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, upper bound (use ascending order. round to two decimal places as needed.)

Explanation:

Step1: Determine the critical value

For a 95% confidence interval and \(n = 61\), the degrees of freedom \(df=n - 1=61 - 1 = 60\). From the t - distribution table, \(t_{\alpha/2}\) (where \(\alpha=1 - 0.95 = 0.05\) and \(\alpha/2=0.025\)) for \(df = 60\) is \(t_{0.025,60}=2.000\)

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\). Given \(s = 4.9\), \(n = 61\), and \(t_{\alpha/2}=2.000\)

$$ E=2.000\times\frac{4.9}{\sqrt{61}}\approx2.000\times\frac{4.9}{7.81}\approx2.000\times0.627\approx1.25 $$

Step3: Calculate the confidence interval

The lower bound \(=\bar{x}-E\) and the upper bound \(=\bar{x}+E\). Given \(\bar{x}=17.6\)
Lower bound: \(17.6-1.25 = 16.35\)
Upper bound: \(17.6 + 1.25=18.85\)

Answer:

Lower bound: \(16.35\), Upper bound: \(18.85\)