Question

find the margin of error for the given values of c, s, and n.
c = 0.98, s = 3.8, n = 15
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

The degrees of freedom \(df=n - 1\). Given \(n = 15\), then \(df=15-1 = 14\).

Step2: Find the \(t\) - value

For a confidence level \(c = 0.98\), the significance level \(\alpha=1 - c=1 - 0.98 = 0.02\). And \(\frac{\alpha}{2}=\frac{0.02}{2}=0.01\).
Looking up the \(t\) - distribution table with \(df = 14\) and \(\frac{\alpha}{2}=0.01\), we get \(t_{\frac{\alpha}{2}}=2.624\).

Step3: Calculate the margin of error

The formula for the margin of error \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\).
Substitute \(t_{\frac{\alpha}{2}} = 2.624\), \(s = 3.8\), and \(n = 15\) into the formula:

$$
LATEXBLOCK0
$$

\(E\approx2.574\)

Answer:

\(2.574\)