Which of the following is correct about the effect of sample size on the width of a confidence interval for the mean when the population standard deviation is known ($\sigma$)?

• A. The sample size has no effect on the width of the confidence interval.
• B. Decreasing the sample size decreases the width of the confidence interval.
• C. Increasing the sample size increases the width of the confidence interval.
• D. Increasing the sample size decreases the width of the confidence interval.

Answer: (D)

Verified step by step guidance

1. Recall the formula for the confidence interval for the mean when the population standard deviation \( \sigma \) is known: \[ CI = \bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( z_{\alpha/2} \) is the critical z-value for the desired confidence level, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
2. Focus on the margin of error (the width component) of the confidence interval, which is: \[ E = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} \]
3. Notice that the margin of error \( E \) is inversely proportional to the square root of the sample size \( n \). This means as \( n \) increases, \( \frac{1}{\sqrt{n}} \) decreases, making the margin of error smaller.
4. Since the width of the confidence interval is twice the margin of error (because it extends \( E \) on both sides of the sample mean), increasing the sample size \( n \) results in a narrower confidence interval.
5. Therefore, the correct interpretation is that increasing the sample size decreases the width of the confidence interval, improving the precision of the estimate.