Question

find the margin of error for the given values of c, s, and n.
c = 0.90, s = 6, n = 9
click the icon to view the t - distribution table.
the margin of error is (round to three decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom

Degrees of freedom \(df=n - 1\). Given \(n = 9\), so \(df=9-1 = 8\).

Step2: Find the \(t\) - value

For a confidence level \(c = 0.90\), the significance level \(\alpha=1 - c=1 - 0.90 = 0.10\), and \(\frac{\alpha}{2}=\frac{0.10}{2}=0.05\).
Using the \(t\) - distribution table with \(df = 8\) and \(\frac{\alpha}{2}=0.05\), the \(t\) - value \(t_{\frac{\alpha}{2}}=1.860\).

Step3: Calculate the margin of error

The formula for the margin of error \(E\) (when population standard deviation \(\sigma\) is unknown) is \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\).
Given \(s = 6\), \(n = 9\), and \(t_{\frac{\alpha}{2}}=1.860\).
Substitute the values into the formula: \(E = 1.860\times\frac{6}{\sqrt{9}}\).
Since \(\sqrt{9}=3\), then \(E=1.860\times\frac{6}{3}\).
\(\frac{6}{3}=2\), so \(E=1.860\times2=3.720\).

Answer:

\(3.720\)