For a specific confidence level, what happens to the width of a confidence interval for the mean as the sample size increases (assuming population standard deviation is known)? The confidence interval for the mean is typically given by $\mu \pm z\bullet \frac{\sigma }{\sqrt{n}}$, where $n$ is the sample size.

• A. The width of the confidence interval stays the same.
• B. The width of the confidence interval increases.
• C. The width of the confidence interval decreases.
• D. The width of the confidence interval first increases, then decreases.

Answer: (C)

Verified step by step guidance

1. Recall the formula for the confidence interval for the mean when the population standard deviation \( \sigma \) is known: \[ \text{CI} = \bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( z_{\alpha/2} \) is the critical z-value for the confidence level, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
2. Focus on the width of the confidence interval, which is twice the margin of error: \[ \text{Width} = 2 \times z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} \]
3. Notice that the width depends on \( \frac{1}{\sqrt{n}} \), meaning it is inversely proportional to the square root of the sample size \( n \).
4. As the sample size \( n \) increases, the denominator \( \sqrt{n} \) increases, which causes the fraction \( \frac{\sigma}{\sqrt{n}} \) to decrease.
5. Therefore, increasing the sample size \( n \) results in a smaller margin of error and thus a narrower confidence interval, meaning the width of the confidence interval decreases.