Final answer:
The statement is false; increasing the confidence level for both averages and proportions will result in wider confidence intervals, not narrower, because a higher confidence level requires a larger z-score or t-score, expanding the margin of error.
Step-by-step explanation:
The statement that increasing the confidence level for proportions results in a narrower confidence interval is false. In reality, whether dealing with averages or proportions, increasing the confidence level will always result in a wider confidence interval.
When calculating confidence intervals for either averages or proportions, the degree of certainty with which we want to estimate the population parameter is the confidence level. Common confidence levels include 90%, 95%, and 99%. The interval is based on the standard error of the mean or proportion and incorporates the z-score or t-score corresponding to the desired confidence level. As the confidence level increases, so does the z-score or t-score value, which is then multiplied by the standard error to calculate the margin of error. A higher confidence level requires a higher z-score or t-score to encapsulate more of the data, leading to a wider margin of error, and thus a wider interval for both averages and proportions.
For example, if we are calculating a confidence interval for a population mean, a 90% confidence level could involve a z-score of approximately 1.645. If this increases to 95%, the z-score rises to about 1.96. This larger z-score multiplied by the standard error results in a larger margin of error, hence a wider interval. The same principle applies when computing confidence intervals for proportions, although the formula for the standard error is different.