Question

construct the indicated confidence interval for the population mean μ using the t - distribution. assume the population is normally distributed.
c = 0.95, \\( \overline { x } = 14.4 \\), s = 2.0, n = 6
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the degrees of freedom

The degrees of freedom \(df=n - 1\). Given \(n = 6\), so \(df=6-1 = 5\).

Step2: Find the critical value \(t_{\alpha/2}\)

The confidence level \(c = 0.95\), so \(\alpha=1 - c=1 - 0.95 = 0.05\) and \(\alpha/2=0.025\).
Looking up the \(t\) - distribution table with \(df = 5\), \(t_{\alpha/2}=2.571\).

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

The formula for the margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\).
Substitute \(t_{\alpha/2}=2.571\), \(s = 2.0\), \(n = 6\) into the formula:
\(E=2.571\times\frac{2.0}{\sqrt{6}}\approx2.571\times0.8165\approx2.1\)

Step4: Construct the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x}+E\).
Substitute \(\bar{x}=14.4\) and \(E = 2.1\) into the formula:
\(14.4-2.1<\mu<14.4 + 2.1\)

Answer:

\((12.3,16.5)\)