Answer:

A 90% confidence interval for the population mean leaf length is \(3.35\text{ cm} \lt \mu \lt 4.30\text{ cm}\).

Since the population is assumed to be normal and the sample is small, we use a \(t\)-interval for the mean.

The sample data are \(3.5, 4.2, 3.8, 5.2, 2.9, 3.7, 4.1, 3.2\), so \(n=8\).

First find the sample mean:

\(\sum x = 30.6\), so

\(\bar{x} = \frac{30.6}{8} = 3.825\).

Now find the sample variance.

\(\sum x^2 = 120.52\).

So

\(s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{1}{7}\left(120.52 - \frac{30.6^2}{8}\right) = \frac{1}{7}(120.52 - 117.045) = \frac{3.475}{7} \approx 0.49643\).

Hence

\(s \approx \sqrt{0.49643} \approx 0.7046\).

For a 90% confidence interval with \(n-1=7\) degrees of freedom, the critical value is \(t=1.895\).

The standard error is

\(\frac{s}{\sqrt{n}} = \sqrt{\frac{0.49643}{8}} \approx 0.2491\).

So the margin of error is

\(1.895 \times 0.2491 \approx 0.472\).

Therefore the 90% confidence interval is

\(\bar{x} \pm t\frac{s}{\sqrt{n}} = 3.825 \pm 0.472\).

This gives

Lower limit \(= 3.825 - 0.472 = 3.353\),

Upper limit \(= 3.825 + 0.472 = 4.297\).

So, to 3 significant figures, the 90% confidence interval is

\(3.35\text{ cm} \lt \mu \lt 4.30\text{ cm}\).