Question

construct the confidence interval for the population mean (mu).
(c = 0.95,overline{x}=6.7,sigma = 0.5), and (n = 48)
a 95% confidence interval for (mu) is ((square,square)). (round to two decimal places
as needed)

Explanation:

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

For a 95% confidence interval (\(c = 0.95\)), \( \alpha=1 - c=1 - 0.95 = 0.05\), and \( \alpha/2=0.025\).
From the standard normal distribution table, \( z_{\alpha/2}=z_{0.025} = 1.96\).

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 \( z_{\alpha/2}=1.96\), \( \sigma = 0.5\), and \( n = 48\) into the formula:
\( E=1.96\times\frac{0.5}{\sqrt{48}}\)
\(=\frac{1.96\times0.5}{6.9282}\)
\(=\frac{0.98}{6.9282}\approx0.14\)

Step3: Construct the confidence interval

The confidence interval for the population mean \( \mu \) is given by \( \bar{x}-E<\mu<\bar{x} + E\).
Substitute \( \bar{x}=6.7\) and \( E = 0.14\) into the formula:
\(6.7- 0.14<\mu<6.7 + 0.14\)
\(6.56<\mu<6.84\)

Answer:

\((6.56,6.84)\)