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. (a) construct a 95% confidence interval about μ if the sample size, n, is 35. lower bound 15.92; upper bound 19.28 (use ascending order. round to two decimal places as needed.) (b) construct a 95% confidence interval about μ if the sample size, n, is 61. lower bound 16.35; upper bound 18.85 (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.

Explanation:

Step1: Recall the formula for margin of error

The margin of error \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\). Here, \(t_{\alpha/2}\) (for 95% confidence level) and \(s\) (sample standard deviation) are constant.

Step2: Analyze the effect of \(n\) on \(E\)

Since \(n\) is in the denominator of the formula \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\), when \(n\) increases (from 35 to 61 in this case), the value of \(\frac{s}{\sqrt{n}}\) decreases. For example, if \(s = 4.9\), when \(n = 35\), \(\frac{4.9}{\sqrt{35}}\approx0.83\); when \(n = 61\), \(\frac{4.9}{\sqrt{61}}\approx0.63\).

Answer:

B. The margin of error decreases