Question

a simple random sample of size n is drawn. the sample mean, x, is found to be 17.6, and the sample standard deviation, s, is found to be 4.9
click the icon to view the table of areas under the t - distribution
lower bound: 10.30, upper bound: 10.60
(use ascending order. round to two decimal places as needed)
how does increasing the sample size affect the margin of error, e?
a. the margin of error does not change
b. the margin of error decreases.
c. the margin of error increases.
(c) construct a 99% confidence interval about μ if the sample size, n, is 35.
lower bound 15.34; upper bound 19.86
(use ascending order. round to two decimal places as needed)
compare the results to those obtained in part (a). how does increasing the level of confidence affect the size of the margin of error, e?
a. the margin of error decreases.
b. the margin of error increases
c. the margin of error does not change

Explanation:

The formula for the margin of error \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\). When the sample size \(n\) increases, the value of \(\frac{1}{\sqrt{n}}\) decreases (since \(n\) is in the denominator). As a result, the entire margin of error \(E\) decreases.

For the confidence level, a higher confidence level (e.g., 99% compared to a lower one) requires a larger \(t_{\alpha/2}\) value. Since \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\), a larger \(t_{\alpha/2}\) leads to a larger margin of error \(E\).

Answer:

• For the effect of increasing sample size on margin of error: B. The margin of error decreases.
• For the effect of increasing confidence level on margin of error: B. The margin of error increases.