Question

find the margin of error for the given values of c, s, and n.
c = 0.80, s = 5, n = 7
click the icon to view the t - distribution table.
the margin of error is \square. (round to three decimal places as needed.)

Explanation:

Step1: Find the degrees of freedom

Degrees of freedom \(df=n - 1\). Given \(n = 7\), so \(df=7-1 = 6\).

Step2: Find the \(t\)-value

For a confidence level \(c = 0.80\), the significance level \(\alpha=1 - c=1 - 0.80 = 0.20\). And \(\frac{\alpha}{2}=\frac{0.20}{2}=0.10\).
Looking up the \(t\)-distribution table with \(df = 6\) and \(\frac{\alpha}{2}=0.10\), we get \(t_{\frac{\alpha}{2}}=1.440\).

Step3: Calculate the margin of error

The formula for the margin of error \(E=t_{\frac{\alpha}{2}}\times\frac{s}{\sqrt{n}}\).
Given \(s = 5\), \(n = 7\), and \(t_{\frac{\alpha}{2}}=1.440\).
First, calculate \(\sqrt{n}=\sqrt{7}\approx2.646\).
Then \(\frac{s}{\sqrt{n}}=\frac{5}{2.646}\approx1.89\).
Finally, \(E = 1.440\times1.89\approx2.722\).

Answer:

\(2.722\)