Question

1. an analyst, using a random sample of n = 500 families, obtained a 90% confidence interval for mean monthly family income for a large population: ($600, $800). if the analyst had used a 99% confidence coefficient instead, the confidence interval would be: (a) narrower and would involve a larger risk of being incorrect (b) wider and would involve a smaller risk of being incorrect (c) narrower and would involve a smaller risk of being incorrect (d) wider and would involve a larger risk of being incorrect (e) wider but it cannot be determined whether the risk of being incorrect would be larger or smaller

Explanation:

• Confidence interval formula: \(\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\).
• When confidence level increases (from \(90\%\) to \(99\%\)), \(z_{\alpha/2}\) value increases.
• Larger \(z_{\alpha/2}\) makes the margin of error \(z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\) larger, so the confidence interval becomes wider.
• Higher confidence level (\(99\%\) compared to \(90\%\)) means smaller risk of being incorrect (because we are more confident that the true population parameter lies within the interval).

Answer:

B. Wider and would involve a smaller risk of being incorrect