Question

find the margin of error for the given values of c, s, and n.
c = 0.99, s = 2.5, n = 14
click the icon to view the t - distribution table.
the margin of error is (round to three decimal places as needed.)

Explanation:

Step1: Calculate the degrees of freedom

Degrees of freedom \(df=n - 1\). Given \(n = 14\), so \(df=14 - 1=13\).

Step2: Find the \(t\)-value

For a confidence level \(c = 0.99\), the significance level \(\alpha=1 - c=1 - 0.99 = 0.01\). And \(\frac{\alpha}{2}=\frac{0.01}{2}=0.005\). Using the \(t\)-distribution table with \(df = 13\) and \(\frac{\alpha}{2}=0.005\), the \(t\)-value \(t_{\frac{\alpha}{2}} = 3.012\).

Step3: Calculate the margin of error

The formula for the margin of error \(E=t_{\frac{\alpha}{2}}\times\frac{s}{\sqrt{n}}\). Given \(s = 2.5\), \(n = 14\), and \(t_{\frac{\alpha}{2}}=3.012\).

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Answer:

\(2.012\)