Question

find the margin of error for the given values of c, s, and n.
c = 0.90, s = 3.4, n = 8
click the icon to view the t - distribution table.
the margin of error is. (round to three decimal places as needed.)

Explanation:

Step1: Find the degrees of freedom

The degrees of freedom \(df=n - 1\). Given \(n = 8\), so \(df=8-1 = 7\).

Step2: Find the \(t\)-value

For a confidence level \(c = 0.90\), the significance level \(\alpha=1 - c=1 - 0.90 = 0.10\). And \(\frac{\alpha}{2}=\frac{0.10}{2}=0.05\).
Looking up the \(t\)-distribution table with \(df = 7\) and \(\frac{\alpha}{2}=0.05\), we get \(t_{\frac{\alpha}{2}}=1.895\).

Step3: Calculate the margin of error

The formula for the margin of error \(E\) (when the population standard deviation \(\sigma\) is unknown) is \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\).
Given \(s = 3.4\), \(n = 8\), and \(t_{\frac{\alpha}{2}}=1.895\).
First, calculate \(\sqrt{n}=\sqrt{8}\approx2.828\).
Then \(E = 1.895\times\frac{3.4}{2.828}\).
\(\frac{3.4}{2.828}\approx1.202\).
\(E=1.895\times1.202\approx2.288\).

Answer:

\(2.288\)