An ecologist randomly samples 17 tulips of a specific variety and measures their heights. He finds that this sample has a mean of 11 cm and a standard deviation of 3 cm. If we assume that height measurements are normally distributed, find a 95% confidence interval for the mean height of all plants of this species.

To calculate a 95% confidence interval for the mean height of all plants of this species, we use the formula for a confidence interval based on the t-distribution, given by:

\[
\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}
\]

where:
- \(\bar{x}\) is the sample mean,
- \(t_{\alpha/2}\) is the t-value for a 95% confidence level with \(n - 1\) degrees of freedom,
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.

### Step-by-Step Solution

1. **Identify the values:**
   - Sample mean, \(\bar{x} = 11\) cm
   - Sample standard deviation, \(s = 3\) cm
   - Sample size, \(n = 17\)

2. **Degrees of freedom (df):**
   - \(df = n - 1 = 17 - 1 = 16\)

3. **Find the critical t-value (\(t_{0.025,16}\)) for a 95% confidence level:**
   - For a 95% confidence interval and 16 degrees of freedom, the t-value is approximately \(t_{0.025,16} \approx 2.12\).

4. **Calculate the margin of error:**
   \[
   \text{Margin of Error} = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} = 2.12 \cdot \frac{3}{\sqrt{17}}
   \]

   Calculating inside the square root and division:
   \[
   = 2.12 \cdot \frac{3}{4.123} \approx 2.12 \cdot 0.727 = 1.542
   \]

5. **Construct the confidence interval:**
   \[
   \bar{x} \pm \text{Margin of Error} = 11 \pm 1.542
   \]

   This gives us:
   \[
   \text{Confidence Interval} = (11 - 1.542, 11 + 1.542) = (9.458, 12.542)
   \]

### Answer:
The 95% confidence interval for the mean height of all plants of this species is approximately:

\[
(9.46, 12.54) \text{ cm}
\]

Would you like further details or have any questions about this solution?

---

1. How would the confidence interval change with a larger sample size?
2. Why do we use the t-distribution instead of the z-distribution for this calculation?
3. What is the effect of a higher confidence level on the confidence interval?
4. How would the interval differ if the sample standard deviation were larger?
5. Can this method be applied if the sample distribution were not normal?

**Tip:** A larger sample size generally narrows the confidence interval, increasing the precision of the estimate.

Determine the 95% confidence interval for the mean height of a tulip species from a random sample of 17 tulips with a mean height of 11 cm and a standard deviation of 3 cm. Understand key statistical concepts like t-distribution and confidence intervals in this solution.

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