Question

construct the confidence interval for the population mean μ.
c = 0.90, \\( \overline { x } = 16.7 \\), \\( \sigma = 9.0 \\), and n = 65
a 90% confidence interval for μ is (). (round to one decimal place as needed.)

Explanation:

Step1: Find the critical value \( z_{\alpha/2} \)

For \( c = 0.90 \), \( \alpha=1 - c=1 - 0.90 = 0.10 \), and \( \alpha/2=0.05 \).
Using the standard normal distribution table or a calculator, \( z_{\alpha/2}=z_{0.05}\approx1.645 \).

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

The formula for the margin of error when the population standard deviation \( \sigma \) is known is \( E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \).
Substitute \( z_{\alpha/2}=1.645 \), \( \sigma = 9.0 \), and \( n = 65 \) into the formula:
\( E=1.645\times\frac{9.0}{\sqrt{65}}\)
\(=\frac{1.645\times9.0}{\sqrt{65}}\approx\frac{14.805}{8.062}\approx1.849\)

Step3: Construct the confidence interval

The confidence interval for the population mean \( \mu \) is \( \bar{x}-E<\mu<\bar{x} + E \).
Substitute \( \bar{x}=16.7 \) and \( E\approx1.849 \) into the formula:
\( 16.7-1.849<\mu<16.7 + 1.849 \)
\( 14.851<\mu<18.549 \)

Answer:

\((14.9,18.5)\)