The width of a confidence interval depends on the standard deviation of the population (σ), the sample size (n), and the confidence level. Specifically, the formula for the margin of error (E) in a confidence interval for a population mean (assuming a normal distribution or a large enough sample size) is:

$E = Z \times \frac{\sigma}{\sqrt{n}}$

where:

• $Z$ is the Z-score corresponding to the desired confidence level.
• $\sigma$ is the population standard deviation.
• $n$ is the sample size.

Given that the population standard deviation ($\sigma$) increases while other factors remain constant, let's analyze its effect on the confidence interval.

When $\sigma$ increases:

• The term $\frac{\sigma}{\sqrt{n}}$ increases.
• Consequently, the margin of error (E) increases.

A larger margin of error means the confidence interval will be wider. Therefore, if the population standard deviation of a sample increases without other changes, the confidence interval is most likely to widen.

The correct answer is:

widens

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. What is the impact of increasing the sample size on the confidence interval?
2. How does changing the confidence level affect the confidence interval width?
3. What is the difference between population standard deviation and sample standard deviation?
4. How do you interpret the Z-score in the context of confidence intervals?
5. What is the role of standard error in constructing a confidence interval?
6. How can you decrease the margin of error in a study?
7. What assumptions must be met to use a Z-distribution for confidence intervals?
8. How do confidence intervals relate to hypothesis testing?

Tip: Remember that increasing the sample size or decreasing the population standard deviation can lead to a narrower confidence interval, providing more precise estimates.