An ecologist randomly samples 12 plants of a specific species and measures their heights. He finds that this sample has a mean of 16 cm and a standard deviation of 1 cm. If we assume that the height measurements are normally distributed, find a 95% confidence interval for the mean height of all plants of this species. Give the lower limit and upper limit of the 95% confidence interval.

Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.

To find the 95% confidence interval for the mean height of all plants of this species, we can use the formula for the confidence interval (CI) for the mean when the sample size is small, and the population standard deviation is unknown. The formula is:

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

Where:
- \(\bar{x}\) is the sample mean,
- \(t_{\alpha/2}\) is the critical value from the t-distribution for a given confidence level and degrees of freedom \(df = n - 1\),
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.

### Given values:
- \(\bar{x} = 16\) cm (sample mean)
- \(s = 1\) cm (sample standard deviation)
- \(n = 12\) (sample size)
- Confidence level = 95%, so \(\alpha = 0.05\) and \(\alpha/2 = 0.025\)
- Degrees of freedom \(df = n - 1 = 12 - 1 = 11\)

Now, let's find the critical value \(t_{\alpha/2}\) for 95% confidence and 11 degrees of freedom.

Using a t-table or calculator, the critical value \(t_{0.025, 11}\) is approximately 2.201.

### Confidence Interval Calculation:
Now, we can calculate the margin of error:

\[
\text{Margin of Error} = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} = 2.201 \cdot \frac{1}{\sqrt{12}}
\]

Let's compute this step-by-step.

\[
\frac{s}{\sqrt{n}} = \frac{1}{\sqrt{12}} \approx 0.289
\]

\[
\text{Margin of Error} = 2.201 \times 0.289 \approx 0.636
\]

Now, calculate the lower and upper limits of the confidence interval:

\[
\text{Lower Limit} = \bar{x} - \text{Margin of Error} = 16 - 0.636 \approx 15.4 \text{ cm}
\]

\[
\text{Upper Limit} = \bar{x} + \text{Margin of Error} = 16 + 0.636 \approx 16.6 \text{ cm}
\]

### Final Answer:
The 95% confidence interval for the mean height of all plants of this species is approximately **15.4 cm to 16.6 cm**.

Would you like any further details or clarifications on this solution?

### Related Questions:
1. How would the confidence interval change if the sample size increased?
2. What is the effect of a larger standard deviation on the confidence interval?
3. How would you calculate a 99% confidence interval instead?
4. Why do we use the t-distribution instead of the normal distribution in this case?
5. How does the choice of confidence level affect the width of the confidence interval?
6. What assumptions are made about the data when using this method?
7. How would the confidence interval change if the sample mean was different?
8. How does the degrees of freedom affect the critical value from the t-distribution?

### Tip:
Always remember that a wider confidence interval reflects greater uncertainty about the population parameter.

Learn how to calculate a 95% confidence interval for the mean height of plants using sample data with a mean of 16 cm and a standard deviation of 1 cm. Understand the application of the t-distribution and sample statistics in statistical inference.

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