Question

a simple random sample of size n is drawn from a population that is normally distributed. the sample mean, x, is found to be 106, and the sample standard deviation, s, is found to be 10
(a) construct a 98% confidence interval about μ if the sample size, n, is 23
(b) construct a 98% confidence interval about μ if the sample size, n, is 19
(c) construct a 96% confidence interval about μ 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

(c) construct a 96% confidence interval about μ if the sample size, n, is 23
lower bound 101.4, upper bound 110.6
(use ascending order. round to one decimal place as needed.)
compare the results to those obtained in part (a). how does decreasing the level of confidence affect the size of the margin of error, e?

Explanation:

Step1: Recall the formula for the margin of error

The formula for the margin of error \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\). When the confidence level decreases, the value of \(t_{\alpha/2}\) decreases (since the critical value \(t_{\alpha/2}\) is related to the confidence level. A lower confidence level means less area in the tails, so a smaller \(t\)-value).

Step2: Analyze the effect on the margin of error

Since \(E\) is directly proportional to \(t_{\alpha/2}\) (while \(s\) and \(n\) are held constant in the comparison between part (a) and part (c) as \(n = 23\) in both cases and \(s=10\) is given), when \(t_{\alpha/2}\) decreases, the value of \(E\) decreases. And the length of the confidence interval \(L=2E\). So when \(E\) decreases, the length of the confidence interval (size of the interval) decreases.

Answer:

C. As the level of confidence decreases, the size of the interval decreases.