Question

construct the indicated confidence interval for the population mean μ using the t - distribution. assume the population is normally distributed.
c = 0.90, \\( \overline { x } = 12.9 \\), s = 0.56, n = 19
(round to one decimal place as needed.)

Explanation:

Step1: Find the degrees of freedom and the critical value \(t_{\alpha/2}\)

The degrees of freedom \(df=n - 1=19 - 1 = 18\).
For a confidence level \(c = 0.90\), \(\alpha=1 - c=1 - 0.90 = 0.10\), and \(\alpha/2=0.05\).
Using the t - distribution table or a calculator, \(t_{\alpha/2,df}=t_{0.05,18}=1.734\).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error when using the t - distribution is \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\).
Substitute \(t_{\alpha/2}=1.734\), \(s = 0.56\), and \(n = 19\) into the formula:
\(E=1.734\times\frac{0.56}{\sqrt{19}}\approx1.734\times\frac{0.56}{4.359}\approx1.734\times0.1285\approx0.223\).

Step3: Construct the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x}+E\).
Substitute \(\bar{x}=12.9\) and \(E = 0.223\) into the formula:
\(12.9-0.223<\mu<12.9 + 0.223\), which is \(12.7<\mu<13.1\).

Answer:

\((12.7,13.1)\)