Question

a simple random sample of size n is drawn from a population that is normally distributed. the sample mean, \\( \bar { x } \\), is found to be 106, and the sample standard deviation, s, is found to be 10(a) construct a 98% confidence interval about \\( \mu \\) if the sample size, n, is 23(b) construct a 98% confidence interval about \\( \mu \\) if the sample size, n, is 19(c) construct a 96% confidence interval about \\( \mu \\) if the sample size, n, is 23(d) could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed?click the icon to view the table of areas under the t - distribution(a) construct a 98% confidence interval about \\( \mu \\) if the sample size, n, is 23lower bound 100.8, upper bound 111.2(use ascending order. round to one decimal place as needed.)(b) construct a 98% confidence interval about \\( \mu \\) if the sample size, n, is 19lower bound 100.1, upper bound 111.9(use ascending order. round to one decimal place as needed.)how does decreasing the sample size affect the margin of error, e?a. as the sample size decreases, the margin of error decreasesb. as the sample size decreases, the margin of error stays the samec. as the sample size decreases, the margin of error increases

Explanation:

Step1: Recall the formula for margin of error

The formula for margin of error \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\). Here, \(s\) (sample standard deviation) is constant (\(s = 10\)), \(t_{\alpha/2}\) changes slightly with \(n\) but the main effect comes from \(\frac{1}{\sqrt{n}}\).

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

As \(n\) decreases, \(\sqrt{n}\) decreases. Since \(E\) is directly proportional to \(\frac{1}{\sqrt{n}}\) (because \(s\) is fixed and \(t_{\alpha/2}\) changes minimally compared to the change in \(\sqrt{n}\) for small - to - moderate changes in \(n\)), when \(n\) goes from \(23\) to \(19\) (a decrease in \(n\)), \(\frac{1}{\sqrt{n}}\) increases. For example, \(\frac{1}{\sqrt{23}}\approx0.208\) and \(\frac{1}{\sqrt{19}}\approx0.229\).

Answer:

C. As the sample size decreases, the margin of error increases