## **Core Concept**
The problem involves calculating the confidence interval for a proportion, specifically the prevalence of Candida glabrata infection in a population. The confidence interval gives a range of values within which the true population parameter is likely to lie. For proportions, the confidence interval can be calculated using the formula: (p pm Z sqrt{frac{p(1-p)}{n}}), where (p) is the sample proportion, (Z) is the Z-score corresponding to the desired confidence level, and (n) is the sample size.

## **Why the Correct Answer is Right**
To calculate the 95% confidence interval for the prevalence of Candida glabrata infection, we first need to know that the sample proportion ((p)) is 80% or 0.8, and the sample size ((n)) is 100. For a 95% confidence level, the Z-score ((Z)) is approximately 1.96. Plugging these values into the formula gives us: (0.8 pm 1.96 sqrt{frac{0.8(1-0.8)}{100}}). Simplifying inside the square root: (0.8 pm 1.96 sqrt{frac{0.8 times 0.2}{100}} = 0.8 pm 1.96 sqrt{frac{0.16}{100}} = 0.8 pm 1.96 sqrt{0.0016} = 0.8 pm 1.96 times 0.04 = 0.8 pm 0.0784). Therefore, the 95% confidence interval is (0.8 pm 0.0784) or (0.7216) to (0.8784), which translates to 72.16% to 87.84%.

## **Why Each Wrong Option is Incorrect**
- **Option A:** This option suggests a range that does not match our calculated confidence interval.
- **Option B:** Similarly, this option does not align with the calculated range of 72.16% to 87.84%.
- **Option D:** This option also does not match the calculated confidence interval.

## **Clinical Pearl / High-Yield Fact**
A key point to remember is that the width of the confidence interval is inversely related to the square root of the sample size. This means that larger sample sizes provide more precise estimates of population parameters. For clinical studies, especially those involving infectious disease prevalence like Candida glabrata, ensuring an adequate sample size is crucial for obtaining reliable estimates.

## **Correct Answer:** C. 72.16% - 87.84%.