Question

find the value of e, the margin of error, for c = 0.95, n = 15 and s = 5.6.
oa 3.101
ob 0.801
oc 2.546
od 3.189

Explanation:

Step1: Find the degrees of freedom

Degrees of freedom \( df=n - 1\). Given \(n = 15\), so \(df=15 - 1=14\).

Step2: Find the \(t\) - value for \(c = 0.95\)

For a confidence level \(c = 0.95\), the significance level \(\alpha=1 - c=1 - 0.95 = 0.05\). And \(\frac{\alpha}{2}=\frac{0.05}{2}=0.025\).
Looking up the \(t\) - distribution table with \(df = 14\) and \(\frac{\alpha}{2}=0.025\), we get \(t_{\frac{\alpha}{2}}=2.145\).

Step3: Calculate the margin of error \(E\)

The formula for the margin of error when the population standard deviation \(\sigma\) is unknown (we use \(s\) as an estimate) is \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\).
Given \(s = 5.6\), \(n = 15\), and \(t_{\frac{\alpha}{2}}=2.145\).
First, calculate \(\frac{s}{\sqrt{n}}=\frac{5.6}{\sqrt{15}}\approx\frac{5.6}{3.873}\approx1.446\).
Then \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}=2.145\times1.446\approx3.101\).

Answer:

A. 3.101