Suppose that a random sample of 16 adult U.S. males has a mean height of 70 inches with a standard deviation of 2 inches. If we assume that the heights of adult males in the U.S. are normally distributed, find a 99% confidence interval for the mean height of all U.S. males. Give the lower limit and upper limit of the 99% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)

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### Educational Content on Confidence Intervals

---

**Problem Statement:**

Suppose that a [random sample](https://mathworld.wolfram.com/RandomSample.html) of 16 adult U.S. males has a [mean](https://mathworld.wolfram.com/ArithmeticMean.html) height of 70 inches with a [standard deviation](https://mathworld.wolfram.com/StandardDeviation.html) of 2 inches. If we assume that the heights of adult males in the U.S. are [normally distributed](https://mathworld.wolfram.com/NormalDistribution.html), find a 99% [confidence interval](https://mathworld.wolfram.com/ConfidenceInterval.html) for the [mean](https://mathworld.wolfram.com/ArithmeticMean.html) height of all U.S. males. Give the lower limit and upper limit of the 99% confidence interval.

---

**Instructions:**

Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a [list of formulas](https://mathworld.wolfram.com/).

---

### Calculation Steps:
1. **Determine the sample mean (\(\bar{x}\)):** 
   - Given: \(\bar{x} = 70\) inches.

2. **Determine the sample standard deviation (s):** 
   - Given: \(s = 2\) inches.

3. **Determine the sample size (n):** 
   - Given: \(n = 16\).

4. **Find the critical value (z) for a 99% confidence interval:**
   - Use a Z-table or standard normal distribution table to find \(z\).
   - For a 99% confidence level, \(z \approx 2.576\).

5. **Calculate the standard error of the mean (SEM):**
   - \( \text{SEM} = \frac{s}{\sqrt{n}} = \frac{2}{\sqrt{16}} = \frac{2}{4} = 0.5 \)

6. **Compute the margin of error (ME):**
   - \( \text{ME} = z \times \text{SEM} = 2.576 \times 0.5 = 1.288 \)

7. **Determine the confidence interval:**
   - \(\bar{x} \pm \text{ME} =
Transcribed Image Text:### Educational Content on Confidence Intervals --- **Problem Statement:** Suppose that a [random sample](https://mathworld.wolfram.com/RandomSample.html) of 16 adult U.S. males has a [mean](https://mathworld.wolfram.com/ArithmeticMean.html) height of 70 inches with a [standard deviation](https://mathworld.wolfram.com/StandardDeviation.html) of 2 inches. If we assume that the heights of adult males in the U.S. are [normally distributed](https://mathworld.wolfram.com/NormalDistribution.html), find a 99% [confidence interval](https://mathworld.wolfram.com/ConfidenceInterval.html) for the [mean](https://mathworld.wolfram.com/ArithmeticMean.html) height of all U.S. males. Give the lower limit and upper limit of the 99% confidence interval. --- **Instructions:** Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a [list of formulas](https://mathworld.wolfram.com/). --- ### Calculation Steps: 1. **Determine the sample mean (\(\bar{x}\)):** - Given: \(\bar{x} = 70\) inches. 2. **Determine the sample standard deviation (s):** - Given: \(s = 2\) inches. 3. **Determine the sample size (n):** - Given: \(n = 16\). 4. **Find the critical value (z) for a 99% confidence interval:** - Use a Z-table or standard normal distribution table to find \(z\). - For a 99% confidence level, \(z \approx 2.576\). 5. **Calculate the standard error of the mean (SEM):** - \( \text{SEM} = \frac{s}{\sqrt{n}} = \frac{2}{\sqrt{16}} = \frac{2}{4} = 0.5 \) 6. **Compute the margin of error (ME):** - \( \text{ME} = z \times \text{SEM} = 2.576 \times 0.5 = 1.288 \) 7. **Determine the confidence interval:** - \(\bar{x} \pm \text{ME} =
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