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 } = 14.4 \\), s = 0.56, n = 17
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the degrees of freedom

Degrees of freedom \(df=n - 1\). Given \(n = 17\), so \(df=17-1 = 16\).

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

Confidence level \(c = 0.90\), then \(\alpha=1 - c=1 - 0.90 = 0.10\) and \(\alpha/2=0.05\). Using the t - distribution table or a calculator, for \(df = 16\) and \(\alpha/2 = 0.05\), \(t_{\alpha/2}=1.746\).

Step3: 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 \(t_{\alpha/2}=1.746\), \(s = 0.56\), and \(n = 17\) into the formula:

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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 \(E = 0.237\) into the formula:

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Answer:

\((14.2,14.6)\)