Statistics: Dissecting a confidence interval question

Suppose the question is like this below 

A random sample of 100 men is taken and their mean height is calculated to be 180 cm. The population variance is 36 cm2. Find the 95% confidence interval for the mean height of the population.

Let’s go step-by-step.

---

**Step 1: Identify given values**

- Sample size \( n = 100 \)

- Sample mean \( \bar{x} = 180 \) cm

- Population variance \( \sigma^2 = 36 \) cm² → Population standard deviation \( \sigma = \sqrt{36} = 6 \) cm

- Confidence level \( 95\% \)

---

**Step 2: Determine the formula for confidence interval**

For a population with known standard deviation \(\sigma\), the \( 95\% \) confidence interval for the population mean \(\mu\) is:

\[

\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}

\]

For \( 95\% \) confidence, \( z_{\alpha/2} = 1.96 \).

---

**Step 3: Compute standard error**

\[

\text{Standard error} = \frac{\sigma}{\sqrt{n}} = \frac{6}{\sqrt{100}} = \frac{6}{10} = 0.6

\]

---

**Step 4: Compute margin of error**

\[

\text{Margin of error} = 1.96 \times 0.6 = 1.176

\]

---

**Step 5: Construct confidence interval**

\[

180 \pm 1.176

\]

\[

(180 - 1.176, \ 180 + 1.176)

\]

\[

(178.824, \ 181.176)

\]

---

**Final answer:**

\[

\boxed{(178.824, 181.176)}

\]

This means we are \( 95\% \) confident that the true population mean height lies between \( 178.824 \) cm and \( 181.176 \) cm.