Question

construct the indicated confidence interval for the population mean $\mu$ using the t - distribution. assume the population is normally distributed.
$c = 0.95$, $\overline{x}=14.4$, $s = 3.0$, $n = 6$
(round to one decimal place as needed.)

Explanation:

Step1: Find the degrees of freedom and critical value

Degrees of freedom \(df=n - 1=6-1 = 5\).
For \(c = 0.95\), the significance level \(\alpha=1 - c=1 - 0.95 = 0.05\), and \(\frac{\alpha}{2}=0.025\).
Using the t - distribution table or a calculator, \(t_{\frac{\alpha}{2},df}=t_{0.025,5}=2.571\).

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\).
Substitute \(t_{\frac{\alpha}{2}} = 2.571\), \(s = 3.0\), and \(n = 6\) into the formula:
\(E=2.571\times\frac{3.0}{\sqrt{6}}\approx2.571\times1.225\approx3.1\).

Step3: 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 \(E = 3.1\) into the formula:
\(14.4-3.1<\mu<14.4 + 3.1\), which is \(11.3<\mu<17.5\).

Answer:

\((11.3,17.5)\)