Question

how does increasing the sample size affect the width of the interval? the width decreases the width does not change the width increases the upper bound is 19.65. (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for the width of a confidence interval

The width of a confidence interval (for example, for a population mean when the population standard deviation \(\sigma\) is known) is \(w = 2z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), where \(z_{\alpha/2}\) is the critical value, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

Step2: Analyze the relationship between \(n\) and \(w\)

We can see that \(w\) and \(\sqrt{n}\) are inversely - related (\(w\propto\frac{1}{\sqrt{n}}\)). As \(n\) (the sample size) increases, the value of \(\frac{\sigma}{\sqrt{n}}\) decreases. Since \(z_{\alpha/2}\) and \(\sigma\) (assuming the population standard deviation is constant) are fixed for a given confidence level and a given population, when \(n\) increases, the product \(2z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\) (the width of the interval) decreases.

Answer:

The width decreases