Question

construct the confidence interval for the population mean $mu$.
$c = 0.90,\bar{x}=6.7,\sigma = 0.6$, and $n = 52$
a $90\\%$ confidence interval for $mu$ is (). (round to two decimal places as needed.)

Explanation:

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

For a confidence level \( c = 0.90 \), \( \alpha=1 - c=1 - 0.90 = 0.10 \). Then \( \alpha/2=0.05 \).
From the standard normal distribution table, \( z_{\alpha/2}=z_{0.05}\approx1.645 \).

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

The formula for the margin of error when the population standard deviation \( \sigma \) is known is \( E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \).
Given \( \sigma = 0.6 \), \( n = 52 \), and \( z_{\alpha/2}=1.645 \).
\( E=1.645\times\frac{0.6}{\sqrt{52}}\)
\(=\frac{1.645\times0.6}{7.211}\)
\(=\frac{0.987}{7.211}\approx0.14 \)

Step3: Construct the confidence interval

The confidence interval for the population mean \( \mu \) is \( \bar{x}-E<\mu<\bar{x} + E \).
Given \( \bar{x}=6.7 \), \( E = 0.14 \).
\( 6.7-0.14<\mu<6.7 + 0.14 \)
\( 6.56<\mu<6.84 \)

Answer:

\((6.56,6.84)\)