Question

(c) if we construct a 90% confidence interval for the percentage of u.s. adults x users who get some news through x, this confidence interval will be wider than a corresponding 99% confidence interval.
true
false
tip: experiment with the simulation page, to see how a different confidence level affects the confidence intervals.
explanation for answer (c) above:
simulation page image: simulating confidence intervals with various settings and results
question help: read

Explanation:

To determine the width of a confidence interval, we consider the relationship between the confidence level and the margin of error. A higher confidence level (e.g., 99%) requires a larger critical value (from the standard normal or t - distribution). The margin of error \( E \) for a proportion (relevant here as we are dealing with a percentage) is given by \( E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \) (for large samples, using the z - distribution). As the confidence level increases, \( z_{\alpha/2} \) (or the corresponding t - value) increases. For a 90% confidence interval, \( \alpha=0.1\), so \( \alpha/2 = 0.05\) and \( z_{0.05}\approx1.645 \). For a 99% confidence interval, \( \alpha = 0.01\), so \( \alpha/2=0.005\) and \( z_{0.005}\approx2.576 \). Since the margin of error is larger for a higher confidence level, the confidence interval (which is \( \hat{p}\pm E \)) will be wider for a higher confidence level. So a 90% confidence interval will be narrower than a 99% confidence interval, meaning the statement is false.

Answer:

false