**: 65 + 0.8225 = 65.8225 (rounded to 65.823) Thus, the 90% confidence interval is: **(64.178, 65.823)** **B. Interpretation:** I am 90% confident that the true population mean height of 15-year-olds falls between 64.178 inches and 65.823 inches. This means that if we were to take many samples and construct confidence intervals in the same manner, we would expect about 90% of those intervals to contain the actual population mean.

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