Question

question 3
which of the following statements is not true?
increasing n makes the confidence interval narrower.
decreasing σ or s makes the confidence interval narrower.
the confidence interval when σ is unknown is generally narrower than when σ is known
decreasing the level of confidence makes the confidence interval narrower.

Explanation:

• For a confidence interval formula (e.g., for the population mean: when \(\sigma\) is known \(CI=\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), when \(\sigma\) is unknown \(CI = \bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(t\) - values have heavier tails than \(z\) - values for small - to - moderate \(n\)).
• Increasing \(n\): As \(n\) increases, the margin of error \(E\) (e.g., \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\) or \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\)) decreases. So, the confidence interval becomes narrower.
• Decreasing \(\sigma\) or \(s\): Since \(\sigma\) (or \(s\)) is in the numerator of the margin - of - error formula, a smaller \(\sigma\) (or \(s\)) leads to a smaller margin of error, and thus a narrower confidence interval.
• Confidence level: A lower confidence level (e.g., from \(99\%\) to \(95\%\)) means a smaller \(z_{\alpha/2}\) (or \(t_{\alpha/2}\)) value. For example, for a \(z\) - distribution, \(z_{0.005}=2.576\) (for \(99\%\) confidence) and \(z_{0.025} = 1.96\) (for \(95\%\) confidence). A smaller critical value (\(z_{\alpha/2}\) or \(t_{\alpha/2}\)) results in a smaller margin of error and a narrower confidence interval.
• \(\sigma\) known vs unknown: When \(\sigma\) is unknown, we use the \(t\) - distribution (for small \(n\), \(t\) - values are larger than \(z\) - values). The margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\) (when \(\sigma\) is unknown) is larger than \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\) (when \(\sigma\) is known) for the same \(\alpha\), \(n\), and sample statistics. So, the confidence interval when \(\sigma\) is unknown is generally wider than when \(\sigma\) is known.

Answer:

The confidence interval when \(\sigma\) is unknown is generally narrower than when \(\sigma\) is known.