Question

a simple random sample of size n is drawn. the sample mean, \\( \overline { 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 \\( \mu \\) if the sample size, n, is 35
lower bound: \\( \square \\); upper bound: \\( \square \\)
(use ascending order. round to two decimal places as needed.)

Explanation:

Step1: Determine the critical value

For a 95% confidence interval and \(n = 35\), the degrees of freedom \(df=n - 1=35 - 1 = 34\). Using the t - distribution table, the critical value \(t_{\alpha/2}\) for a 95% confidence interval (\(\alpha=1 - 0.95 = 0.05\), \(\alpha/2=0.025\)) is approximately \(t_{0.025,34}\approx 2.032\).

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\alpha/2}\times\frac{s}{\sqrt{n}}\). Given \(s = 4.9\), \(n = 35\), and \(t_{\alpha/2}=2.032\), we have \(E = 2.032\times\frac{4.9}{\sqrt{35}}\).
First, calculate \(\frac{4.9}{\sqrt{35}}\approx\frac{4.9}{5.916}\approx0.828\). Then \(E=2.032\times0.828\approx1.683\).

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x}+E\). Given \(\bar{x}=17.6\), the lower bound is \(17.6-1.683 = 15.92\) and the upper bound is \(17.6 + 1.683=19.28\).

Answer:

Lower bound: \(15.92\); Upper bound: \(19.28\)