A public health researcher records the number of antenatal visits for 14 women. The sample mean is 6.5 visits and the sample standard deviation is 1.8 visits. Calculate the 95% confidence interval for the population mean number of visits.

We are given:

• Sample mean ($\bar{x}$) = 6.5 visits
• Sample standard deviation ($s$) = 1.8 visits
• Sample size ($n$) = 14 women
• Confidence level = 95%

Since the sample size is small ($n < 30$) and the population standard deviation is unknown, we use the t-distribution to calculate the confidence interval for the population mean ($\mu$). The formula is:

$$\bar{x} \pm t_{\alpha/2, df} \cdot \frac{s}{\sqrt{n}}$$

The degrees of freedom ($df$) are $n - 1$.

$$df = 14 - 1 = 13$$

For a 95% confidence interval, $\alpha = 1 - 0.95 = 0.05$. So, $\alpha/2 = 0.025$.
Using a t-distribution table for $df = 13$ and $\alpha/2 = 0.025$, the critical t-value ($t_{0.025, 13}$) is approximately 2.160.

The standard error ($SE$) is $\frac{s}{\sqrt{n}}$.

$$SE = \frac{1.8}{\sqrt{14}}$$ $$SE \approx \frac{1.8}{3.741657}$$ $$SE \approx 0.4811$$

The margin of error ($ME$) is $t_{\alpha/2, df} \cdot SE$.

$$ME = 2.160 \cdot 0.4811$$ $$ME \approx 1.039176$$

The confidence interval is $\bar{x} \pm ME$.
Lower bound: $6.5 - 1.039176 \approx 5.460824$
Upper bound: $6.5 + 1.039176 \approx 7.539176$

Rounding to two decimal places, the 95% confidence interval is:

$$(5.46, 7.54)$$

The 95% confidence interval for the population mean number of visits is $\boxed{\text{(5.46, 7.54) visits}}$.